C program: Why am I not getting Segmentation Fault? Is it UB?

[abhishekm71]

I am reading the book: Computer Systems, A Programmer's Perspective.

I wrote the following code just for my own understanding:

#include <stdio.h>

int main()
{
    int A[10];
    int i;
    for (i = 0; i < 14; i++)
        A[i] = i;
    printf("%d\n",A[11]);
    return 0;
}

This code does not give segmentation fault. However if I extend the termination condition of the for loop to "< 15", then it results in seg fault.

Is it UB that I am experiencing or is there a valid reason for such behaviour?

Does it have anything to do with an int taking 4 bytes of memory space vis-a-vis a pointer taking 8 bytes on a 64 bit machine?

Thanks!

[Computergeek01]

Your 'A' array is 10 units large and you are writing to the 15th element. By all rights this should seg fault to, it's just a matter of chance that it does not. Moving this to the beginners section will provide more help if you need it.

[cire]

Is it UB that I am experiencing or is there a valid reason for such behaviour?

It is undefined behavior that you are experiencing and there is a valid reason for such behavior. The valid reason is that it is undefined behavior and undefined behavior is.. well, undefined. It could do anything and have conforming behavior.

[abhishekm71]

ok, thanks for the responses!