Free an assigned pointer

Question

Does the following code free the memory that was allocated for x?

int main()
{
    char *x = (char*)calloc(100, sizeof(char));
    char *y = x;
    free(y);
}

Answer 1

Yes

When you do

char *y = x;

you make y point to the location where x points to. Since y points to a memory location returned by calloc,

free(y);

is perfectly valid. As @haccks commented, this would not work if you make y point to another memory location, provided that this memory location wasn't returned by malloc/calloc/realloc.

---

In C, you should not cast the result of malloc/calloc/realloc. Also, checking the return value of calloc to see if it was successful is good. calloc will return NULL on failure.

Answer 2

The original version of the question had

int *y = x;
free(y);

I.e. assigning the char pointer to an int pointer and then invoke free() on the int pointer. The signature of free() is void free(void *ptr); so irrespective of char * vs. int *, memory would be released.

The edited (and current) version of the question has

char *y = x;
free(y);

Here, both y and x points to the same memory. This is known as pointer aliasing. Upon the call to free(), that memory would certainly be released.

A problem, however, is that after free(y), the pointer x would be dangling, i.e. the memory it is pointing to is no longer valid.

Note, it is neither necessary nor recommended to cast the return value from calloc.