3 ms·
It'll only evaluate the pointer once. It's possible to make this a function though, that might be preferable
by dromtrund 6y ago
It'll only evaluate the pointer once. It's possible to make this a function though, that might be preferable
- btrask 6y agoI 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.
- Thorrez 6y agoGood 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);
- btrask 6y agoTaking 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.