[Voted into WP at April, 2006 meeting.]

Proposed resolution (April, 2005):

Change the second sentence of 7.6.1.8 [expr.typeid] paragraph 4 as follows:

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.

There is an inconsistency between the normative text in section 7.6.1.8 [expr.typeid] and the example that follows.

Here is the relevant passage (starting with paragraph 4):

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 type, the result of the typeid expression refers to a std::type_info object representing the referenced type.

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

and the example:

    typeid(D) == typeid(const D&); // yields true

The second paragraph above says the “type-id that is the operand”. This would be const D&. In this case, the const is not at the top-level (i.e., applied to the operand itself).

By a strict reading, the above should yield false.

My proposal is that the strict reading of the normative test is correct. The example is wrong. Different compilers here give different answers.