I need a type in C++ that is literally a member, as opposed to a nested type

Question

I need something that will work with the following:

struct thing *s;
printf("the size of %s is %d", s->type, (int) sizeof(s->type));

I thought I had at last found a solution with typedef, but no:

struct thing {
  typedef int type;
}

expects syntax of the form thing::type. I suspect the solution still involves typedef, but I need the type to actually be a member, as opposed to a nested type.

Edit: Whoops, looks like the functionality I needed was just a string after all. Sorry, thanks for helping!

Types (and typedefs) themselves are cannot be values. The closest you get is `std::type_info` returned from `typeid`. — goji, Oct 08 '13 at 21:10
The code above looks more like you need a *string* for the type... Other than that, it is unclear why you need a *member* and not a *nested* type — David Rodríguez - dribeas, Oct 08 '13 at 21:12
The idea wouldn't even work as `printf("%s %d", int, sizeof(int))` so why do you expect it to work with a "member type"? — MSalters, Oct 08 '13 at 22:23
I need it because of the specific and arbitrary whims of a professor. I guess I'll complain to him to see what he had in mind. — Elliot Way, Oct 08 '13 at 22:59

score 1 · Accepted Answer · answered Oct 08 '13 at 21:14

I think this is likely the closest you can do at the moment, although the type name will likely be mangled (not found an implementation that isn't yet):

struct thing {
  int i;
};

thing t;
std::cout << "the size of " << typeid(t.i).name() 
          << " is " << sizeof(t.i) << "\n";

I need a type in C++ that is literally a member, as opposed to a nested type

1 Answers1