Does offsetof require pointer derefence?

Question

I am wondering whether a simple macro offset_of_ requires a pointer dereference of not. For example, a C++ (means that this code will be compiled using a C++ compiler) struct which is declared with packed attribute

struct A {
  int x;
  int y;
} __attribute__(packed);

Suppose that sizeof(int) == 4 in this case, to calculate the offset of y inside A , I wrote a simple macro:

#define offset_of_y (ptrdiff_t)(&(((A*)0)->y))

Can we be sure that the value of this macro is always 4 ? Is there any pointer dereference from 0 (which may be a UB) when the macro is invoked?

Many thanks for any response.

Answer 1

Accessing something through a null pointer is UB, period.

^{(Unless the context is unevaluated, but in this case it IS evaluated.)}

You probably want offsetof() .

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In practice your code probably could work, buf formally it's undefined.