Question

In the code snippet, I expected a segmentation fault as soon as trying to assign a value to count[1]. However the code continues and executes the second for-loop, only indicating a segmentation fault when the program terminates.

#include <stdio.h>

int main()
{
        int count[1];
        int i;

        for(i = 0; i < 100; i++)
        {
                count[i] = i;
        }
        for(i = 0; i < 100; i++)
        {
                printf("%d\n", count[i]);
        }
        return 0;
}

Could someone explain what is happening?

Reasons for edit:
Improved the example code as per comments of users,
int count[0] -> int count[1],
too avoid flame wars.

Answer 1

You’re writing beyond the bounds of the array. That doesn’t mean you’re going to get a segmentation fault. It just means that you have undefined behavior. Your program’s behavior is no longer constrained by the C standard. Anything could happen (including the program seeming to work) — a segfault is just one possible outcome.

In practice, a segmentation fault occurs when you try to access a memory page that is not mapped to your process by the OS. Each page is 4KB on a typical x86 PC, so basically, your process is given access to memory in 4KB chunks, and you only get a segfault if you write outside the current chunk.

With the small indices you’re using, you’re still staying within the current memory page, which is allocated to your process, and so the CPU doesn’t detect that you’re accessing memory out of bounds.