Type of a struct member at compile time

Question

Is there a way of knowing the type of a struct member at compile time? Something analogous to offsetof(), but for types.

Eg. something like:

typedef struct{
  int  b;
  char c;
}a_t;

typeof(a_t,b) a_get_b(void* data){
  return *(typeof(a_t,b)*)(data + offsetof(a_t,b));
}

Answer 1

If you're willing to use typeof (which is currently a very common nonstandard C extension slated for inclusion in the next version of the standard), you can apply it to a member obtained from a compound literal as in typeof((a_t){0}.b):

typedef struct{ int  b; char c; }a_t;

typeof((a_t){0}.b) a_get_b(void* data){ return (a_t*){data}->b; }

(Given a type a_t, (a_t){0} is a reliable way to get an instance of it. Because of how initialization works in C, the 0 will initialize a deepest first elementary member and elementary types are scalars and therefore 0-initializable.)

As for the obtaining the member from a void pointer pointing to the container, you could do:

*(typeof(&(a_t){0}.b)((char*)data + offsetof(a_t,b))

but that's just an awfully long-winded way to do:

(a_t*){data}->b

(which is 100% equivalent to the former as long as the effective type of data is indeed a_t*).

Answer 2

Another way (other than Jerry Jeremiah's) is:

#define struct_get(STRUCT,ELEM) *(typeof(STRUCT.ELEM)*) (STRUCT+offsetof(typeof(STRUCT),ELEM))