Question

void getFree(void *ptr)
{
    if(ptr != NULL)
    {
        free(ptr);
        ptr = NULL;
    }
    return;
}
int main()
{
char *a;
a=malloc(10);
getFree(a);
if(a==NULL)
    printf("it is null");
else
    printf("not null");
}

Why is the output of this program not NULL?

Answer 1

Because the pointer is copied by value to your function. You are assigning NULL to the local copy of the variable (ptr). This does not assign it to the original copy.

The memory will still be freed, so you can no longer safely access it, but your original pointer will not be NULL.

This the same as if you were passing an int to a function instead. You wouldn’t expect the original int to be edited by that function, unless you were passing a pointer to it.

void setInt(int someValue) {
    someValue = 5;
}

int main() {
    int someOtherValue = 7;
    setInt(someOtherValue);
    printf("%i\n", someOtherValue); // You'd expect this to print 7, not 5...
    return 0;
}

If you want to null the original pointer, you’ll have to pass a pointer-to-pointer:

void getFree(void** ptr) {
    /* Note we are dereferencing the outer pointer,
    so we're directly editing the original pointer */

    if (*ptr != NULL) {
        /* The C standard guarantees that free() safely handles NULL,
           but I'm leaving the NULL check to make the example more clear.
           Remove the "if" check above, in your own code */
        free(*ptr);
        *ptr = NULL;
    }

    return;
}

int main() {
    char *a;
    a = malloc(10);

    getFree(&a); /* Pass a pointer-to-pointer */

    if (a == NULL) {
        printf("it is null");
    } else {
        printf("not null");
    }

    return 0;
}