Question

I came across this code accidentally:

#include<stdio.h>
int main()
{
    int i;
    int array[3];
    for(i=0;i<=3;i++)
        array[i]=0;
    return 0;
}

On running this code my terminal gets hanged – the code is not terminating.

When I replace 3 by 2 code runs successfully and terminates without a problem.
In C there is no bound checking on arrays, so what’s the problem with the above code that is causing it to not terminate?

Platform – Ubuntu 10.04
Compiler – gcc

Answer 1

Just because there’s no bound checking doesn’t mean that there are no consequences to writing out of bounds. Doing so invokes Undefined Behavior, so there’s no telling what may happen.

This time, on this compiler, on this architecture, it happens that when you write to array[3], you actually set i to zero, because i was positioned right after array on the stack.