typeid doesn't return correct type

Question

cout << typeid(int&).name();  

This, in my opinion, should return int& as a type, not an int, but on GCC 4.5.1 and on VS2010 SP1 beta it returns int. Why is this?

Answer 1

This is how typeid is supposed to work. When you apply typeid to a type-id of a reference type, the type_info object refers to the referenced type.

ISO/IEC 14882:2003, 5.2.8 / 4 [expr.typeid]:

When typeid is applied to a type-id, the result refers to a type_info object representing the type of the type-id. If the type of the type-id is a reference type, the result of the typeid expression refers to a type_info object representing the referenced type. If the type of the type-id is a class type or a reference to a class type, the class shall be completely-defined. Types shall not be defined in the type-id.

Answer 2

Your first mistake is expecting anything useful from std::type_info::name(). From the standard:

• §18.5.1/1: "The names, encoding rule, and collating sequence for types are all unspecified and may differ between programs."
• §18.5.1/7: "const char* name() const; Returns: an implementation-defined NTBS."

If you want a portable solution for meaningful (through not necessarily consistent) type names, I recommend using Boost.TypeIndex's boost::typeindex::type_id_with_cvr<>().pretty_name() (reference).

Answer 3

The C++ spec does not guarantee that type_info::name actually hands back the name of the type as it appears in the C++ source code; in fact, the spec, §18.5.1/7, only guarantees that the function hand back "an implementation-defined NTBS."

Consequently, there's no reason to assume that using typeid to get the name of a type will actually hand back the name of the type as you'd expect it to.

The reason you're seeing the type of int and not int& is that the definition of typeid says that it ignores references. To quote the spec, §5.2.8/4:

When typeid is applied to a type-id, the result refers to a type_info object representing the type of the type-id. If the type of the type-id is a reference type, the result of the typeid expression refers to a type_info object representing the referenced type.

(My emphasis)

This means that typeid(int&) and typeid(int) are completely identical to one another, hence the output being int and not int& or something related to it.