C ++类型限定符和相等性

Question

Are int& and int the same type? if I use is_same<int,int&>::value i get false but typeid(int).name() == typeid(int&).name() are the same?

secondly the same question for int and const int ?

Thirdly int and int* ?

I can understand if int and int* are not as one actually stores the address of another object and works differently but I would have thought int& and int are as one is just an alias for another.

Keen to get some good commentary on this.

Answer 1

From Paragraph 5.2.7/4 of the C++11 Standard:

When typeid is applied to a type-id, the result refers to a std::type_info object representing the type of the type-id. If the type of the type-id is a reference to a possibly cv-qualified type, the result of the typeid expression refers to a std::type_info object representing the cv-unqualified 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.

Thus, typeid(int) and typeid(int&) will give the same result, although the two types are definitely different. Similarly, for the type system int and int const are different types, but the typeid operator ignores the const qualification. From Paragraph 5.2.7/5 of the C++11 Standard:

The top-level cv-qualifiers of the glvalue expression or the type-id that is the operand of typeid are always ignored.

Finally, int and int* are again different types for the type system, and the typeid operator returns different results for them.

Answer 2

Type qualifiers, ( const and volatile ), create different types. int is a different type from const int .

So do references, pointers, and arrays. For example:

int , int& , int[10] and int* are all different types.

T is a different type from std::remove_reference<T>::type if T is a reference.

The <typeinfo> output of typeid(int).name() is platform-dependent and doesn't have to distinguish between reference/non-reference types. However, the C++ type system definitely distinguishes between T and T& , as you've discovered through type_traits .

Answer 3

std::type_info::name says nothing about identity. If you insist on using typeid to test for identity, try the following:

assert(typeid(T) != typeid(U));

This is using the defined equality comparison operator on the type_info objects. But prepare for disappointment: the above assertion will fail for T = int and U = int& because of §5.2.7/4 (see Andy's anser).