Break out of a loop

Question

Now that we have over a hundred ways to get stuck in a loop, the next challenge is to get out.

Specifically, write a piece of code with a (seemingly) infinite loop, but somehow manage to break out.

The "infinite loop" can be a statement like while true, for (;1>0;) or anything else which obviously lasts forever. Recursion does not count (since stack overflows will quickly occur).

To break out, you need to execute an instruction written just after the loop. You can not just exit the program or leave the method.

Valid answer (if only it somehow exited the loop):

while (true)
    print("I'm stuck!")

print("I escaped!")

Invalid answer:

define loop() {
    while (true)
        print("I'm stuck!")
}

loop()
print("I escaped!")

Answer 1

GolfScript

"0:1 Argentina - Germany":party~
{party puts 1}do
"Party is over."

Try it online.

How it works

"0:1 Argentina - Germany" # Push party string.
:party                    # Save string in party variable.
~                         # Discard the string from the stack.
{                         #
  party                   # Push party string.
  puts                    # Print string followed by a newline.
  1                       # Push a truthy conditional on the stack.
}do                       # Pop conditional from the stack and repeat the loop if truthy.
"Party is over."          # Should never happen.

Why is the party over?

~ not only discards he string; it evaluates it. Notably, it executes 0:1. Since 1 is a valid identifier (which happens to have the default value 1), this saves the value 0 in the variable 1. Therefore, the loop's conditional will be falsy, so it is executed only once.

Answer 2

C

Classic integer overflow.

#include<stdlib.h>

int main() {
    unsigned a = 0;
    unsigned b = 1000;
    while (a<b) {
        a++;
        b++;
    }
    puts("Hi");
    return 0;
}

Answer 3

C/C++

The following program outputs

I'm stuck at 41...
I'm stuck at 18467...
I'm stuck at 6334...
I'm stuck at 26500...
I'm stuck at 19169...
I'm stuck at 15724...
I'm stuck at 11478...
I escaped at 29358!

(the output on your system may be different)

#include <stdio.h>
#include <stdlib.h>
#include <setjmp.h>

jmp_buf b;

int my_location()
{
    int result = rand();
    if (result % 7 == 0)
        longjmp(b, result);
    return result;
}

int main()
{
    int location = setjmp(b);

    if (location)
        return printf("I escaped at %d!\n", location);

    for (;;)
    {
        printf("I'm stuck at %d...\n", my_location());
    }
}

The "infinite" loop at the end of the main function is broken

by the longjmp function, which is called when a random number becomes divisible by 7.

This is pretty standard... For people who are old enough to remember that trick (which was once pretty much the only way to implement this behavior in C).

Answer 4

Befunge-98

This program breaks out of its infinite loop by randomly modifying itself

"ITRH"4(v
        >222Sac*%Se3*%S5%S.pp

Answer 5

C

Patricide

#include <stdio.h>
#include <unistd.h>

int main(void) {
    pid_t id = fork();

    if (id != 0) {
        while(1) {
            wait(1);
            printf("Apples\n");
        }
    }
    else {
        wait(5);
        kill(getppid(), 9);
    }

    printf("Bananas\n");
}

Fork a child process and enter an endless loop: child process terminates the parent process with a KILL signal.

Answer 6

Java with SnakeYAML library used

Know your libraries :)

import java.util.ArrayList;
import java.util.List;
import org.yaml.snakeyaml.Yaml;
public class UnLoop
{
    final static List<Object> queue = new ArrayList<>();
    public static void main(String[] args)
    {
        //running serializer nonstatic
        new Thread(()->{
        for(;;){
            if(queue.size()!=0)
            System.out.println(new Yaml().dump(
                    queue.remove(0)));

            }}).start();

        queue.add("DERP");
        queue.add(new String[]{"1","2","3","4"});
        queue.add(new java.awt.Point(1, 1));

    }
}

Snakeyaml hates to serialize certain objects due to the way it works, java.awt.point is one of then, so it stackoverflows out

to satisfy the rule the loop is now a serializing thread and the data inerted from another thread

Answer 7

C

#include <stdio.h>
int i;
int main()
{
    ++i;
    while(i)
    {
        i*=1000;
        printf("%d",i);
    }
}