CODEWITHSUNDEEP
c++templatesc++11language-lawyer
T.C.
T.C.

Reputation: 137310

'typename' and alias templates

The following code compiles using both Clang and GCC, even though Foo_t<T>::Bar doesn't have typename in front of it:

struct Foo {
    using Bar = int;
};

template<class...>
using Foo_t = Foo;

template<class T>
void f(){
    Foo_t<T>::Bar b; // No typename!
}

int main(){
    f<int>();
}

Should it compile?

Upvotes: 24

Views: 2471

Answers (2)

T.C.
T.C.

Reputation: 137310

With some more digging, this is CWG issue 1390.

The issue description is

According to 14.6.2.1 [temp.dep.type] paragraph 8, a type is dependent (among other things) if it is

a simple-template-id in which either the template name is a template parameter or any of the template arguments is a dependent type or an expression that is type-dependent or value-dependent

This applies to alias template specializations, even if the resulting type does not depend on the template argument:

struct B { typedef int type; };
template<typename> using foo = B;
template<typename T> void f() {
  foo<T>::type * x;  //error: typename required
}

Is a change to the rules for cases like this warranted?

And there's a note in that issue:

Notes from the October, 2012 meeting:

CWG agreed that no typename should be required in this case. In some ways, an alias template specialization is like the current instantiation and can be known at template definition time.

The issue is still in "drafting" status, but it looks like the compiler vendors are already implementing the intended resolution.

Upvotes: 9

Filip Ros&#233;en
Filip Ros&#233;en

Reputation: 63797

Introduction

Foo_t<T>::Bar might look like a dependent-name, but it isn't since the template-arguments passed to the alias-declaration are not used when determining what the qualified-id Bar is referring to.

The code is well-formed.

What does the Standard (N3337) say?

14.5.7/2 Alias templates [temp.alias]

When a template-id refers to the specialization of an alias template, it is equivalent to the associated type obtained by substitution of its template-arguments for the template-parameters in the type-id of the alias template.

A.6 Declarations [gram.dcl]

alias-declaration:
  using identifier attribute-specifier-seq_opt = type-id ;

What is the Standard really saying?

Since there are no template-parameters in the type-id of Foo_t, the template alias-declaration is always directly equivalent to Foo, no matter what template-arguments we pass to it.

template<class... Ts>
using Foo_t = Foo;
//            ^--- "Foo" = type-id

Replacing the usage of Foo_t<T> with the equivalence of the template alias-declaration leaves us with the following:

template<class T>
void f(){
    Foo::Bar b; // ok, nothing here depends on `T`
}

Upvotes: 16

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