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));
}
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*).
Another way it works:
#include <stdio.h>
#define typeof_element(_struct,el) typeof(((_struct *)(0))->el)
typedef struct{
int a;
int b;
}Row;
int main()
{
typeof_element(Row, a) value_a = 10;
printf("%d\n", value_a);
return 0;
}
Another way (other than Jerry Jeremiah's) is:
#define struct_get(STRUCT,ELEM) *(typeof(STRUCT.ELEM)*) (STRUCT+offsetof(typeof(STRUCT),ELEM))