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by dromtrund 2039 days ago
It'll only evaluate the pointer once. It's possible to make this a function though, that might be preferable
2 comments

Good point. But it seems like it would require usage like this:

    int* p = malloc(sizeof(int));
    FREE(&p);
What if we instead define the macro like this:

    #define FREE(ptr) do { \
        __typeof__(ptr)* const __x = &(ptr); \
        free(*__x); *__x = NULL; \
    } while(0)
Then make usage slightly shorter, as well as more similar to free():

    int* p = malloc(sizeof(int));
    FREE(p);
Taking a pointer-to-pointer is intentional to make it clear that the pointer will be modified. That's actually the most important difference from nn3's version IMHO.
I tried making it a plain function at one point but ran into some weirdness around using void * * with certain arguments (const buffers?). You don't want to accept plain void * because it's too easy to pass a pointer instead of a pointer to a pointer. Using a macro is (ironically) more type safe.

Maybe someone else could figure out how to do it properly, since I'd definitely prefer a function.