text:
Kind of meta: get the longest of the shortest answers
Your task - if you accept it - is to write a program which helps understanding my proposal on meta by calculating the winner of a code-golf-reversed competition. Of course, answers to this question will be treated as proposed, so your program (if correct) can calculate whether your answer will become the accepted answer.
Rules

The program reads a file with multiple lines of the following format (see example below): [Language] TAB [NumberOfCharacters] TAB [LinkToAnswer]
The file name is passed as an argument to your program or the file is redirected to standard input of your program. It's your choice, please mention the method when giving the answer
It's expected that the input format is correct. There's no need for error handling.
The number of characters is positive. Your program must handle lengths up to 65535. 64k should be enough for everybody :-)
The program outputs those lines on standard output which meet the idea of the meta proposal, that is

the shortest code of a particular programming language wins (reduction phase)
the longest code among all programming languages wins (sorting phase)
in case of a draw, all answers with the same length shall be printed

The order of the output is not important
Although the longest code wins, this is not code-bowling. Your code must be as short as possible for your programming language.
Answers on seldom programming languages which are not attempting to shorten the code deserve a downvote, because they try to bypass the intention of this kind of question. If there's only one answer for a specific programming language, it would be considered as a winner candidate, so you could start blowing its code.

Example input file (separated by single tabs if there should be an issue with formatting):
GolfScript  34  http://short.url/answer/ags
GolfScript  42  http://short.url/answer/gsq
C#  210 http://short.url/answer/cs2
Java    208 http://short.url/answer/jav
C#  208 http://short.url/answer/poi
J   23  http://short.url/answer/jsh
Ruby    67  http://short.url/answer/rub
C#  208 http://short.url/answer/yac
GolfScript  210 http://short.url/answer/210

Expected output (order is not important):
C#  208 http://short.url/answer/poi
C#  208 http://short.url/answer/yac
Java    208 http://short.url/answer/jav

Update
Some programs rely on the fact that there is a single maximum (like the C# 210 character program). Derived from reality, someone can also write a GolfScript program with 210 characters. The output would remain the same. I have added such a GolfScript to the input.
Update 2
As suggested I have retagged (still code-golf as well) and the deadline is 2014-03-06 (which looks like an arbitrary date, but I'll be back to Germany from travelling then).
Final results
I decided to vote like the following:

Answers where the number of characters cannot be confirmed get a comment to explain the count.
Answers which can easily be reduced get a comment, an edit suggestion and go into the result with the lower count value. (Hopefully I have seen that in advance).
Answers which do not compile get a downvote. (Quite a hard task as it turns out).
Answers which are not golfed get a downvote (as described in the rules already).
Answers which produce expected output get an upvote. Due to some answers which do not work as expected, I use 4 different input files and check against the expected result.

Finally, the winner is determined by providing the qualifying answers table as input to my reference program (plus double checking the result manually). If my own answer would be the winning one, I'd exclude it from the list.
In case of several winners, I would have to pick only one. Therefore, some bonuses can be earned:

answers which accept more input than expected (e.g. outside the defined ranges)
answers which use a clever idea of making it short

I have taken a snapshot of the answers on 6th of March 2014, 19:45 UTC+1. The analysis is ongoing. Checking all the answers is harder than expected...

Shouldn't this be tagged [tag:code-challenge] for now? Also when is deadline?

Alt-name for this is Code-Blackjack

Java - 556
import java.util.*;class G{public static void main(String[]x){TreeMap<?,TreeMap>m=new TreeMap();try{Scanner s=new Scanner(System.in);for(;;){String[]a=s.nextLine().split("\t");a(new Long(a[1]),a(a[0],m)).put(a[2],a);}}catch(Exception e){}TreeMap<?,Map<?,String[]>>n=new TreeMap();for(TreeMap o:m.values())a(o.firstEntry().getKey(),n).putAll((Map)o.firstEntry().getValue());for(String[]o:n.lastEntry().getValue().values())System.out.println(o[0]+"\t"+o[1]+"\t"+o[2]);}static<T>Map a(T t,Map m){if(m.get(t)==null)m.put(t,new TreeMap());return(Map)m.get(t);}}

Program will read from STDIN.
import java.util.Map;
import java.util.Scanner;
import java.util.TreeMap;

class G {
    public static void main(String[] x) {
        TreeMap<?, TreeMap> m = new TreeMap();
        try {
            Scanner s = new Scanner(System.in);
            for (; ; ) {
                String[] a = s.nextLine().split("\t");
                a(new Long(a[1]), a(a[0], m)).put(a[2], a);
            }
        } catch (Exception e) {
        }
        TreeMap<?, Map<?, String[]>> n = new TreeMap();
        for (TreeMap o : m.values())
            a(o.firstEntry().getKey(), n).putAll((Map) o.firstEntry().getValue());
        for (String[] o : n.lastEntry().getValue().values())
            System.out.println(o[0] + "\t" + o[1] + "\t" + o[2]);
    }

    static <T> Map a(T t, Map m) {
        if (m.get(t) == null)
            m.put(t, new TreeMap());
        return (Map) m.get(t);
    }
}

Programm will read line by line until an exception occours (either ArrayIndexOutOfBoundsException when a blank line is encountered or NoSuchElementException if input ends without trailing new line). Each line read is added to the TreeMap m, which could have been defined as TreeMap<String, TreeMap<Long, TreeMap<String,String[]>>> (left-to-right: language, code-size, URL, input).
Then a result-TreeSet<Long, TreeSet<String, String[]>> n (left-to-right: code-size, URL, input) is build where the contents of every languages firstEntry() are aggregated.
lastEntry() of the aggregated TreeMap contains our result - we only need to print it.

Try on ideone.com (switched last two lines of input to show that all lines are read)

Had to add an variable to store BufferedReader -.-

Maybe Java wins this time, because it doesn't have a var keyword...

@ThomasW. just realized I left some unnecessary {} inside the golfed version -.-

Also just relaized I was using Integer. While int is shorter than long, Integer shurely should be golfed to Long -.-

Perl, 195 bytes
while(<>){/(\S+)\t(\d+)\t(.+)/;push@{$a{$1}},$3if$2==$l{$1};$l{$1}=$2,$a{$1}=[$3]if $2<($l{$1}//65536)}$m=(sort{$b<=>$a}values%l)[0];map{$l=$_;map{print"$l\t$m\t$_\n"if$l{$l}==$m}@{$a{$l}}}keys%l

Input is expected in STDIN, result is written to STDOUT:
C#      208     http://short.url/answer/poi
C#      208     http://short.url/answer/yac
Java    208     http://short.url/answer/jav

Ungolfed version
#!/usr/bin/env perl
use strict;
$^W=1;

# hash %language remembers the minimum count for a language
# %language: <language> => <minimum count>
my %language;
# hash %array remembers the URLs for the minimum count of the language
# %array: <language> => [<url>, <url>, ....]
my %array;

while(<>){
    # parse input line (no error checking)
    /(\S+)\t(\d+)\t(.+)/;
    my ($lang, $count, $url) = ($1, $2, $3);
    # add URL, if the count is the current minimum for the language
    if ($count == ($language{$lang}//0)) {
    # better, but longer version:
    # if (defined $language{$lang} and $count == $language{$lang}) {
        push @{$array{$lang}}, $url;
    }
    # create a new entry for the language, if there is a new minimum
    if ($count < ($language{$lang}//65536)) {
    # better, but longer version:
    # if (not defined $language{$lang} or $count < $language{$lang}) {
        $language{$lang} = $count;
        $array{$lang} = [$url];
    }
}

# Sort the minimal values in numerical descending order and
# get the first entry as maximum.
my $maximum = (sort { $b <=> $a } values %language)[0];

# Loop over all URLs of minimal answers for the language,
# but print only the entries for the languages with the largest
# minima.
foreach my $lang (keys %language) {
    foreach my $url (@{$array{$lang}}) {
        if ($language{$lang} == $maximum) {
            print "$lang\t$maximum\t$url\n";
        }
    }
}
__END__

Heiko, as @grovesNL pointed out, some programs might rely on the fact that there is a single maximum. Maybe you can check whether your program is affected. Just add a GolfScript  210 http://short.url/answer/210 line to the input and see if the output remains the same. Actually I think yours is not affected, because you're using [0] for the maximum, but I don't have Perl available at the moment to give it a try.

@ThomasW.: It is not affected. I have added the line and the output remains the same. After the first part that reads the file, data structure %l/%language contains the languages and their minimal values. Data structure %a/%array contains only those language/URL pairs, whose value is the minimum for this language. Then the minimum values are sorted in descending order and the first one is uses as global maximum and as filter condition for %a/%array.

Python 378 377 372
import sys
d=__import__("collections").defaultdict(list)
o={}
x=int
n="\n"
for i,l,u in[a.split()for a in sys.stdin.read().strip().split(n)]:d[i]+=[(l,u)]
for e,b in d.items():o[e]=[i for i in b if i[0]==str(min([x(i[0])for i in b]))]
print("".join(n.join("\t".join([u,s[0],s[1]])for s in y if x(s[0])==max(x(i[0][0])for i in o.values()))+n for u,y in o.items()).strip())

Input on the stdin:
C:\Users\gcq\Documents\python>type m.txt | python test.py
C#      208     http://short.url/answer/poi
C#      208     http://short.url/answer/yac
Java    208     http://short.url/answer/jav

And this is what i had before starting to compress it down, at 551 chars:
from collections import defaultdict
import sys
d = defaultdict(list)

for language, length, url in [a.split() for a in sys.stdin.read().strip().split("\n")]:
    d[language].append((length, url))

o = {}
for language, data in d.items():
    winval = data[0][0]
    for i in data:
        if int(i[0]) < int(winval):
            winval = i[0]
    o[language] = [i for i in data if i[0] == winval]

maxlen = max(int(i[0][0]) for i in o.values())

for language, dataa in o.items():
    for data in dataa:
        if int(data[0]) == maxlen:
            print("\t".join([language, data[0], data[1]]))

dg - 286 281 260 251 218 bytes
import '/sys'
d=dict!
for(a,b,c)in(map str.split$(sys.stdin.read!).splitlines!)=>d!!a=(d.get a list!)+(list'(int b,c))
for(i,l)in(d.items!)=>for(s,u)in l=>s==(max$map(i->fst$min i)d.values!)=>print$i+' '+(str s)+' '+u

Example:
$ cat langs.txt | dg langs.dg
C# 208 http://short.url/answer/poi
C# 208 http://short.url/answer/yac
Java 208 http://short.url/answer/jav

Ungolfed version:
import '/sys'

s = sys.stdin.read!
d = dict!
# convert the string into a list of tuples (name, score, url)
u = map str.split $ s.splitlines!
# add all the values to the dict (converting the score to an integer)
for (a, b, c) in u =>
  d!!a = (d.get a list!) + (list' (int b, c))
# computes the maximum value amongst the mins
m = max $ map (i -> fst $ min i) d.values!
for (i, l) in (d.items!) =>
  for (s, u) in l =>
    # if the score equals the maximum then print all the line
    s == m => print $ i + ' ' + (str s) + ' ' + u  # actually here .format()
                                                   # would be better

Q: What the heck is dg?
A: A programming language that compiles to CPython bytecode, much like Scala compiles to JVM's. That essentially means that dg is an alternative syntax for Python 3. It allows you to use all of the existing libraries, too.
More info here (even a tutorial!): https://pyos.github.io/dg

If I put that into a file, it has only 217 bytes (Linux line endings)

@ThomasW. Weird! Using cat langs.dg | wc -c I get 218!

Maybe trailing newline vs. no trailing newline?

@TheConstructor it could be, although it shouldn't save it with a newline oO

C# - 628
Here's a longer alternative of yours that uses DataTable:
using Microsoft.VisualBasic.FileIO;namespace System{using T=Data.DataTable;using R=Data.DataRow;using V=Data.DataView;using C=Data.DataColumn;class p{static void Main(string[] a){var I=typeof(Int32);T t=new T();t.Columns.AddRange(new[]{new C("a"),new C("b",I),new C("c"),new C("d",I)});var f=new TextFieldParser(a[0]);f.SetDelimiters("\t");while(!f.EndOfData){var r=t.NewRow();r.ItemArray=f.ReadFields();t.Rows.Add(r);}foreach(R r in t.Rows){r[3]=t.Compute("min(b)","a='"+r[0]+"'");}V v=new V(t);T s=v.ToTable();foreach(R r in s.Select("b='"+t.Compute("max(d)","")+"'")){Console.WriteLine(String.Join("\t",r[0],r[1],r[2]));}}}}

I originally thought I may have gained slight code reduction from using max/min with DataTable, but the types required to build the DataTable (rows/columns/view) add a lot of length, unfortunately. I'm new to code golfing so maybe somebody would be able to reduce it further. Still a fun challenge.

Rebol - 314
d: map[]foreach r read/lines to-file system/script/args[r: split r tab p: take r r/1: to-integer r/1 r/2: reduce[r/2]either none? d/:p[repend d[p r]][case[d/:p/1 > r/1[d/:p: r]d/:p/1 = r/1[append d/:p/2 r/2]]]]l: 0 foreach[k v]d[l: max l v/1]foreach[k v]d[if l = v/1[foreach n v/2[print rejoin[k tab v/1 tab n]]]]

un-golfed
d: map []

foreach r read/lines to-file system/script/args [
    r: split r tab
    p: take r
    r/1: to-integer r/1
    r/2: reduce [r/2]
    either none? d/:p [repend d [p r]] [
        case [
            d/:p/1 > r/1 [d/:p: r]
            d/:p/1 = r/1 [append d/:p/2 r/2]
        ]
    ]
]

l: 0 foreach [k v] d [l: max l v/1]
foreach [k v] d [
    if l = v/1 [
        foreach n v/2 [print rejoin [k tab v/1 tab n]]
    ]
]

Usage example:
$ rebol script.reb data.txt
C#    208   http://short.url/answer/poi
C#    208   http://short.url/answer/yac
Java  208   http://short.url/answer/jav

C# - 515
Expects a file name as argument
using System.Collections.Generic;namespace N{using S=SortedList<int,T>;class T:List<string>{static void Main(string[]a){var d=new Dictionary<string,S>();int n,m=0;T w=new T();foreach(var l in System.IO.File.ReadAllLines(a[0])){var p=(a=l.Split('\t'))[0];n=int.Parse(a[1]);if(!d.ContainsKey(p))d.Add(p,new S());if(!d[p].ContainsKey(n))d[p].Add(n,new T());d[p][n].Add(l);}foreach(var e in d){n=e.Value.Keys[0];if(n==m)w.AddRange(e.Value[n]);if(n>m)w=e.Value[m=n];}foreach(var e in w)System.Console.WriteLine(e);}}}

First I designed my C# program to be straight forward, because I wanted to have a kind of reference program. But then I decided to also jump into contest myself and golfed it. This is one of the earlier versions of the code + some comments:
// N: namespace
// P: Program
// S: type definition: sorted dictionary
// a: arguments
// d: data container
// r: lines read from text file
// l: single line from r
// t: tabbed part of l after splitting
// p: programming language name
// n: character count
// m: maximum character count
// w: list of winners
// e: entry in data container
// c: winner candidate
using System.Collections.Generic;
namespace N
{
    using S = SortedList<int, P>;
    public class P : List<string>
    {
        public static void Main(string[] a)
        {
            var r = System.IO.File.ReadAllLines(a[0]);
            // Make it a data structure
            var d = new Dictionary<string, S>();
            foreach (var l in r)
            {
                var t = l.Split('\t');
                var p = t[0];
                var n = int.Parse(t[1]);
                if (!d.ContainsKey(p)) d.Add(p, new S());
                if (!d[p].ContainsKey(n)) d[p].Add(n, new P());
                d[p][n].Add(l);
            }
            // Get the maximum values
            var m = 0;
            P w = null;
            foreach (var e in d)
            {
                foreach (var s in e.Value.Keys)
                {
                    if (s > m)
                    {
                        w = e.Value[s];
                        m = s;
                    }
                    else if (s == m)
                    {
                        w.AddRange(e.Value[s]);
                    }
                    break; // Break here to get the shortest solution per language
                }
            }
            // Print everything on console
            foreach (var e in w)
            {
                System.Console.WriteLine(e);
            }
        }
    }
}

It seems 515 file size on my disk included a byte order mark. Copy/paste from here is only 512 bytes.

C# - 460 359
After realizing how bulky my DataTable solution was, I created the following example using Linq. It uses the same methodology as my previous solution.
Golfed
namespace System{using Linq;using IO;class p{static void Main(string[]i){var l=(from f in File.ReadAllLines(i[0])let s=f.Split('\t')select new Tuple<string,int,string>(s[0],Convert.ToInt16(s[1]),f)).ToList();foreach(var f in l.Where(a=>a.Item2==l.Where(b=>b.Item1==l.Single(c=>c.Item2==l.Max(d=>d.Item2)).Item1).Min(e=>e.Item2)))Console.WriteLine(f.Item3);}}}

Ungolfed
namespace System
{
    using Linq;
    using IO;
    class p
    {
        static void Main(string[]i)
        {
            var l=(from f in File.ReadAllLines(i[0])
                   let s=f.Split('\t')
                   select new Tuple<string, int, string>(s[0],Convert.ToInt16(s[1]),f)).ToList();
            foreach(var f in l.
                Where(a=>a.Item2==l.
                    Where(b=>b.Item1==l.
                        Single(c=>c.Item2==l.
                            Max(d=>d.Item2)).Item1).
                                Min(e=>e.Item2)))
            Console.WriteLine(f.Item3);
        }
    }
}

I'm still fairly new to Linq so I'm almost positive those expressions can be reduced further.
From your question, it's not clear if there is a single maximum length solution. For my answers I've used the assumption that there is a single maximum point (ie. if there was a GolfScript maximum of 210 as well, it may fail based on the Single maximum record returned). Heiko's solution would have the same issue. To fix this, we would have to add another step that contained a list of tied maxima to check minima for each language.

It seems to me that you could save f as Item3 (instead of s[2]) and then write f.Item3 to the Console without the need to recombine Item1 and Item2.

Thanks for the clarification of the requirements. Derived from reality I would say it could happen that someone (unexperienced, obviously) writes a GolfScript with 210 characters.

If you convert the first foreach-loop to LINQ, it becomes even shorter: namespace System{using Linq;using IO;class p{static void Main(string[]i){var l=(from f in File.ReadAllLines(i[0])let s=f.Split('\t') select new Tuple<string, int, string>(s[0],Convert.ToInt16(s[1]),f)).ToList();foreach(var f in l.Where(a=>a.Item2==l.Where(b=>b.Item1==l.Single(c=>c.Item2==l.Max(d=>d.Item2)).Item1).Min(e=>e.Item2)))Console.WriteLine(f.Item3);}}}

@ThomasW.: Good idea for the split. Thanks for the foreach loop conversion, I had thought about it adding that originally but I didn't know I could use let that way.

I think I begin to understand what your query does :) and it will also fail for simpler input like this: Golfscript 100 ..., C# 1 ..., C# 200 .... This probably needs some rework

@ThomasW. That input should still be fine. I was talking about the case where you may have two or more identical maximums, ie. GolfScript 200 ... C# 200 ... Java 200 ... C# 100 .... In this case it's not clear that C# 100 ... wins because of the maximum ties. There are some better approaches to that but it would likely require more code (I wouldn't be able to use .Single() in the manner I do here).

C++ - 535
Will only output the answers tied for longest position after selecting only the shortest answers of each language as potential winners.
#include<fstream>
#include<iostream>
#include<string>
#include<vector>
using namespace std;
int main(){

    string s;
    vector<string>l;
    vector<int>v;
    vector<string>u;
    cin>>s;

    ifstream i(s.c_str());

    do{
        int n;
        i>>s;
        if(i.eof())break;
        l.push_back(s);
        i>>n;
        v.push_back(n);
        i>>s;
        u.push_back(s);
    }while(1);

    for(int i=0;i<l.size();i++){
        for(int j=0;j<l.size();j++){
            if(l[j]==l[i]){
                if(v[i]>v[j])l[i]="";
                else if(v[i]<v[j])l[j]="";
            }
        }
    }
    int n=0;
    for(int i=0;i<v.size();i++)
        if(n<v[i]&l[i]!="")n=v[i];

    for(int i=0;i<v.size();i++)
        if(v[i]==n)cout<<l[i]<<'\t'<<v[i]<<'\t'<<u[i]<<endl;
}

Golfed (not as unreadable as some languages):
#include<fstream>
#include<iostream>
#include<string>
#include<vector>
using namespace std;
int main(){string s;vector<string>l;vector<int>v;vector<string>u;cin>>s;ifstream i(s.c_str());do{int n;i>>s;if(i.eof())break;l.push_back(s);i>>n;v.push_back(n);i>>s;u.push_back(s);}while(1);for(int i=0;i<l.size();i++)for(int j=0;j<l.size();j++)if(l[j]==l[i]){if(v[i]>v[j])l[i]="";else if(v[i]<v[j])l[j]="";}int n=0;for(int i=0;i<v.size();i++)if(n<v[i]&l[i]!="")n=v[i];for(int i=0;i<v.size();i++)if(v[i]==n)cout<<l[i]<<'\t'<<v[i]<<'\t'<<u[i]<<endl;}

Have you checked your output against the expected output? With 8 lines of input, it shall give only 3 lines as output. Your program seems to output everything.

@ThomasW. Oh, I misunderstood that part.  I will remove it for now.

@ThomasW. Does this look right?

If I put that into a file I get 540 bytes, not 535. Could you double check the length, please?

It is 535 characters.  We are both correct.  I can change it if you want.

I can compile your program, but it seems you have chosen to read the file name from standard input, which was not an option. Even if I enter the file name there, it seems stuck in an endless loop.

So the name is to be hard-coded in?

The file must be in the same directory as the .cpp file.

The two options were: a) get the file name as command line parameter (e.g. Cpp.exe input.txt) or b) read the file content from standard input (e.g. Cpp.exe < input.txt)

I have specified a full path and file name. I hope that would work for ifstream as well. BTW: the path of the CPP file doesn't matter. If a path matters then it's the path of the compiled executable.

I can't get it to work through the command prompt.  I have just been running it through Visual Studio.


text:
Align a horizontally-aligned, right-angled ASCII shape along a diagonal
Your program will receive an ascii shape consisting of only underscores _, pipes |, spaces  and newlines \n, which, together, represent a horizontally-aligned, right-angled shape. Together with this shape, on a new line, you will also receive either a positive or a negative integer, whose absolute value will be either one of 45, 135, 225 or 315 — signifying the degree that the original image should rotate. For instance:
 _
| |_
|___|
-45

Your program's job is to rotate this shape the provided amount of degrees. Following are the example outputs, for all possible degrees, given the above provided example shape:

for 45° or -315°:
 /\
/ /
\ \
 \/

for 135° or -225°:
 /\
/  \
\/\/

for 225° or -135°:
/\
\ \
/ /
\/

for 315° or -45°:
/\/\
\  /
 \/

Remember though: the original shape above is just an example. Your program should be able to handle any shape constructed with said characters (within reason, of course — let's say: with a maximum of 100 x 100 characters).
The output of your program should be the rotated image, constructed with forward slashes /, backward slashes \, spaces and newlines.
Shortest code, after one week, wins.
Note: although this challenge offers many similarities with these [1] [2] two challenges, I think this one offers just enough variation (because the original image is not supposed to be rotated at 90° angles) for it to generate interesting new results.

This question is ill-defined. How should we rotate this shape? __ |  | |__| (sp _ _ sp, | sp sp |, | _ _ |) This is a 1.5 by 2 box, judging by your example above.

Python 2 - 285
import sys
l=list(sys.stdin)
L=len;R=range
r=L(l)-2
c=L(max(l,key=L))-2
k=r+c+1
d=int(l[-1])/90%4
a=d/2;x=a^d%2
z=1-x*2;t=1-a*2
m=[k*[' ']for i in R(k)]
for i in R(r+1):
 for j in R(L(l[i])-1):m[a*c+x*r+z*i+t*j][r-a*r+x*c-t*i+z*j]='  \/'[ord(l[i][j])/40^d%2]
for s in m:print''.join(s)

I strongly suspect it can be golfed more, tips are welcome.
Examples:
% python r45.py <<< "||
_____
45"
 /
\ /
 \
  \
   \
    \

% python r45.py <<< " _
| |_
|___|
-45"

    / \
 / \ /
    /
 \ /
  \

If the 2nd one looks misaligned, that's because of some issues with the question:
First, the underscore is at the bottom of the character "cell", while the pipe, slash and backslash are all in the middle. It would be fairer to use a dash (-) instead of underscore.
Second, with the rotation of the heart in the question, you couldn't represent it in any reasonable way if the first line was ___ (3 underscores) instead of
 _, as the 1st and 3rd underscore would be "capping" the pipes in the 2nd line.
Here's a better heart:
% python r45.py <<< " _
| |
|  _
|   |
 ___
-45"

 / \ / \

 \     /
  \   /
   \ /

Mathematica <225> 27
Second Version
Strange as it may seem, both the input and output are ASCII,
"constructed with forward slashes /, backward slashes \, spaces and newlines"!
The command works by Rotateing the text within the Mathematica environment, while leaving the string in ASCII.  The output is directly editable, as is, with no additional keystrokes or commands required.  However, it will not export directly as ASCII.  It can, however be exported directly as an image, as I have done here.
inputString =
 " _
  | |_
  |___|";

The following works by rotating the ASCII string on-screen.
s_~f~d_ := Rotate[s, -d Degree]

Test Cases
f[s, 45]
f[s, -315]

f[s, 135]
f[s, -225]

f[s, -135]
f[s, 225]

f[s, 315]
f[s, -45]

Your output is not "constructed with forward slashes /, backward slashes , spaces and newlines"

@aditsu.  You were correct.  However, it is now so constructed. They are rotated, but they remain ASCII characters.

Nope, those are still underscores and pipes (drawn at an angle)

-1 because you're not following the rules

HTML/CSS/Javascript
FIDDLE : http://jsfiddle.net/c5h2d/1/
<!DOCTYPE HTML>

<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<title>Untitled Document</title>

<textarea id=t cols="30" rows="10"></textarea>
<input type="text" id=i /><button onclick='rotate()'>Rotate</button>

Even in browsers that support transform:rotate directly (chrome/chromium does not), your output is not "constructed with forward slashes /, backward slashes , spaces and newlines". Basically you're doing the same thing as David Carraher.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Strange ways to keep an endless high CPU usage
There are some ways which cause an endless high CPU usage such as catching recursion exception. e.g. the following script in Python causes that problem:
def main():
    try:
        main()
    except:
        main()

main()

This is an obvious and simple example, but there might be some strange ways to arise such problems.

possible duplicate of Minimal code CPU stress-tester…

@ace not really. The other one is a code-golf.

@ace - I don't think so...

As per the tag info for popularity contest, we require challenges to have objective validity criteria. There is no such criteria for this challenge (how high is "high CPU usage"?), so I'm closing it.

JavaScript
Even JavaScript beginners should understand the code below!
// JavaScript for Beginners
// Variables, for loops

// The line below will set a global variable called "x" to 90
// Note: never use this style of code. It is evil!
х = 90;

// The line below will set a local variable called "x" to 90
var x = 90;

for (var х=0; х<10; x++) {
  console.log('JS for beginners: this is iteration #' + х);
}

Warning

 It produces an infinite loop
 Don't run it in your browser unless you know how to kill the tab/process or restart the machine :)

How does it work?

 There are two kinds of variable names in the code:
 1. the normal 'x' which can be found on the keyboard
 2. a cyrillic 'х': http://en.wikipedia.org/wiki/%D0%A5

That would be impossible to catch... Nice one.

Reminds me of http://en.wikipedia.org/wiki/IDN_homograph_attack

@aditsu In fact, I copied the cyrillic 'x' from that page :)

@impinball You can tell, because the SE code parser colors both xs differently. The second and fifth x are slightly darker than the others.

@Timtech I'm sorry, but this method does not work anymore :P (I've set the language hint for the formatter to 'none'.)

@ComFreek Oh no! $-/

Java
This may look like a silly but harmless program that should finish in an instant. However, appearances can be deceiving...
public class Cpu {
    public static void main(final String... args) {
        // let's print a bunch of empty lines
        int i = 0; // the loop counter
        while (i < 10) {
            // the unicode representation of a line feed is \u000A  /*/// loop start
            char c = 0x000A;
            System.out.print(c); // print the line feed
            i++; // increment and repeat  /*/// loop end
        }
    }
}

And another solution:
public class Cpu2 {
    public static void main(final String... args) {
        // now let's add the numbers from 0 to 255
        int sum = 0;
        for (byte i = 0; i < 255; i++) {
            sum += i;
        }
        System.out.println("The sum is: " + sum);
    }
}

Spoilers:

 1. The "\u000A" inside the comment generates an actual line break in the source code, so all the instructions in the loop are commented out. Without the "\u000A" (or even without the backslash before "u"), the loop would just run 10 times and finish.
 2. The 'byte' type is signed, so 'i' never reaches 255, but wraps around from 127 to -128

Is it the String args... that crashes it?

The second one has an implicit cast, if I'm reading it properly. If not, then I would be surprised if it compiles.

@impinball there is no String args... and no crash. Both programs compile and run with no errors and no warnings.

Ok. Then I'm lost.

The first has the contents of the loop commented out. That's why there

@impinball well, you're not wrong, but you haven't really explained it :) I added spoilers now.

I forgot about byte ending at 127 instead of 255 (all numbers are signed in Java)

JavaScript
var i = 0;
while (i < 10) {
  console.log("Hi #" + i + "!");
  // Let the user know we said hi and now     \
  // continue the loop                        \
  i++;
}

This looks rather harmless... (run at your own risk)
Spoiler:

 Look at the commenting style. It will never get to 10 because of the escaped line breaks, which hold in comments. The problem would disappear completely if there was a line break after the final comment.

Nice use of escaping newlines. That is the closest thing to php's heredocs and nowdocs.

This doesn't work on newer browsers, which don't recognise line continuations in comments.

It's the same solution I used in this question, but this time it's 100% fair.
Bash, 12 bytes
:(){ :|:&};:

As found on Wikipedia.
WARNING: harmful code, don't run it on your computer!
It produces an endless 100% CPU load (until system crashes, but there's nowhere stated that it mustn't be harmful!);

Your answer actually has a slight error: http://stackoverflow.com/questions/21186724/why-is-whitespace-needed-for-this-to-be-parsed

Drop the one space and it works in zsh ;-)

As @shiona pointed out, without the space it won't work in Bash :P

So I opened top in a second tab and ran this. It didn't actually generate any CPU load. Not even a single percent on my system. I suspect this is precisely because of the process limit.

C++
Try calculating the first 100 weird numbers by brute-force:
#include<iostream>
#include<vector>
using namespace std;

vector<int> divisors(int i) {

    vector<int> divs;
    for(int k = 1; k < i; k++)
        if(i%k==0)
            divs.push_back(k);
    return divs;
}

bool u(vector<int>::const_iterator vi, vector<int>::const_iterator end, int s) {

    if(s == 0) return 0;
    if(vi == end) return 1;
    return u(vi + 1, end, s - *vi) & u(vi + 1, end, s);
}

bool t(vector<int>&d, int i) {

    bool b = u(d.begin(), d.end(), i);
    if(b) cout<< i << endl;
    return b;
}

int main() {

    vector<int> divs;
    int n;
    cin>>n;

    for(int i = 2, j = 0; j < n; i++) {
        divs = divisors(i);

        int sum_divs = 0;
        for(int k = 0; k < divs.size(); k++)
            sum_divs += divs[k];

        if(sum_divs > i)
            if(t(divs, i))
                j++;
    }
}

I think you will find your CPU usage will spike, although it may not crash your computer.

Vodoo in java
public class W
{
    public static void main(String[]a)
    {
        int o=0,о=o;
        while(o<a.length)
        {
            System.out.println(a[o]);
            о++;
        }
    }
}

cyrillic letters work in java, too :P that o=o isnt a o=o but 2 variables

EcmaScript 6:
z=z=>{while(1)z()};_=i=>(i+=1,i-=1,i++,i--,--i,++i,i<<=2,i>>=2,i+=0|Math.round(1+Math.random())&1|0,z(x=>setInterval(x=>z(x=>new Worker('data:text/javascript,'+_.toSource()),5))));setInterval(x=>z(x=>_(...Array(i=9e3).map((x,z)=>z*3/2*2/4*4e2>>2<<2))),5)

This doesn't always use 100% of the CPU, but at least with Firefox it has the added bonus that Firefox keeps using up more and more memory, the whole interface locks up and the only way to stop it is to kill Firefox.

BASH
This is a simple bash script and i'm sure it will be down voted for not being "strange" enough, but i found this fun. It's not elegant and it's not some fancy way to do it but it does the job in a fun way. Plus i have not seen anyone attempt to eat CPU usage by hashing out the sha256sum recently. This pipes its output to the file "~/junk" and thus eats up diskspace too (just for the giggles). It spawns itself over and over again. crashed my machine shortly after starting it.
THIS CODE IS DANGEROUS
Enjoy!
Name it songthatneverends.sh
#!/bin/bash
x=1
exec 1> ~/junk 2>&1

        while [ $x -le 1 ]
                do
                        song="This is the song that never ends, yes it goes on and on my friend. Some people started singing it, not knowing what it was, and they'll continue singing it forever just because..."
                        echo $song
                        $x"_"$song | sha256sum
                        x= $(( $x - 1 ))
                        sh songthatneverends.sh
                done

No part of this script does what it's intended to. 1. exec 1> ~/junk 2>&1 truncates the junk file. It never gets bigger than a few hundred bytes. 2. $x"_"$song produces an error, since the command 1_This does not exist. 3. x= $(( $x - 1 )) produces another error. Because of the space, the variable x gets set to the empty string before executing -1. 4. Because of the process limit, it doesn't actually generate any CPU load. Not even a single percent on my system.

Javascript:
(function(a){a=arguments.callee;try{a();}catch(e){a();}})();

How does this work?
This runs itself until it reaches the maximum recursion level. When this happens, the last function exits and runs itself again!
On my CPU, I got all 4 cores around 10-40%, using Opera 12.16.
To use it, simply press F12 on your keyboard, paste this in the console and done!
To stop, close the console.

C#
using System;
using System.Threading;

// without a program, we wouldn't have much, would we?
class Program
{
    // somehow, houdini always gets out of his locks
    static Guid houdini = Guid.NewGuid();

    // this is the main part
    static void Main()
    {
        // let's make some threads
        Thread[] threads = new Thread[20];
        // all of them
        for (int i = 0; i < threads.Length; i++)
        {
            // new threads are always better than stale ones
            threads[i] = new Thread(DoThing);
            // started threads are better than stopped ones
            threads[i].Start();
        }
    }

    // this does the thing
    static void DoThing()
    {
        // this does it lots of times
        while (true)
        {
            // this lock appears to limit CPU usage
            lock ((object)houdini)
            {
                // this line indicates I cannot think of anything clever to put here
                Thread.SpinWait(1000000);
            }
        }
    }
}

Spoiler:

 lock ((object)houdini) creates a new reference to houdini and locks on that, so each lock gets a different reference, and the locks do not affect the other threads. This is why you should never use a value type for locking.

Ruby
Evals itself until it can't, then starts over.
begin;eval `cat spin.rb`;rescue;`ruby spin.rb`;end

Must be in a file named "spin.rb"

CMD 12 bytes (counting the important newline):
:a
a&goto a

This works because it has to look for the command a, and it has to spend time to produce an error message.
Then it goes to a: and repeats!
The newline is IMPORTANT and counts as 2 bytes under windows.
Since this is a Windows solution, the byte count has to (unfortunately) be made in Windows' way.
To stop it, close the window or press +c

DOS Batch
%0|%0

This is actually another way to code the fork bomb.
Equivalent in POSIX
$0|$0

Spoiler:

 It runs the script itself, and attempts to pipe the output to the same script. Another way to write an endless loop of this is simply %0 or $0. The difference is that it doesn't nearly as much strain the CPU because it is simple recursion.

I totally used this as a golf for a similar question.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Shortest code to post a string to WebServer
In bash, this code posts String to post to http://www.example.com/target:
curl -d "String to post" "http://www.example.com/target"

NOTE: You should use http://www.example.com/target as the URL to POST to, without any change.
I think all other languages that I have knowledge in need a lot more code than this. I'm not blaming other languages. Let's see the interesting answers that reduce the effort.
Your task is to write the shortest code possible to send a POST request to a sample server, http://www.example.com/target, and the request data should be the string String to post.

It's not clear that the title and the body are asking for the same thing, let alone what you consider to be sufficient and necessary to qualify as a solution.

Dont you understand the question? this is the best i can explain. the title and body are clear for me

@Peter Taylor I have edited the question to try to make it more clear.

Another one of the same size: POST http://www.example.com/target<<<'String to send'. (POST is either a stand-alone script or symlink to lwp-request, a command line tool delivered with the LWP perl module.)

@manatwork why don't you post it as an answer?

@ace, I felt it's not valuable enough for an answer.

Is it sufficient that the request be permitted by the HTTP spec (and if so, which?), or is there a set of additional minimal headers to make it useful to the server?

I've reduced my answer by 3 bytes. Can you check it?

Yeah, updated... :)

I've reduced my answer by 9 characters!

@toothbrush Yeah updaetd that :)

Some of the answers are posting to example.com, and some to www.example.com. You may wish to clarify that requirement, one way or the other.

@ViswanathL Yes, I suggest you clarify.

Yeah i updated the url, Please change and update th answers accordingly

@ViswanathL OK. I've updated my solution, too. Do we really need the http: part, too?

@toothbrush. The http: part is the protocol. I'd presume that if your language infers the protocol, that's fine. But there is a real difference between example.com and www.example.com: they're two different names.

Bash: 90 characters
No external tools, just pure bash.
echo 'POST /target HTTP/1.0
Content-length:14

String to post'>/dev/tcp/www.example.com/80

That's Bash running on Linux.  I have a Bash commandline for my windows box but do not have /dev

Not sure whether it depends on /dev/, but /dev/tcp/ doesn't exist anywhere – it is just Bash's own syntax for opening a TCP connection. Personally I would expect it to work regardless the presence of /dev/. It certainly works in Cygwin, but I have a feeling that you use MinGW. Could you please @JerryJeremiah post us some details about this? Like Bash's version, the package from which it comes, what actually happens when you run my posted code. (By the way, should not happen anything visible – Bash just connects, transfers, disconnects, exits with 0.)

I'm astonished to announce this works - even without a /dev/ directory!  I saw the /dev/ and jumped to a stupid conclusion.  I am running the bash version 3.1.17 from the bmatzelle/gow github repo.

For enybody else that never knew you could do that, look here: http://www.linuxjournal.com/content/more-using-bashs-built-devtcp-file-tcpip

Rebol (51 chars)
write http://www.example.com/target"String to post"

Above works in Rebol 3.  Below is a Rebol 2 version (63 chars):
read/custom http://www.example.com/target[post"String to post"]

Javascript, 84
with(XMLHttpRequest())open('POST','//www.example.com/target'),send('String to post')

Try in the console of any page served over plain HTTP.

I think you can replace http://www.example.com/target with ://.example.com/target

@ClydeLobo. Can you get rid of the www? I don't see how that's legitimate.

@TRiG OK. It's now fully compliant with the question.

Powershell v3 - 58
iwr http://www.example.com/target -b "String to post"-me 3

Lua (texlua), 73 bytes
require("socket.http").request("www.example.com/target","String to post")

The question said to POST to http://www.example.com/target.

@toothbrush: Thanks, I missed www.. I have updated the answer and also changed .org to .com. http:// can be omitted, because it is the default.

Great! I know that you can omit the http:// part - I wasn't saying to include it.

Smalltalk (Smalltalk/X), 64 71
HTTPInterface post:'http://www.example.com/target'with:'String to post'

this contains 71 characters

sorry - you're right - I tested (and counted) another string to post. Embarrased!

newLISP - 58 characters
(post-url "http://www.example.com/target""String to post")

Bash - 63 chars
wget --post-data="String to post" http://www.example.com/target

PHP - 100 bytes
<?fwrite(fsockopen('www.example.com',80),'POST /target HTTP/1.0
Content-length:14

String to post');

I apologize for any similarities to manatwork's bash solution, but I believe that this is the shortest way to do it.

There is a solution I wasn't expecting. Well done!

You need to use www.example.com, not example.com.

@toothbrush they do resolve to the same IP. But I agree that  for fairness, everyone should be using exactly the same URL (and, that the question was modified after this was posted to make that explicit).

@primo Yes, that's true. However, the question did originally say that the code should POST to http://www.example.com/target.

Tcl - 85 chars
package require http
http::geturl "//www.example.com/target" -query "String to post"

VBScript: 114 characters
dim x:set x=createobject("MSXML2.ServerXMLHTTP")
x.open"post","http://example.com/target"
x.send"String to post"

Slightly smaller: With CreateObject("MSXML2.ServerXMLHTTP"):.Open"POST","http://example.com/target":.Send"String to post":End With

Yes, you're right. Thank you for your comment!

The question now states that you need to POST to http://www.example.com/target. Note the www. part.

Game Maker Language (GMStudio), 66
http_post_string("http://www.example.com/target","String to post")

Python 2: 80 chars
Just using the standard library:
import urllib2 as u
u.urlopen("http://www.example.com/target","String to post")

For an extra char, you can make the import less unusual.
import urllib2
urllib2.urlopen("http://www.example.com/target","String to post")

If you want to gain another char: import('urllib2').urlopen("http://www.example.com/target","String to post")

Sneaky. I like it

jQuery, 58
$.post("http://www.example.com/target", "String to post");

jQuery isn't a programming language...

, 47 chars / 51 bytes (noncompetitive)
ɟŏ`//www.example.com/target⍪String to post

Try it here (Firefox only).
Uses jQuery's post function.

PHP, 198 bytes
This is shamefully large...:
<?fopen('http://www.example.com/target','rb',0,stream_context_create(array('http'=>array('method'=>'POST','header'=>'Content-Type: application/x-www-form-urlencoded','content'=>'String to post'))));

At least I answered!

The second one complains that the HTTP wrapper does not support writable connections. If it were to do anything, it would attempt to overwrite the contents of the http request with the given string.

That's why I said that I'm not sure if it works. But I will remove the second answer.

The answer was removed, but still can be seen on the edit option.

But surely the Content-Type is text/plain, not application/x-www-form-urlencoded?

application/x-www-form-urlencoded is the default mime-type to send data to a server, text/plain is the default mime-type to receive any non-binary text without a declared mime-type.

Bash, 69 bytes
lwp-request -m POST -b http://www.example.com/target "String to post"

You can read about the lwp-request command here: http://linux.about.com/library/cmd/blcmdl1_POST.htm
I think that the answer given by @manatwork is a much more neat and compatible version.

Actionscript 3, 201
var aRV:URLVariables = new URLVariables();
aRV.STRING = "my string";
var aR:URLRequest = new URLRequest("http://www.example.com/target");
aR.method = URLRequestMethod.POST;
aR.data = aRV;
var aL = new URLLoader();
aL.load(aR);

no error handling

Batch/Powershell - 118 Chars
If you need to post from a Windows batch file or avoid changing ExecutionPolicy:
powershell -Command "(New-Object Net.WebClient).UploadString('http://www.example.com/target', 'p1=String%20to%20Post')"


text:
Calculate the cube root of a number
The goal of this code golf is to create a program or function that calculates and outputs the cube root of a number that's given as input.
The rules:

No external resources
No use of built-in cube root functions.
No use of methods/operators that can raise a number to a power (that includes square root, 4th root, etc.).
Your function/program must be able to accept floating-point numbers and negative numbers as input.
If the cube root is a floating-point number, then round it to 4 numbers after the decimal point.
This is a code golf, the shortest code in bytes wins.

Test cases:
27 --> 3
64 --> 4
1  --> 1
18.609625 --> 2.65
3652264 --> 154
0.001 --> 0.1
7  --> 1.9129

You can use all test cases above to test negative numbers (-27 --> -3, -64 --> -4 ...)

damn, if you allowed only numbers with precise cube, I would have a nice golf

Judging from your test cases I assume the program only needs to deal with real numbers?

@ace add complex and I change 2 letters in my code ;)

Is rounding to 4 digits after the decimal point a strong requirement? Or could it be something like "you aren't required to show more than 4 digits after the decimal point"?

With reference to my answer using Exp(ln(x)/3) (and several clones of it) please clarify if Exp is allowed. I assume pow(x,1/3) is not (even though it is technically not a cube root function.)

@ace: Yes, only real numbers.

Is a built-in for square-root allowed? By this, I mean that the builtin for square root is a single byte command and not a builtin that raises a number to the .5 power

@KritixiLithos No, it isn't. I understand that this wasn't really clear, but my intent was to disallow them. And I'm not invalidating any answers with this, so I updated the question.

J: 16 characters
Loose translation of the Haskell answer:
-:@((%*~)+])^:_~

Test cases:
   -:@((%*~)+])^:_~27
3
   -:@((%*~)+])^:_~64
4
   -:@((%*~)+])^:_~1
1
   -:@((%*~)+])^:_~18.609625
2.65
   -:@((%*~)+])^:_~3652264
154
   -:@((%*~)+])^:_~0.001
0.1
   -:@((%*~)+])^:_~7
1.91293

It works like this:
     (-:@((% *~) + ])^:_)~ 27
↔ 27 (-:@((% *~) + ])^:_) 27
↔ 27 (-:@((% *~) + ])^:_) 27 (-:@((% *~) + ])) 27
↔ 27 (-:@((% *~) + ])^:_) -: ((27 % 27 * 27) + 27)
↔ 27 (-:@((% *~) + ])^:_) 13.5185
↔ 27 (-:@((% *~) + ])^:_) 27 (-:@((% *~) + ])) 13.5185
↔ 27 (-:@((% *~) + ])^:_) -: ((27 % 13.5185 * 13.5185) + 13.5185)
↔ 27 (-:@((% *~) + ])^:_) 6.83313
...

In words:
half =. -:
of =. @
divideBy =. %
times =. *
add =. +
right =. ]
iterate =. ^:
infinite =. _
fixpoint =. iterate infinite
by_self =. ~

-:@((%*~)+])^:_~ ↔ half of ((divideBy times by_self) add right) fixpoint by_self

Not one of the best wordy translations, since there's a dyadic fork and a ~ right at the end.

Haskell - 35
c n=(iterate(\x->(x+n/x/x)/2)n)!!99

Example runs:
c 27  =>  3.0
c 64  =>  4.0
c 1  =>  1.0
c 18.609625  =>  2.6500000000000004  # only first 4 digits are important, right?
c 3652264  =>  154.0
c 0.001  =>  0.1
c 7  =>  1.9129311827723892
c (-27)  =>  -3.0
c (-64)  =>  -4.0

Moreover, if you import Data.Complex, it even works on complex numbers, it returns one of the roots of the number (there are 3):
c (18:+26)  =>  3.0 :+ 1.0

The :+ operator should be read as 'plus i times'

This deserves a +1. I've been refactoring generalized nth root algs for the last hour, and I just now arrived at the same result. Bravo.

@primo I instantly recalled all n'th root approximation algorithms, and after giving up on Taylor/Maclaurin series in APL I used this.

Using Newton method I got x=(2*x+n/x/x)/3, can you explain why you can use x=(x+n/x/x)/2 ? It converges slower but I can't explain why it converges...

@Michael because if you take x=cbrt(n), then x=(x+n/x/x)/2 is true. So is it true for your expression

@Michael I got there this way: http://codepad.org/gwMWniZB

@mniip Instantly recalled all of them? That's just a handful; there are, of course, infinitely many.

@primo those are formulas, algorithms are more generic

@mniip Hmm, I suppose you're right, those were all generated with the same method, really. My primary concern at the time was convergence rate.

SageMath, (69) 62 bytes
However, don't ever believe it will give you the result, it's very difficult to go randomly through all the numbers:
def r(x):
 y=0
 while y*y*y-x:y=RR.random_element()
 return "%.4f"%y

if you didn't insist on truncating:
def r(x):
 y=0
 while y*y*y-x:y=RR.random_element()
 return y

SageMath, 12 bytes, if exp is allowed
Works for all stuff: positive, negative, zero, complex, ...
exp(ln(x)/3)

I believe you are using an operator that can raise a number to a power.

ah ok, right, edited

+1 for a monumentally stupid algorithm that still satisfies the requirements.

@Mechanicalsnail Thanks. I hope it's obvious that what I do is a sort of recession :D However, if exp is allowed, I'm down to 12 and not being stupid at all :)

Considering that exp is short for "exponential function", which is "a function whose value is a constant raised to the power of the argument, especially the function where the constant is e.", and there is "No use of methods/operators that can raise a number to a power", exp is not allowed.

@mbo well, the problem statement is not quite clear regarding this; it explicitly mentions only power functions as examples of what's forbidden, but no exponentials.

An exponential is a power function. I quoted the definition right there. It raises a number to a power.

Python - 62 bytes
x=v=input()
exec"x*=(2.*v+x*x*x)/(v+2*x*x*x or 1);"*99;print x

Evaluates to full floating point precision. The method used is Halley's method. As each iteration produces 3 times as many correct digits as the last, 99 iterations is a bit of overkill.
Input/output:
27 -> 3.0
64 -> 4.0
1 -> 1.0
18.609625 -> 2.65
3652264 -> 154.0
0.001 -> 0.1
7 -> 1.91293118277
0 -> 1.57772181044e-30
-2 -> -1.25992104989

How does this work?

@justhalf I think this is the Newton's method of approximation basically.

Btw, fails on 0

Fails on -2, sorry for that.

@tohecz fixed for all cases.

You can save space using powers (x*x*x == x**3), also this allows you to save a space on the or (it follows a number and Python knows the number can't be the beginning of an identifier). exec"x*=(2.*v+x**3)/(v+2*x**3or 1);"*99;print x :)

@plg The problem description forbids the use of any exponential function, otherwise v**(1/.3) would be a sure winner.

Oh right, oops! :p

Javascript (55)
function f(n){for(i=x=99;i--;)x=(2*x+n/x/x)/3;return x}
BONUS, General formulation for all roots
function f(n,p){for(i=x=99;i--;)x=x-(x-n/Math.pow(x,p-1))/p;return x}
For cube root, just use f(n,3), square root f(n,2), etc...
Example : f(1024,10) returns 2.
Explanation
Based on Newton method :
Find : f(x) = x^3 - n = 0, the solution is n = x^3
The derivation : f'(x) = 3*x^2
Iterate :
x(i+1) = x(i) - f(x(i))/f'(x(i)) = x(i) + (2/3)*x + (1/3)*n/x^2
Tests
[27,64,1,18.609625,3652264,0.001,7].forEach(function(n){console.log(n + ' (' + -n + ') => ' + f(n) + ' ('+ f(-n) +')')})

27 (-27) => 3 (-3)
64 (-64) => 4 (-4)
1 (-1) => 1 (-1)
18.609625 (-18.609625) => 2.65 (-2.65)
3652264 (-3652264) => 154 (-154)
0.001 (-0.001) => 0.09999999999999999 (-0.09999999999999999)
7 (-7) => 1.912931182772389 (-1.912931182772389)

One character shorter: function f(n){for(i=x=99;i--;)x-=(x-n/x/x)/3;return x}

Can be reduced to 47 bytes f=(n)=>eval('for(i=x=99;i--;)x=(2*x+n/x/x)/3')

Perl, 92 bytes
sub a{$x=1;while($d=($x-$_[0]/$x/$x)/3,abs$d>1e-9){$x-=$d}$_=sprintf'%.4f',$x;s/\.?0*$//;$_}

The function a returns a string with the number without
an unnecessary fraction part or insignificant zeroes at the right end.

Result:
              27 --> 3
             -27 --> -3
              64 --> 4
             -64 --> -4
               1 --> 1
              -1 --> -1
       18.609625 --> 2.65
      -18.609625 --> -2.65
         3652264 --> 154
        -3652264 --> -154
           0.001 --> 0.1
          -0.001 --> -0.1
               7 --> 1.9129
              -7 --> -1.9129
 0.0000000000002 --> 0.0001
-0.0000000000002 --> -0.0001
               0 --> 0
              -0 --> 0

Generated by
sub test{
    my $a = shift;
    printf "%16s --> %s\n", $a, a($a);
    printf "%16s --> %s\n", "-$a", a(-$a);
}
test 27;
test 64;
test 1;
test 18.609625;
test 3652264;
test 0.001;
test 7;
test "0.0000000000002";
test 0;

The calculation is based on Newton's method:

PHP - 81 bytes
Iterative solution:
$i=0;while(($y=abs($x=$argv[1]))-$i*$i*$i>1e-4)$i+=1e-5;@print $y/$x*round($i,4);

What happens if it tries to calculate the cube root of zero?

It will just output "0" (thanks to the error suppression operator - "@").

0.0001 can be replaced by 1e-4 and 0.00001 by 1e.5.

This requires PHP<7 (0/0 gives NAN in PHP 7). $i=0; is unnecessary (-5 bytes. If it wasn´t, for would save one byte.) The space after print is not required (-1 byte). -R can save 3 bytes with $argn.

Save a pair of parantheses with while(1e-4+$i*$i*$i<$y=abs($x=$argn)) (-2 bytes).

Javascript: 73/72 characters
This algorithm is lame, and exploits the fact that this question is limited to 4 digits after the decimal point. It is a modified version of the algorithm that I suggested in the sandbox for the purpose of reworking the question. It counts from zero to the infinite while h*h*h<a, just with a multiplication and division trick to handle the 4 decimal digits pecision.
function g(a){if(a<0)return-g(-a);for(h=0;h*h*h<1e12*a;h++);return h/1e4}

Edit, 4 years later: As suggested by Luis felipe De jesus Munoz, by using ** the code is shorter, but that feature was not available back in 2014 when I wrote this answer. Anyway, by using it, we shave an extra character:
function g(a){if(a<0)return-g(-a);for(h=0;h**3<1e12*a;h++);return h/1e4}

Instead h*h*h you can do h**3 and save 1 byte

@LuisfelipeDejesusMunoz This answer is from 2014. The ** operator was proposed in 2015 and was accepted as part of ECMAScript 7 in 2016. So, at the time that I wrote that, there was no ** in the language.

Java, 207 182 181
Sometimes when I play golf I have two many beers and play really really bad
class n{public static void main(String[]a){double d=Double.valueOf(a[0]);double i=d;for(int j=0;j<99;j++)i=(d/(i*i)+(2.0*i))/3.0;System.out.println((double)Math.round(i*1e4)/1e4);}}

Iterative Newton's Method of Approximation, runs 99 iterations.
Here is the unGolfed:
class n{
    public static void main(String a[]){
        //assuming the input value is the first parameter of the input
        //arguments as a String, get the Double value of it
        double d=Double.valueOf(a[0]);
        //Newton's method needs a guess at a starting point for the
        //iterative approximation, there are much better ways at
        //going about this, but this is by far the simplest. Given
        //the nature of the problem, it should suffice fine with 99 iterations
        double i=d;

        //make successive better approximations, do it 99 times
        for(int j=0;j<99;j++){
            i=( (d/(i*i)) + (2.0*i) ) / 3.0;
        }
        //print out the answer to standard out
        //also need to round off the double to meet the requirements
        //of the problem.  Short and sweet method of rounding:
        System.out.println( (double)Math.round(i*10000.0) / 10000.0 );
    }
}

You may rename the args variable to something like z, reducing 6 chars. You may remove the space and the curly braces in the body of the for-loop, reducing 3 chars. You may replace 10000.0 by 1e4, reducing 6 chars. The class does not needs to be public, so you can reduce more 7 chars. This way it will be reduced to 185 characters.

Is the cast at the end really necessary? It does not for me.

@Victor Thanks for the good eye, the use of E notation for the 10000.0 double was a spectacularly good idea.  By the design of the question, I think it is legit to make this a method instead of a functioning cli class, which would reduce the size considerably.  With Java, I didn't think I had a chance, so I erred on the side of functional.

Welcome to CodeGolf! Don't forget to add an in-answer explanation of how this works!

@Quincunx, Thanks, made recommended change.

you can drop the (double) and .0s

You can save a few more chars: flip the location of the [] in the main method, and use Double instead of double for d, change #.0 to #, use System.out.print: class n{public static void main(String[]a){Double d=0;d=d.valueOf(a[0]);double i=d;for(int j=0;j<99;j++)i=(d/(i*i)+(2*i))/3;System.out.print((double)Math.round(i*1e4)/1e4);}} Also, consider using Floats instead of Doubles.

Except I need to debug it....

@Quincunx: Thanks, lot of good ideas here.  I think it would need to be Double d=0.0;d=d.valueOf which is longer than double d=Double.valueOf.  I didn't know that the space could be ommitted from the array decleration in String[]a, awesome tip!

@md_rasler The standard for declaring arrays is with the array on they type, not on the variable. Also, don't forget that you can leave off the 0 after the 0.

TI-Basic, 26 24 bytes
Input :1:For(I,1,9:2Ans/3+X/(3AnsAns:End

That directly uses the ^ operator, doesn't it. It is forbidden by the rules

@mniip: Is e^ is a single operator on the TI-83 series? I don't remember. Either way, it's violating the spirit of the rules.

@Mechanicalsnail It doesn't matter I would say. In most languages you could just do exp(ln(x)/3) or e^(ln(x/3)) if you allowed any of these two. But somehow I understand to exp(ln(x)/a) as too much equivalent to x^(1/a) to be allowed by the rules :-/

Exponential function: "a function whose value is a constant raised to the power of the argument, especially the function where the constant is e." ... "No use of methods/operators that can raise a number to a power"

Thanks for the catch @mbomb007, I wrote this answer more than 3 years ago and I will fix it to comply now.

APL - 31
(×X)×+/1,(×\99⍴(⍟|X←⎕)÷3)÷×\⍳99

Uses the fact that cbrt(x)=e^(ln(x)/3), but instead of doing naive ⋆ exponentiation, it computes e^x using Taylor/Maclaurin series.
Sample runs:
⎕: 27
3
⎕: 64
4
⎕: 1
1
⎕: 18.609625
2.65
⎕: 3652264
154
⎕: 0.001
0.1
⎕: 7
1.912931183
⎕: ¯27
¯3
⎕: ¯7
¯1.912931183

Seeing as there's a J answer in 16 characters, I must be really terrible at APL...

Js 57 bytes
f=(x)=>eval('for(w=0;w**3<1e12*x;w++);x<0?-f(-x):w/1e4')

f=(x)=>eval('for(w=0;w**3<1e12*x;w++);x<0?-f(-x):w/1e4')
document.getElementById('div').innerHTML += f(-27) + '
'
document.getElementById('div').innerHTML += f(-64) + '
'
document.getElementById('div').innerHTML += f(-1) + '
'
document.getElementById('div').innerHTML += f(-18.609625) + '
'
document.getElementById('div').innerHTML += f(-3652264) + '
'
document.getElementById('div').innerHTML += f(-0.001) + '
'
document.getElementById('div').innerHTML += f(-7) + '

'
document.getElementById('div').innerHTML += f(27) + '
'
document.getElementById('div').innerHTML += f(64) + '
'
document.getElementById('div').innerHTML += f(1) + '
'
document.getElementById('div').innerHTML += f(18.609625) + '
'
document.getElementById('div').innerHTML += f(3652264) + '
'
document.getElementById('div').innerHTML += f(0.001) + '
'
document.getElementById('div').innerHTML += f(7) + '
'

Javascript - 157 characters
This function:

Handle negative numbers.
Handle floating-pointing numbers.
Execute quickly for any input number.
Has the maximum precision allowed for javascript floating-point numbers.

function f(a){if(p=q=a<=1)return a<0?-f(-a):a==0|a==1?a:1/f(1/a);for(v=u=1;v*v*va)v=k;else u=k}return u}

Ungolfed explained version:
function f(a) {
  if (p = q = a <= 1) return a < 0 ? -f(-a)      // if a < 0, it is the negative of the positive cube root.
                           : a == 0 | a == 1 ? a // if a is 0 or 1, its cube root is too.
                           : 1 / f (1 / a);      // if a < 1 (and a > 0) invert the number and return the inverse of the result.

  // Now, we only need to handle positive numbers > 1.

  // Start u and v with 1, and double v until it becomes a power of 2 greater than the given number.
  for (v = u = 1; v * v * v  a)
      v=k;
    else
      u=k
  }

  // At this point u <= cbrt(a) and v >= cbrt(a) and they are the closest that is possible to the true result that is possible using javascript-floating point precision.
  // If u == v then we have an exact cube root.
  // Return u because if u != v, u < cbrt(a), i.e. it is rounded towards zero.
  return u
}

PHP, 61
Based on Newton's method. Slightly modified version of Michael's answer:
for($i=$x=1;$i++<99;)$x=(2*$x+$n/$x/$x)/3;echo round($x,14);

It works with negative numbers, can handle floating point numbers, and rounds the result to 4 numbers after the decimal point if the result is a floating point number.
Working demo

You can save two bytes with for($x=1;++$i<100;).... But using predefined variables as input is generally frowned upon. Better use $argv[1] or $argn.

GameMaker Language, 51 bytes
for(i=x=1;i++<99;1)x=(2*x+argument0/x/x)/3;return x

Befunge 98 - Work in progress
This language does not support floating point numbers; this attempts to emulate them. It currently works for positive numbers that do not start with 0 after the decimal point (mostly). However, it only outputs to 2 decimal places.
&5ka5k*&+00pv
:::**00g`!jv>1+
/.'.,aa*%.@>1-:aa*

It works by inputting the part before the decimal point, multiplying that by 100000, then inputting the part after the point and adding the two numbers together. The second line does a counter until the cube is greater than the inputted number. Then the third line extracts the decimal number from the integer.
If anyone can tell me why the third line only divides by 100 to get the correct values, please tell me.
IOs:
27.0       3 .0
64.0       4 .0
1.0        1 .0
18.609625  2 .65
0.001      0 .1
7.0        1 .91

0.1        0 .1

Smalltalk, 37
credit goes to mniip for the algorithm; Smalltalk version of his code:
input in n; output in x:
1to:(x:=99)do:[:i|x:=2*x+(n/x/x)/3.0]

or, as a block
[:n|1to:(x:=99)do:[:i|x:=2*x+(n/x/x)/3.0].x]

Haskell: 99C
Can't beat @mniip in cleverness. I just went with a binary search.
c x=d 0 x x
d l h x
 |abs(x-c)<=t=m
 |c < x=d m h x
 |True=d l m x
 where m=(l+h)/2;c=m*m*m;t=1e-4

Ungolfed:
-- just calls the helper function below
cubeRoot x = cubeRoot' 0 x x

cubeRoot' lo hi x
    | abs(x-c) <= tol = mid           -- if our guess is within the tolerance, accept it
    | c < x = cubeRoot' mid hi x      -- shot too low, narrow our search space to upper end
    | otherwise = cubeRoot' lo mid x  -- shot too high, narrow search space to lower end
    where
        mid = (lo+hi)/2
        cubed = mid*mid*mid
        tol = 0.0001

You can use an infix operator for d (like (l#h)x) to save a byte for each call. c then becomes id>>=(0#).

You can remove spaces around c < x.

You can use 1>0 instead of True.

J 28
*@[*(3%~+:@]+(%*~@]))^:_&|&1

Using Newtons method, finding the root of x^3 - X the update step is x - (x^3 - C)/(3*x^2), where x is the current guess, and C the input. Doing the maths on this one yields the ridiculously simple expression of (2*x+C/x^2) /3 . Care has to be taken for negative numbers.
Implemented in J, from right to left:

| Take abs of both arguments, pass them on
^:_ Do until convergence
(%*~@]) is C / x^2 (*~ y is equivalent to y * y)
+:@] is 2 x
3%~ divide by three. This yields the positive root
*@[ * positive_root multiplies positive root with the signum of C.

Test run:
   NB. give it a name:
   c=: *@[*(3%~+:@]+(%*~@]))^:_&|&1
   c 27 64 1 18.609625 3652264 0.001 7
3 4 1 2.65 154 0.1 1.91293

AWK, 53 bytes
{for(n=x=$1;y-x;){y=x;x=(2*x+n/x/x)/3}printf"%.4g",y}

Example usage:
$ awk '{for(n=x=$1;y-x;){y=x;x=(2*x+n/x/x)/3}printf"%.4g",y}' <<< 18.609625
2.65$

Thanks go to @Mig for the JavaScript solution which this is derived from. It runs surprisingly quickly given that the for loop requires the iteration to stop changing.

C, 69 bytes
i;float g(float x){for(float y=x;++i%999;x=x*2/3+y/3/x/x);return x;}

Just another implementation of Newton's method. Try it online!

Stax, 10 bytesCP437
╘♀┘A╕äO¶∩'

Run and debug online!
Explanation
Uses the unpacked version to explain.
gpJux*_+h4je
gp              Iterate until a fixed point is found, output the fix point
  Ju            Inverse of square
    x*          Multiplied by input
      _+h       Average of the value computed by last command and the value at current iteration
         4je    Round to 4 decimal digits

JAVA Solution
public BigDecimal cubeRoot(BigDecimal number) {
    if(number == null || number.intValue() == 0) return BigDecimal.ZERO;
    BigDecimal absNum = number.abs();
    BigDecimal t;
    BigDecimal root =  absNum.divide(BigDecimal.valueOf(3), MathContext.DECIMAL128);

    do {

        t = root;
        root = root.multiply(BigDecimal.valueOf(2))
                .add(absNum.divide(root.multiply(root), MathContext.DECIMAL128))
                .divide(BigDecimal.valueOf(3), MathContext.DECIMAL128);

    } while (t.toBigInteger().subtract(root.toBigInteger()).intValue() != 0);

    return root.multiply(number.divide(absNum), MathContext.DECIMAL128);
}

Welcome to PPCG! This is a [tag:code-golf] challenge, which means the goal is to solve the challenge is as little code as possible (counted in bytes of the source file). You should show some effort towards optimising your solution towards that goal and include the byte count in your answer.

Python Solution
def cube_root(num):
    if num == 0:
        return 0

    t = 0
    absNum = abs(num)
    root = absNum/3

    while (t - root) != 0:
        t = root
        root = (1/3) * ((2 * root) + absNum/(root * root))

    return root * (num / absNum)

MMIX, 44 bytes (11 instrs)
E0013FF0 48000002 E8018000 C1FF0100
1001FFFF 14010001 040101FF E401FFF0
11FF01FF 53FFFFFA F8020000

Disassembly
cbrt    SETH  $1,#3FF0      // y = 1.
        BNN   $0,0F         // if(x >= 0) goto loop (ensuring sign is correct)
        ORH   $1,#8000      // y = fnabs(y) (bit tricks!)
0H      SET   $255,$1       // loop: prev = y
        FMUL  $1,$255,$255  // y = prev *. prev
        FDIV  $1,$0,$1      // y = x /. y
        FADD  $1,$1,$255    // y = y +. prev
        INCH  $1,#FFF0      // y = y /. 2. (bit tricks again)
        FCMPE $255,$1,$255  // prev = comp_eps(y, prev)
        PBNZ  $255,0B       // if(prev) goto loop
        POP   2,0           // return {y,x}

It is up to the caller to set a suitable value of rE. Much faster convergence may be arranged by essentially setting y up by dividing the exponent by 3, but that costs another four instructions (though in time it pays really quickly, since three FDIVs cost about the same amount as two DIVs):
cbrt    SLU  $1,$0,1    // 3B010001 shift left to erase sign
        INCH $1,#8020   // E4018020 unbias exponent
        DIV  $1,$1,3    // 1D010103 divide by 3 as integer; approximate cube root
        INCH $1,#7FE0   // E4017FE8 rebias exponent
        SRU  $1,$1,1    // 3F010101 shift right again

and then continue from the second line in the original.
If mispredicted branches are particularly expensive, I would also replace the second and third lines of the original with:
        ZSN $255,$0,1       // 71020001 $255 = ($0 neg? 1 : 0)
        SLU $255,$255,63    // 3B02023F $255 <<= 63
        OR  $1,$1,$255      // C0010102 $1 |= $255

This costs one more instruction, but is branchless.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Code to display a string without calling any methods or functions
Write a code to display "I Love You".
Without using any built-in tools or functions like echo, System.out.println, printf, ...
You are not allowed to directly call any function throughout the program execution, but the system environment or the compiler-generated code might do. This way, you can put your code in a main function or method that is called by the running environment, but your program should not further call any functions.
By the way, a function is not an instruction, nor a command nor an operator. So calling sum(a, b) is not valid, but making a = b + c is, even if the compiler reduces this to internal function calls.
This is a popularity-contest, so the valid answer with the most upvotes wins.

how about treating stdout as a file and writing to it?

You are not allowed to call any function

So you are excluding c and many other languages.

People here tend not to accept answers so early, since usually accepting an answer means that the contest is over.

You can use cout <<< 'text here' in c

And php echo is not a function, but a language construction. (http://pt1.php.net/echo < documentation)

What is so bad about this challenge?  Most languages won't be able to compete, but some are free to do so. For example, GolfScript works fine.  PHP also works fine (paste text into blank text file).  There are many other languages that can do this challenge, so why are there so many downvotes and close votes?

@Quincunx This got 2 close votes as "too broad" and 2 as "primary opinion based". I do not think that this is particularly a great challenge (it is boring for me, but I really liked the brainfuck answer). However, I do not think that it should be closed (and I will vote for a reopen if it get closed). Being not much interesting and/or boring is not a close reason, and frankly, I am seeing people closing questions just for the sake of being closed using some lame excuse just because they find them boring or impossible to do in their favorite language.

@Victor IMHO it is unclear what you are asking, especially seeing all the discussions whether . is a function in Brainfuck. IMHO it is not, but as long as you don't say what is a function, it can't be verified.

-1 because what's considered a function call is a slippery slope. Everything printed to the screen eventually goes through a system call to the OS.

Let the system make a function call, but only YOU are not allowed to call any function throughout the program execution.

@tohecz, it is not me who are asking this. But anyway, if people think that the question has problems and should be closed, normally it is adequate to drop a comment explaining why (like you and Cruncher, but none of the closers, did). Otherwise, there would be no real point in closing the question as no one would be saying to the OP what should be improved to make the question be reopened. Further, if the question has a minor problem or unclear point, people should ask or inform the OP first. In most questions, the OP fixes or clarifies the minor point quickly and no close-voting is needed.

@Victor I agree (well, I come from TeX.SE, and we prefer comment->(improve or close) over downvote -> maybe comment -> close -> improve -> reopen), so I agree that it's wrong to cast a close vote without a comment if the reason for closure is not 100% obvious (like duplicate, spam, ...).

@tohecz Yeah, and I believe that "fixing the question if you might do" is much better than just "close for some minor unimportant problem". By the way, I think that this question is boring, but I took the freedom to edit it anyway and now I think that it addresses the informed problems.

@Cruncher I edited the question. Is it better now?

@vengets this is a popularity contest so please accept the answer with the most upvotes.

Html-asp- xhtml-tex-php.....
I didn't use any functions
I Love You

That is a polyglot. It's valid in php, html, xhtml, asp ...

valid in plain TeX as well, btw. And certainly doesn't use any functions, since there're no functions in the language at all.

@IsmaelMiguel I agree but I'm not the one you should say that to. You should tell the OP.

@IsmaelMiguel There! I fixed it.

Brainfuck
    ++++     ++++
  ++[>+++   ++++>++
 +>+++++++ ++++<<<-]
>+++.>++.<+++.>>+.+++
 ++++.--------------
   ---.<.<++++++++
     +++++.>>+++
      +++++++.+
        +++++
          .

The heart looks horribly thin here, but it is a lot better on a text editor / terminal

Here's a nicely indented version:
+++++ +++++ [
    > +++++ ++
    > +++
    > +++++ +++++ +
    <<< -
]
> +++ .
> ++ .
< +++ .
>> + .
+++++ ++ .
----- ----- ----- -- .
< .
< +++++ +++++ +++ .
>> +++++ +++++ .
+++++ + .

There's a much shorter one

@mniip Length is not a concern in this question since it is not code golf. So I'd go for a slightly longer one, which makes it easier to make it look like a heart.

nicely indented version:) LOL

aren't "." equivalent to putchar, which is in terms of brainfuck like echo?

@V-X the OP said "no functions", and I'm quite certain there aren't any functions in brainfuck

isn't that build-in tool - command?

@V-X I don't know, but the OP seems to be okay with this

Javascript:
var l='I love you.';
if(window)window.status=l;
if(document)if(document.readyState=='complete')document.innerHTML=document.title=l;
else document.onload=function(){document.innerHTML=document.title=l;};
throw new l;

The part where it says 'function' is as ASSIGNMENT!
There is nothing in the question saying I can't assign functions and trigger events.
And the 'throw new l;' will say "I love you" in the error message.
This will show 'I love you' in the console, document, title and in the status bar.

C
void main( )
{
  char *string="I Love You";
    volatile char *video = (volatile char*)0xB8000;
    while( *string != 0 )
    {
        *video++ = *string++;
        *video++;
    }
}

you are using the main function.

We are not allowed to call any function, Without main() how will you write a pgm ?

In this case gLibc calls main()

I don't know you wrote the question not me.

Nope, Read the question once again.. In this code no function call has been made. That's it..

@vengets int main(){ printf("I love you"); } I don't do any function call, that's the processor who does. (Almost like: "It's not my fault, it's the knive who killed that guy!")

printf() is a function, you are using it.. Problem here is u should not use it. (Let the processor call ..It doesnt matter..) Got it ??

Mathematica, GolfScript, Python Interpreter, bc
This outputs the sentence:
"I love you"

Polyglot; works in GolfScript and Python Interpreter too. Works in PHP but outputs with the quotes.

Also works in bc.

@Quincunx, manatwork, I included the languages you mentioned in the title.

PHP
Quite rule bending maybe...
<? ?>I Love You

HTA - HyperText Application
I Love You

Microsoft says any plain text is valid HTA. They even mention in their official documentation that their "Hello, World." can be as simple as:
Hello, World.

Although most examples that you see on the web go the "official" route.

Shell Script
mkdir "I Love You"
cd I\ Love\ You/
pwd | grep -o '/I Love You' | cut -c2-

Smalltalk
as there is a read-eval-print-loop, and strings are self evaluating, this is trivial. From the command line, type this:
stx -P "'I love you'
A workspace, which is a basically a graphical wrapper for the eval, type in a string and select the "printIt" menu function.

DOS Command Prompt
C:> prompt I Love You
I Love You

I am wondering why my bold text is not as big and bold as others'.

Because they are not using a bold text,they are using headers. Try adding a # before the title. (remove the *)

Or put ======== under the title. Only one is enough, but I usually put the same number of = as the length of the title

is prompt not an external function?

Prompt is a command, but it's purpose is NOT to print strings.


text:
Print n weird numbers
A weird number is a number that the sum of proper divisors is greater than the number itself and no subset of proper divisors sum to that number.
Examples:

70 is a weird number because its proper divisors (1, 2, 5, 7, 10, 14, and 35) sum to 74, which is greater than 70, and no combination of these numbers sum to 70.
18 is not a weird number because its proper divisors (1, 2, 3, 4, 6, 9) sum to 25, which is greater than 18, but 3, 6, and 9 sum to 18.

Your task is to write the shortest program that inputs through std-in any number n and calculates and prints to a file or std-out the first n weird numbers with newline separation.  No hard coding of the answers is allowed (sorry for not specifying this in the beginning).
For more examples, see this page: http://mathworld.wolfram.com/WeirdNumber.html

When this question was in the sandbox, I didn't comment that you should add a "no hard-coding" rule because it's there already in the word "calculate". I encourage people to downvote and flag as not-an-answer or low-quality any answers which don't attempt to calculate the result. (Relevant meta discussion).

 Mathematica 99 94 87
Spaces not needed. Slow!:
j = i = 0;
While[j<#, i++; If[Union@Sign[Tr /@ Subsets@Most@Divisors@i-i]=={-1, 1}, j++; Print@i]]&

At the expense of a few chars this is a faster version that checks only even numbers and skips multiples of 6 that are never weird:
j = i = 0;
While[j < #,
      i += 2; If[Mod[i, 6] != 0 && Union@Sign[Tr /@ Subsets@Most@Divisors@i - i] == {-1, 1},
                 j++; Print@i]] &@3

it's still too slow for any useful purpose. Finds the first two in a few seconds but gets slower and slower as the number of divisors increase.

I was fiddling with a similar solution exploiting the fact that weird numbers are not pseudoperfect, but you got it much golfier than I had yet managed to. Very nice!

Haskell - 129
I'm sure there's lots to golf here, but since the competition seems low for now I'll throw this in.
Don't try running this though, I managed to wait only the two first elements, third will start taking minutes.
(%)=filter
w n=take n$e%[1..]
e x=let d=((==0).mod x)%[1..x-1]in sum d>x&&all((/=x).sum)(i d)
i[]=[[]]
i(y:z)=map(y:)(i z)++(i z)

That's the second time someone in Haskell is better than me in Sage, damn :D

Ruby - 152
x=2;gets.to_i.times{x+=1 while((a=(1..x/2).find_all{|y|x%y==0}).reduce(:+)<=x||(1..a.size).any?{|b|a.combination(b).any?{|c|c.reduce(:+)==x}});p x;x+=1}

Ruby With ActiveSupport - 138
x=2;gets.to_i.times{x+=1 while((a=(1..x/2).find_all{|y|x%y==0}).sum<=x||(1..a.size).any?{|b|a.combination(b).any?{|c|c.sum==x}});p x;x+=1}

Really slow and I'm almost sure there is still room for golfing...

Python 2.7 (255 bytes)
import itertools as t
a=int(raw_input())
n=1
while a>0:
    d=[i for i in range(1,n/2+1) if not n%i]
    if all([n not in map(sum,t.combinations(d,i)) for i in range(len(d))]+[sum(d)>n]):
        print n
        a-=1
    n+=1

PHP, 267 bytes
$n=$x=0;while($n<$argv[1]){$x++;for($i=1,$s=0,$d=array();$i<$x;$i++){if($x%$i){continue;}$s+=$i;$d[]=$i;}if($s<$x){continue;}$t=pow(2,$m=count($d));for($i=0;$i<$t;$i++){for($j=0,$s=0;$j<$m;$j++){if(pow(2,$j)&$i){$s+=$d[$j];}}if($s==$x){continue 2;}}$n++;print"$x\n";}

And here's the original source code:
$n = 0;
$x = 0;

while ($n < $argv[1]) {
    $x++;

    for ($i = 1, $sum = 0, $divisors = array(); $i < $x; $i++) {
        if ($x % $i) {
            continue;
        }

        $sum += $i;
        $divisors[] = $i;
    }

    if ($sum < $x) {
        continue;
    }

    $num = count($divisors);
    $total = pow(2, $num);

    for ($i = 0; $i < $total; $i++) {
        for ($j = 0, $sum = 0; $j < $num; $j++) {
            if (pow(2, $j) & $i) {
                $sum += $divisors[$j];
            }
        }

        if ($sum == $x) {
            continue 2;
        }
    }

    print "$x\n";
}

You will note that it takes some time to output the numbers as it is performing a brute-force verification (you should get to 70 pretty fast, though).

R, 164
r=0;x=1;n=scan();while(r<n){i=which(!x%%(2:x-1));if(sum(i)-1&&!any(unlist(lapply(2:sum(i|T),function(o)colSums(combn(i,o))==x)))&sum(i)>x){r=r+1;cat(x,"\n")};x=x+1}

Un-golfed version:
r = 0
x = 1
n = scan()
while(r < n) {
  i = which(!x %% (2:x - 1))
  if( sum(i) - 1 &&
       !any(unlist(lapply(2:sum(i | T),
                          function(o) colSums(combn(i, o)) == x))) &
       sum(i) > x
     ){ r = r + 1
        cat(x, "\n")
  }
  x = x + 1
}

This takes some time due to brute-force.

Perl, 176 174 bytes
$n=<>;$i=9;X:while($n){@d=grep{!($i%$_)}1..$i-1;$l=0;map{$a=$_;$s=0;$s+=$d[$_]for grep{2**$_&$a}0..@d-1;$i++,next X if$s==$i;$l=1 if$s>$i}0..2**@d-1;$n--,print$i,$/if$l;$i++}

The number of weird numbers are expected in STDIN and the found numbers are printed to STDOUT.
Ungolfed version
#!/usr/bin/env perl
use strict;
$^W=1;

# read number from STDIN
my $n=<>;
# $i is the loop variable that is tested for weirdness
my $i=9; # better start point is 70, the smallest weird number
# $n is the count of numbers to find
X: while ($n) {
    # find divisors and put them in array @divisors
    my @divisors = grep{ !($i % $_) } 1 .. $i-1; # better: 1 .. int sqrt $i
    # $large remembers, if we have found a divisor sum greater than the number
    my $large = 0;
    # looping through all subsets. The subset of divisors is encoded as
    # bit mask for the divisors array.
    map {
        my $subset = $_;
        # calculate the sum for the current subset of divisors
        my $sum = 0;
        map { $sum += $divisors[$_] }
            grep { 2**$_ & $subset }
                0 .. @divisors-1;
        # try next number, if the current number is pseudoperfect
        $i++, next X if $sum == $i; # better: $i+=2 to skip even numbers
        $large = 1 if $sum > $i;
    } 0 .. 2**@divisors - 1;
    # print weird number, if we have found one
    $n--, print "$i\n" if $large;
    $i++; # better: $i+=2 to skip even numbers
}
__END__

Limitations

Slow, brute force.
The count of divisors for a number is limited to the "bitness" of integers in Perl.

C++ - 458
This is not all my solution as I had to ask on SO for help calculating the sum of the subsets, but everything else is mine:
#include<iostream>
#include<vector>
using namespace std;
#define v vector<int>
#define r return
#define c const_iterator
v x(int i){v d;for(int k=1;k<i;k++)if(i%k==0)d.push_back(k);r d;}bool u(v::c i,v::c e,int s){if(s==0)r 0;if(i==e)r 1;r u(i+1,e,s-*i)&u(i+1,e,s);}bool t(v&d,int i){bool b=u(d.begin(),d.end(),i);if(b)cout<<i<<endl;r b;}int main(){v d;int n;cin>>n;for(int i=2,j=0;j<n;i++){d=x(i);int l=0;for(int k=0;k<d.size();k++)l+=d[k];if(l>i)if(t(d,i))j++;}}

Long version:
#include<iostream>
#include<vector>
using namespace std;

vector<int> divisors(int i) {

    vector<int> divs;
    for(int k = 1; k < i; k++)
        if(i%k==0)
            divs.push_back(k);
    return divs;
}

bool u(vector<int>::const_iterator vi, vector<int>::const_iterator end, int s) {

    if(s == 0) return 0;
    if(vi == end) return 1;
    return u(vi + 1, end, s - *vi) & u(vi + 1, end, s);
}

bool t(vector<int>&d, int i) {

    bool b = u(d.begin(), d.end(), i);
    if(b) cout<< i << endl;
    return b;
}

int main() {

    vector<int> divs;
    int n;
    cin>>n;

    for(int i = 2, j = 0; j < n; i++) {
        divs = divisors(i);

        int sum_divs = 0;
        for(int k = 0; k < divs.size(); k++)
            sum_divs += divs[k];

        if(sum_divs > i)
            if(t(divs, i))
                j++;
    }
}

It has currently only calculated the first two (70 and 836).  I killed it after that.

It would be nice to post a readable version as well, especially since you make it as a one-liner ;)

@tohecz Sure, let me edit it.

@tohecz I'm done.

SageMath: 143 131 bytes
x=1
def w():
 l=x.divisors()
 return 2*x>=sum(l)or max(2*x==sum(i)for i in subsets(l))
while n:
 while w():x+=1
 print x;n-=1;x+=1

It's moreorless not even golfed, there's not too much to golf anyways in the code. The biggest thing is that you should do the test 2*x>=sum(l) first, it would save a lot of computation time. One has to realize that max on booleans is or Second thing is that w(x) is False for weird numbers and True for non-weird numbers. Ungolfed version:
def w(x) :
 Divisors = x.divisors()
 return 2*x >= sum(Divisors) or max ( sum(SubS) == 2*x for SubS in subsets(Divisors) )

x=1

for k in xrange(n) :
 while w(x) : x += 1
 print x
 x += 1

Smalltalk, 143
((1to:(Integer readFrom:Stdin))reject:[:n||d|d:=(1to:n//2)select:[:d|(n\\d)=0].d sum<n or:[(PowerSet for:d)contains:[:s|s sum=n]]])map:#printCR

input:
1000

output:
70
836

Perl, 173
Let me add another useless solution. This solution is so slow that it can't even output anything past the first weird number. I dare say it is the slowest of all the solution here.
$n=<>;$i=2;while($n){$b=qr/^(?=(.+)\1{2}$)((.+)(?=.*(?(2)(?=\2$)\3.+$|(?=\1$)\3.+$))(?=.*(?=\1$)\3+$))+/;$_='x'x3x$i;if(/$b/&&($+[0]>$i)&&!/$b\1{2}$/){print"$i\n";$n--}$i++}

Demo
The same code written in Java (which I am more comfortable with) can't even recognize the 2nd weird number (836), and I have already fed the number directly to the checking method (instead of looping and checking every number).
The core of this solution lies in the regex:
^(?=(.+)\1{2}$)((.+)(?=.*(?(2)(?=\2$)\3.+$|(?=\1$)\3.+$))(?=.*(?=\1$)\3+$))+

And how the string is set up to be 3 times the number that we are checking.
The length of the string is set up to be 3 times the number that we are checking i: the first 2 i is for matching summation of factors and the last 1 i is reserved for checking whether a number is a factor of i.
(?=(.+)\1{2}$) is used to capture the number that we are checking.
((.+)(?=.*(?(2)(?=\2$)\3.+$|(?=\1$)\3.+$))(?=.*(?=\1$)\3+$))+ matches the factors of the number. Later iteration will match a smaller factor than an earlier iteration.

We can see that these 2 parts (.+) and (?=.*(?=\1$)\3+$) together selects a factor of the number being checked.
(?=.*(?(2)(?=\2$)\3.+$|(?=\1$)\3.+$)) makes sure that the factor selected is smaller than the number being checked in the first iteration, and is smaller than previous factor in subsequent iterations.

The regex tries to match as many factors of the number as it can within the limit of 2 i. But we don't care about the actual value of sum of divisors, we only care whether the number is abundant.
Then the 2nd regex, which is the first regex with \1{2}$ added. As a result, the regex makes sure the sum of (some) factors of the number being checked is equal to the number itself:
^(?=(.+)\1{2}$)((.+)(?=.*(?(2)(?=\2$)\3.+$|(?=\1$)\3.+$))(?=.*(?=\1$)\3+$))+\1{2}$

The constraint added will cause the regex engine to perform a backtracking search on all possible subsets of factors, so it is going to be extremely slow.

Jelly, 15 bytes
ÆḌŒP§e@=ÆḌS<Ɗµ#

Try it online!
Takes input via STDIN.
Insanely inefficient, times out for \$n > 1\$ on TIO, as it checks each \$k = 1, 2, \dots\$ until it finds \$n\$ values that fit, and for each \$k\$ it generates \$2^{\sigma(k)-k}\$ sublists (where \$\sigma(k)\$ is the sum of divisors function)
How it works
ÆḌŒP§e@=ÆḌS<Ɗµ# - Main link, takes no arguments
             µ# - Take an integer from STDIN, n, and run the previous code over each integer, k, from 0 until n truthy values are found:
ÆḌŒP§e@         -   First check for weirdness:
ÆḌ              -     Take the proper divisors of k
  ŒP            -     Find the powerset of the proper divisors
    §           -     Get the sum of each
                -       This gives the sum of each subset of proper divisors
     e@         -     Is k equal to one of the sums?
                -     This gives 0 if no combination of these numbers sum to k and 1 otherwise
        ÆḌS<Ɗ   -   Second check for weirdness:
            Ɗ   -     Group the previous 3 commands together:
        ÆḌS     -       Is the proper divisor sum of k...
           <    -       Less than k?
                -     This gives 0 if the proper divisor sum of k is greater than k and 1 otherwise
      =         -   Are the two checks equal?
                -     There are three possible values that the two checks could output:
                -        0, 1: k cannot be expressed as a sum of a subset of its proper divisors and k is less than or equal to its proper divisor sum, therefore k is not weird
                -        1, 0: k can be expressed as a sum of a subset of its proper divisors and k is greater than its proper divisor sum, therefore k is not weird
                -        0, 0: k cannot be expressed as a sum of a subset of its proper divisors and k is greater than its proper divisor sum, therefore k is weird
                -     1, 1 cannot be a possibility as that would require that k can be expressed as a sum of a subset of its proper divisors and that k is greater than its proper divisor sum, which are contradictory
                -     Therefore, checking whether the two checks result in the same value suffices to check for weirdness

Python 2, 100 bytes
n=input()
i=0
while n:
 i+=1;s=1;d=1
 while d<i:s|=s<<d*(i%d<1);d+=1
 s>>=i
 if s>1>s&1:print i;n-=1

Attempt This Online!
I use the integer s as a bitset, telling me the sums of divisors of i. The k-th bit tells me whether k is a sum of the divisors found.


text:
Create a flaky test
Create a piece of code and a genuinely-looking test for it. However, the test must be flaky: when run multiple times, it must sometimes fail, sometimes succeed.
Be creative and try to hide the flakiness as much as possible. The most popular answer in (at least) 10 days wins.

I'm voting to close this question as off-topic because [tag:underhanded] questions are off-topic.

JavaScript
Here is a very simple function that divides two given numbers:
function div(x, y) {
    return x / y;
}

And a test function that for 100 test cycles generates two random numbers from 0 to 10 and checks if the function returns the same result of division as provided in the test:
function test() {
    for (var i = 0; i < 100; i++) {
        var x = Math.floor(Math.random() * 10),
            y = Math.floor(Math.random() * 10);

        if (x / y === div(x, y)) {
            console.log('Test OK');
        } else {
            console.log('Test FAILED');
            break;
        }
    }
}

 This example doesn't relate to the problem of math with floating numbers or any type casting tricks, as you may probably decide at first sight. Test will fail when x and y equal to 0. In this case JavaScript will always return NaN, which itself never equals to NaN.

C#, Thread-safe?
OK, let's write a kind of Dictionary that supports increasing and summing of values. Oh, and it's thread-safe. It even says so in its name:
public class ThreadSafeCounter<TKey>
{
    private Dictionary<TKey, int> _dictionary = new Dictionary<TKey, int>();
    private object _lock = new object();

    public bool ContainsKey(TKey key)
    {
        lock (_lock) { return _dictionary.ContainsKey(key); }
    }

    public void Add(TKey key, int value)
    {
        lock (_lock) { _dictionary[key] = value; }
    }

    public void Inc(TKey key)
    {
        lock (_lock) { _dictionary[key]++; }
    }

    public int GetSum()
    {
        lock (_lock) { return _dictionary.Values.Sum(); }
    }
}

So, since it's obviously thread-safe, let's fill it using several threads. What could possibly go wrong?
private void FillCounter(ThreadSafeCounter<int> counter)
{
    for (int i = 0; i < 10; i++)
    {
        if (counter.ContainsKey(i))
        {
            counter.Inc(i);
        }
        else
        {
            counter.Add(i, 1);
        }
    }
}

public void Test()
{
    var counter = new ThreadSafeCounter<int>();

    var thread = new Thread(() => FillCounter(counter));
    thread.Start();
    FillCounter(counter);
    thread.Join();

    int sum = counter.GetSum();
    if (sum != 20)
    {
        Console.WriteLine("What? Test failed!");
    }
}

If you know a bit about multi-threading, the problem in the test should be boringly obvious (Is that a word? "boringly"). But it might fool novices.

 While each operation of ThreadSafeCounter is indeed thread-safe, the way the counter is used is not. Between checking for the existence of the key and adding it, the other thread might do the same. So, both threads end up setting the counter to 1 and none actually increases it. A classical race condition.

public bool IsLeapYear(int year)
{
    return year % 4 == 0 && year % 100 != 0 || year % 400 == 0;
}

[Test]
public void TestIsLeapYear()
{
    Assert.Equals(DateTime.IsLeapYear(DateTime.Now.Year), IsLeapYear(DateTime.Now.Year));
}

 The logic is sound, but the test isn't. Run the test a few times at 2015.12.31. 23:59:59. The error is in getting the current time twice. It will fail about twice every four years (at the start and end of leap years, when there is a transition), so it's very well hidden.

I'm sure there are even better examples in PHP, but here's one:
<?php
// Computes the sum of two numbers
function sum($a, $b) {
    return $a + $b;
}

function test_add_numbers() {
    // Perform 100 tests
    for ($i = 0; $i < 100; $i++) {
        // Get two numbers between 0.1 and 0.9
        $a = rand(1, 9) / 10;
        $b = rand(1, 9) / 10;

        // Compute the sum of the two numbers: 0.2 .. 1.8
        $sum = $a + $b;

        // Make it integer: 2 .. 18
        $sum = $sum * 10;

        if (sum($a * 10, $b * 10) == (int)$sum) {
            print "Test OK\n";
        }
        else {
            print "Test FAILED\n";
            die;
        }
    }
}

test_add_numbers();

Click here for DEMO.

 This exploits the fact that (int)((0.1 + 0.7)*10) evaluates to 7 (rather than 8 as expected). More details here.