Why isn't it ending? [closed]

Question

Your task: To write a program that should obviously terminate, but it never (to the extent of a computer crash) does. Make it look like it should perform a simple task: adding numbers, printing something,... But it just gets caught in an infinite loop.

Try to make your program very clear and simple, while it actually will get stuck in an unforeseen loop. Voters: judge the answers on how "underhanded" they are!

This is a popularity contest: Be creative!

Answer 1

Javascript

This once bit me years ago:

function outerloop(){
  for(i=0;i<10;i++){
    innerloop();
  }
}
function innerloop(){
  for(i=0;i<5;i++){
    console.log(i)
  }
}
outerloop();

The trick here is that when one uses for( i=0 ..., the variable i is created in the global scope (even if they're in separate functions), so it gets overwritten every time by the inner loop. This is an easy pitfall for those coming from C++ or PHP. The fix is to use var to declare the variable in local scope (for (var i...)

Answer 2

PowerShell

$i = 1

do {
    $i++
} until( $i > 10 )

The > is not the greater-than operator (which is -gt); it is the redirection operator. Since nothing is returned, PowerShell evaluates the expression as false and continues.

Answer 3

It is a GUI program that will show an empty frame titled Hello, world! and hide the frame in 1 second.

import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;

import javax.swing.JFrame;
import javax.swing.SwingUtilities;
import javax.swing.Timer;

public class Main {
    public static void main(String[] args) {
        SwingUtilities.invokeLater(new Runnable() {
            public void run() {
                // Create a frame titled "Hello, world!".
                final JFrame frame = new JFrame("Hello, world!");
                frame.setVisible(true);

                // After 1 second...
                Timer timer = new Timer(1000, new ActionListener() {
                    @Override
                    public void actionPerformed(ActionEvent e) {
                        // Hide the frame.
                        frame.setVisible(false);
                    }
                });
                timer.setRepeats(false);
                timer.start();
            }
        });
    }
}

The frame is hidden, but it is not disposed, so the virtual machine would not terminate. Reference: http://docs.oracle.com/javase/7/docs/api/java/awt/doc-files/AWTThreadIssues.html#Autoshutdown

Answer 4

C

int main(void) {
  float x = 0;
  while(x != 1) x+=0.1;
}

lots of nooses for a C developer....

Answer 5

C

This was tested on a 64 bit machine with gcc using the compiler flag -fomit-frame-pointer.

int main() {
    long x = (long)main;
    asm("sub %0,%%rsp"::"i"(sizeof(x)));
}

This makes the return address the address of main. Which makes an infinite loop. The assembly line is needed to not run out of stack.

Answer 6

Java

Spawns a thread that will sleep for 1 second. The program should effectively sleep for a second while waiting for that thread to quit, and then end.

import java.lang.*;

class A {
    static boolean done;
    public static void main(String args[]) {
        done = false;
        new Thread() {
            public void run() {
            try {
                Thread.sleep(1000); // dummy work load
            } catch (Exception e) {
                done = true;
            }
            done = true;
            }
        }.start();
        while (!done);
        System.out.println("bye");
    }
}

Why it doesn't end

Answer 7

JavaScript

while(true)
    if(NaN == NaN)
        break;
    else
        console.log("Batman");

A funny property of the global NaN (Not a Number) object in JavaScript is that it is not equal to itsself. To test if a number is NaN one must use isNaN(). It's also amusing that typeof(NaN) returns "number".

Answer 8

C

This prints numbers from 0 to 9. Or at least it should...

#include <stdio.h>

int main() {
    int i;
    char line[12]; // "Iteration: x"

    for(i = 0; i < 10; i++) {
        sprintf(line, "Iteration: %d", i);
        puts(line);
    }

    return 0;
}

This is a classic off-by-one error triggering a stack overflow. The string is indeed 12 characters long, but the programmer forgot about the NULL terminator inserted by sprintf! On GCC, local variables are pushed to the stack in declaration order. Other compilers may order the variables differently. Since the stack grows backwards, this means i is after line in memory. The null byte overwrites the least significant byte of i. At the end of the block, i is always 0. i is incremented and it loops forever with i = 1.

Answer 9

C

int main(void)
{
        float count = 0;
        while(count < 33554432L)
        {
                count++;
        }
        return 0;
}

Floating-point numbers have limited precision. Under IEEE 754 floating-point math, a single-precision float only has 24 bits of precision: the largest integer that can be represented exactly is 16,777,216. 16777216 + 1 = 16777216.

Answer 10

C#

Perhaps too obvious, but

class Program
{
    static void Main()
    {
        int x = 0;
        while (x * x != 1000)
            x++;
        System.Console.WriteLine("Square root of 1000 is " + x);
    }
}

The square root of 1000 is not an integer, so it will never find it. It's just going to loop forever, assuming overflows aren't checked (the default is unchecked). So like I said, perhaps too obvious. But when I first saw it, I was pretty sure it was going to terminate - 1000 looks as though it should have an integer square root - to me, anyway.

Answer 11

Postscript

hello.ps:

/hello
{
    (Hello, world\n) print
    quit
}
dup 2 3 index cvx put def

hello

This prints "Hello, world" forever. To try it with ghostscript:

gs -sDEVICE=nullpage hello.ps

dup 2 3 index cvx put replaces the procedure's call to quit with a recursive call to procedure hello. Since postscript does tail call optimization, this runs forever without overflowing the stack.

Answer 12

TeX (Plain): Typesetting The Song That Never Ends

Compile with pdftex. It is not obfuscated (besides, well, being written in TeX) and actually does end, as pdftex stops after somewhat about 2.6 million pages...

\font\bigrm=cmr12 at 14pt
\centerline{\bigrm The Song That Never Ends}
\bigskip
\def\songthatneverend{
\noindent This is the song that never ends.\hfil\break
Yes, it goes on and on my friends.\hfil\break
Some people started singing it not knowing what it was,\hfil\break
and they'll continue singing it forever just because\smallskip

\expandafter\songthatneverend}
\songthatneverend
\bye

Answer 13

Ruby

if p
  //misses
  raise 'Unreachable'
end

while p //nonsense, Kernel#p always returns nil, which is falsey, so this will immediately terminate.
  puts 'Looping'
end

puts 'Done looping'

Answer 14

Batch

This loops forever outputting Hello.

@echo off
for /l %%a in (0,0,1) do echo Hello.

The for loop tries to count from 0 to 0 incrementing by 1 - (0,0,1). So it never reaches zero.

Or if you want to cause the prompt to crash.

@echo off
for /l %%a in (0,1,1) do call %%%%a
:1
echo END

--

h:\uprof>end.bat
******  B A T C H   R E C U R S I O N  exceeds STACK limits ******
Recursion Count=477, Stack Usage=90 percent
******       B A T C H   PROCESSING IS   A B O R T E D      ******

Answer 15

Javascript

Here's a typical beginner's programming exercise.

Continually repeat a sequence of 1..n1, until you've output a total of n2 numbers. For example, looping through 1..3 where the total quantity is 10 would produce this:

1 2 3 1 2 3 1 2 3 1

Below is a solution that uses a currying function. It's simple, elegant, and will hang your browser :-)

function output(msg) {
    $('<span></span>').text(msg).appendTo($('#output'));
}

function loopThroughSequence(max) {
    i = 0;
    return function() {
        if(i >= max) {
            i = 0;
        }
        return ++i;
    }
}

var getNextInSequence = loopThroughSequence(3);
for(var i = 0; i < 10; ++i) {
    output(' ' + getNextInSequence());
}

Note that loopThroughSequence forgets to declare i in scope. (It needed to preceed 'i' with the 'var' keyword.) Instead, it uses the i value declared below it in the for loop, and continually resets it back to zero.

Answer 16

C++

#include<fstream.h>
int main()
{ 
 char ch[80];
 ifstream fin("evil.txt");
 while(fin)
 {
  fin.get(ch,50); //read by lines
  cout<<ch;
 } //nothing wrong, right? wrong.
 return 0;
}

evil.txt contains following data :

hello.
I am awesome.

fin.get() leaves a '\n' in input stream. The rest is Elementary, my dear.

Answer 17

How about this...

#include<stdio.h>

void main(){
while((10*5/3)!=(10/3*5))
printf("In a loop\n");
}

Mathematically (10/3)*5==(10*5)/3;

Answer 18

Haskell

This is just a simple program to illustrate return statements. Since the return statement ends the function, it doesn't run anything after it.

sayHello :: IO String
sayHello = do
    return "Hello!"
    sayHello

main = do
    hello <- sayHello
    putStrLn hello

Return statements don't end functions, unlike in other languages.

Answer 19

Whitespace

(actual working code follows)

Readable version:

FSSSSSTTFFSFSSSSSTTFF

Where F is linefeed, S is space and T is tab

Answer 20

Java

public class Main {
    public static void main(String[] args) {
        try{
        double i=1.0/0.0;
            while(i==i+1){  //weired condition should never be true
                System.out.println("This should never be printed");
            }
        }catch (ArithmeticException e){
            System.out.println("Caught divide by 0 exception");

        }
    }
}

Answer 21

C

int main(void) {
  int x;
  while(x<=10) x+=1;
}

from my experience this thing loops indefinitely only after 4 years of stable production at 3AM on a Saturday and once it loops for first time no system reboot can make it go away and within 15 minutes it happens to all high availability servers, dumping the entire production stack offline for a day.

Answer 22

Ruby

This is a super efficient piece of code that simulates a dice.

def almost_dice
  rand(7) # will output a number between 0 and 6. 
end

#I want a dice that outputs a number between 1 and 6, so I have no other choice, right?
def proper_dice
  d = almost_dice
  while d = 0
    d = almost_dice #Eventually, I'll get d != 0
  end
  return d
end

puts proper_dice

In ruby what could look like a test : d = 0 is actually an assignment. Since the assignement always succeed, then the test always returns true, hence the infinite loop

Answer 23

C#

Let's take a page from this answer: https://codegolf.stackexchange.com/a/24376/16729 but do the computation on a background thread in the class constructor!

class BackgroundSummer
{
  static int sum;
  static BackgroundSummer() 
  {
    // Let's compute the sum on another thread!
    var thread = new System.Threading.Thread(ComputeSum);
    thread.Start();
    thread.Join();
  }

  static void ComputeSum() 
  { /* omitted: compute the sum here */ }

  static void Main() 
  { }
}

An explanation of why this runs forever is here: http://ericlippert.com/2013/01/31/the-no-lock-deadlock/

Answer 24

Scala

var break = false
for(i <- Stream.from(0) if !break) {
  break = i >= 10
}
println("Well done!")

Infinite streams will continue to be evaluated even if break is true.

Or the second one.

var g = 6
def counter: Int = {
  g -= 2
  g
}
def test() = {
  val t = ListBuffer('s','t','a','r','t',' ','e','m','p','t','y')
  var i = t mkString // returns "start empty"
  var l = ListBuffer[Int]()
  while(i != "42") i = (l += counter) mkString // returns "42 after two iterations"
  s"$i should equal 42"
}
println(test())

mkString captures the second argument as a separator, here s"$i should equal 42". So the value of $i is never "42" but goes indefinitely.

Answer 25

wxpython

"Hello World" example that fails to exit.

import sys
import time
import wx
from wx.lib.pubsub import setupkwargs
from wx.lib.pubsub import pub

class LoggerWindow(wx.Frame):
    def __init__(self):
        wx.Frame.__init__(self, None, title="Logger")
        self._log = wx.TextCtrl(self, style=wx.TE_MULTILINE)
        pub.subscribe(self.OnLogMessage, "LOG_MESSAGE")

    def OnLogMessage(self, msg):
        self._log.AppendText(msg)


class GUI(wx.Frame):
    def __init__(self):         
        wx.Frame.__init__(self, None, title="Demo GUI - Hello World")  
        self._logger = LoggerWindow()
        pub.sendMessage("LOG_MESSAGE", msg="GUI Starting")

        # Set up the Hello World message
        self._display = wx.TextCtrl(self, -1, size=(400,20), style=wx.TE_MULTILINE)
        self._display.AppendText("Hello World!")

        # Show it for a while
        self.Show()
        time.sleep(10)

        # Done
        self.Destroy()


App = wx.App()
Gui = GUI()
App.MainLoop()

The log window, while not showing, keeps existing, so the "GUI" window disappears but the app keeps running. I had intended for this to be another example of "not using a loop", but obviously I am using a loop!

Answer 26

Python

def sum(first, add=[]):
    add.append(first)
    sum = 0
    for el in add:
        sum+= el
    print sum
    return add

if __name__ == '__main__':
    #init
    add = sum(1)
    for el in add:
        sum(el)

this will loop forever due to the fact that expressions in default arguments are calculated when the function is defined, not when it’s called. So the list will grow until the user terminate the script.

Answer 27

JAVASCRIPT

found this while doing my project

function weirdCode(){
  var myarray= ["test",123,false];
  var myarray2 = myarray;
  myarray2[2] = !myarray2[2];
  while(myarray.toString() == myarray2.toString()){
    console.log("shouldn't myarray[2] == false? but it is "+myarray[2]+". but i only edit myarray2, which is "+myarray2[2]);
  }
  console.log("this will never called");
}
weirdCode();

at line "var myarray2 = myarray;" it should be "var myarray2 = myarray.slice()", cause it will be referring the var to myarray, like sharing a var

Answer 28

JAVASCRIPT

 var x = 1; while(x != 0) x -= 0.1;

x = 1;
x - 0.1 = 0.9;
x - 0.1 = 0.8;
x - 0.1 = 0.7000000000000001
WHAT!!!!!

Now more about it

Answer 29

Java, equal but not?

import java.util.HashMap;
import java.util.Map;

public class Endless {
    static class A {
        public A(int a) {
            b = a;
        }

        int b;
        @Override
        public boolean equals(Object obj) {
            return obj != null && (obj instanceof A) && ((A)obj).b == b;
        }
    }

    public static void main(String args[]) {
        Map<A,String> map = new HashMap<A, String>();
        A a = new A(1);
        A sameA = new A(1);

        System.out.println("a is sameA?" + a.equals( sameA ));

        map.put(a, "look ma, no hands!");

        while (!map.containsKey(sameA)){
            //we never get here, do we?
        }
        System.out.println("Done!");
    }
}

Answer 30

How about a fork bomb?

:(){ :|: &};: