P0634R1
2017-10-15

Nina Ranns (dinka.ranns@gmail.com)
Daveed Vandevoorde (daveed@edg.com)

Core Working Group (CWG)

Down with typename!

Summary

If X::Y — where T is a template parameter — is to denote a type, it must be preceded by the keyword typename; otherwise, it is assumed to denote a name producing an expression. There are currently two notable exceptions to this rule: base-specifiers and mem-initializer-ids. For example:

template<class T>
struct D: T::B { //
No typename required here.
};
Clearly, no typename is needed for this base-specifier because nothing but a type is possible in that context. However, there are several other places where we know only a type is possible and asking programmers to nonetheless specify the typename keyword feels like a waste of source code space (and is detrimental to readability).

We therefore propose we make typename optional in a number of commonplace contexts that are known to only permit type names.

A cursory read through some common standard library headers suggests that by-far most occurrences of typename for the purpose of disambiguating type names from other names can be eliminated with the new rules.

The EDG front end has an `implicit typename` mode to emulate pre-C++98 compilers that didn't parse templates in their generic form. Although that mode doesn't exactly cover the contexts where we are proposing to make typename optional, the implementation effort is similar (and not excessively expensive).

This proposal was approved in Toronto (July 2017) by EWG:

SF: 14 | F: 8 | N: 1 | A: 0 | SA: 0

Wording changes

Hide deleted text

Change in [temp.res] paragraph 3:

When a qualified-id is intended to refer to a type that is not a member of the current instantiation (17.7.2.1) and its nested-name-specifier refers to a dependent type, it shall be prefixed by the keyword typename, forming a typename-specifier. If the qualified-id in a typename-specifier does not denote a type or a class template, the program is ill-formed.unless: A qualified name is said to be in an type-id-only context if it appears in a type-id, new-type-id, or defining-type-id and the smallest enclosing type-id, new-type-id, or defining-type-id is a
[ Example:
template<class T> T::R f(T::P); // OK, global scope
template<class T> struct S {
using Ptr = PtrTraits::Ptr; //
OK, in a defining-type-id
T::R f(T::P p) { //
OK
return static_cast<T::R> //
OK, type-id of a static_cast
}
auto g() -> S<T*>::Ptr;  //
OK, trailing-return-type
};
template<typename T> void f() {
void (*pf)(T::X); //
Variable pf of type void* initialized with T::X
void g(T::X); //
Error: T::X (not at class scope) does not denote a type
};

— end example ]

A qualified name prefixed with typename in this way forms a typename-specifier:
typename-specifier:
typename nested-name-specifier identifier
typename nested-name-specifier templateopt simple-template-id
If the qualified-id in a typename-specifier does not denote a type or a class template, the program is ill-formed.