what happens to array elements after the original array is reallocated?

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

int main()
{

    int *a;
    a = (int *)malloc(100*sizeof(int));
    int i=0;


    for (i=0;i<100;i++)
    {
        a[i] = i+1;
        printf("a[%d] = %d \n " , i,a[i]);
    }

    a = (int*)realloc(a,75*sizeof(int));

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

    free(a);

    return 0;

}

In this program I expected the program to give me a segmentation fault because im trying to access an element of an array which is freed using realloc() . But then the output is pretty much the same except for a few final elements !

So my doubt is whether the memory is actually getting freed ? What exactly is happening ?

Solution

The way realloc works is that it guarantees that a[0]..a[74] will have the same values after the realloc as they did before it.

However, the moment you try to access a[75] after the realloc, you have undefined behaviour. This means that the program is free to behave in any way it pleases, including segfaulting, printing out the original values, printing out some random values, not printing anything at all, launching a nuclear strike, etc. There is no requirement for it to segfault.

So my doubt is whether the memory is actually getting freed?

There is absolutely no reason to think that realloc is not doing its job here.

What exactly is happening?

Most likely, the memory is getting freed by shrinking the original memory block and not wiping out the now unused final 25 array elements. As a result, the undefined behaviour manifests itself my printing out the original values. It is worth noting that even the slightest changes to the code, the compiler, the runtime library, the OS etc could make the undefined behaviour manifest itself differently.