Yeah, my bad. I missed the point of the example (ie that the size argument might be too small). I assumed the point was memset writing to padding bytes. It is

Yeah, my bad. I missed the point of the example (ie that the size argument might be too small). I assumed the point was memset writing to padding bytes. It is indeed true that y might not get zeroed in principle (though I doubt that'll happen in practice for this particular example as padding gets introduced to maintain alignment, and alignment factors a type's size, which is 1 in case of char). A better example would be struct boxed_value { uint16_t flags; double value; }; struct boxed_value v; memset(&v, 0, sizeof (uint16_t) + sizeof (double)); That's of course a braindead way to write it instead of the more convenient memset(&v, 0, sizeof v); That said, neither case will invoke undefined behaviour.