typeid doesn't return correct type

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

This, in my opinion, sh开发者_开发技巧ould 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?

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.

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).

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.