Question

In the following code, will freeing Dairy free Yogurt as well?
Both point to the same address as far as I know.

Also, is this style of coding a bad practice? Say if I only kept pointers to the Dairy and indirectly freed the Yogurt and Cheese as well?

#include <stdlib.h>

typedef struct {
    int calcium;
    int protein;
} Dairy;

typedef struct {
    Dairy dairy;
    int sugar;
    int color;
} Yogurt;

int main () {
    Yogurt* yogurt = malloc(sizeof(Yogurt));

    Dairy* dairy = &yogurt->dairy;

    free(dairy); // Will this free yogurt?
}

Answer 1

Yes the behavior is well-defined.

As per the C Standard You need to pass the exact same address to deallocation function free as was returned by an malloc or other allocation functions.

Reference:

7.22.3.3 The free function

Synopsis:

§1 #include <stdlib.h> void free(void *ptr);

Description:

§2 The free function causes the space pointed to by ptr to be deallocated, that is, made
available for further allocation. If ptr is a null pointer, no action occurs. Otherwise, if the argument does not match a pointer earlier returned by a memory management
function, or if the space has been deallocated by a call to free or realloc, the
behavior is undefined.

So, In this case the behavior is guaranteed to be well defined if and only if pointer to the structure is same as pointer to it’s first member.
And that is guaranteed by the standard in:

6.7.2.1 Structure and union specifiers
§15

Within a structure object, the non-bit-field members and the units in which bit-fields
reside have addresses that increase in the order in which they are declared. A pointer to a structure object, suitably converted, points to its initial member (or if that member is a bit-field, then to the unit in which it resides), and vice versa. There may be unnamed
padding within a structure object, but not at its beginning.