typeid for polymorphic types

Question

I expected this code to print 'Same 1' and 'Same2', but it prints only 'Same1':

#include <iostream>
#include <typeinfo>
using namespace std;

struct C{virtual ~C(){}};
struct D : C{};
int main(){
   D d;
   C c, &cr1 = d;
   if(typeid(cr1) == typeid(D)) cout << "Same1";
   if(typeid(&cr1) == typeid(D*)) cout << "Same2";
}

Both §5.2.8/2 and §5.3.1/3 seem to suggest to me that 'Same2' should be printed.

What and where is the catch?

@dark_charlie: That's basically just moving the address-of operator two lines up. You'd get the same result, even if you wrote typeid((C*) &d). — MSalters

MSalters MSalters · Accepted Answer · 2010-11-17T09:08:04

Pointers aren't polymorphic types. They don't have virtual members. In fact, they have no members whatsoever. They also cannot derive from other types, nor be used as base classes. Hence, the static and dynamic type of a T* is always T*.

In your "Same2" line, you're comparing the typeid of a pointer, not the pointed-to object. The compiler therefore only looks at the static types C* and D*. They're obviously not the same, and must have distinct type_info objects.

typeid for polymorphic types

1 Answers