"typename" allowed/required 在 C++11 中使用声明吗？

Question

以下代码可在 g++ 和 clang 中正确编译：

template<typename T>
struct foo
{
    class iterator;
    using bar = foo::iterator;
};

int main() {}

但是 MSVC 2013 给出了以下错误：

foo.cpp(9): error C2061: syntax error : identifier 'iterator'
          foo.cpp(10) : see reference to class template instantiation 'foo<T>' being compiled
foo.cpp(9): error C2238: unexpected token(s) preceding ';'

如果我将该行更改为：

using bar = typename foo::iterator;

然后三个编译器都编译成功了。原始版本是否正确？（即这是一个 MSVC 错误，还是 gcc/clang 扩展）

Answer 1

[temp.res]/p3:

When a qualified-id is intended to refer to a type that is not a member of the current instantiation (14.6.2.1) and its nested-name-specifier refers to a dependent type, it shall be prefixed by the keyword typename, forming a typename-specifier.

[temp.dep.type]/p1:

A name refers to the current instantiation if it is

in the definition of a class template, a nested class of a class template, a member of a class template, or a member of a nested class of a class template, the injected-class-name (Clause 9) of the class template or nested class,

[...]

[temp.dep.type]/p4:

A name is a member of the current instantiation if it is

An unqualified name that, when looked up, refers to at least one member of a class that is the current instantiation or a non-dependent base class thereof. [ Note: This can only occur when looking up a name in a scope enclosed by the definition of a class template. —end note ]

A qualified-id in which the nested-name-specifier refers to the current instantiation and that, when looked up, refers to at least one member of a class that is the current instantiation or a non-dependent base class thereof. [ Note: if no such member is found, and the current instantiation has any dependent base classes, then the qualified-id is a member of an unknown specialization; see below. —end note ]

[...]

foo 是当前实例化。 foo::iterator 是一个 qualified-id，其中 nested-name-specifier (foo::) 指的是当前实例化，当查找时，"refers to at least one member of a class that is the current instantiation or a non-dependent base class thereof"；因此它是当前实例化的成员。因此，[temp.res]/p3 不适用，不需要 typename。您仍然可以添加一个 - 或者直接使用 iterator 不合格。

Answer 2

来自标准：

14.6.2.1 Dependent types [temp.dep.type]

1 A name refers to the current instantiation if it is

__ in the definition of a class template, a nested class of a class template, a member of a class template, or a member of a nested class of a class template, the injected-class-name (Clause 9) of the class template or nested class,

名称foo指的是当前实例化，这是显而易见的。

由于iterator在模板定义中声明为嵌套的class，因此iterator指的是foo当前实例化中的名称。 foo::iterator 等同于 iterator.

using bar = foo::iterator;

以及

using bar = iterator;

应该可以。

在我看来，您运行存在 MSVC 缺陷。

"typename" allowed/required 在 C++11 中使用声明吗？

Is "typename" allowed/required in C++11 using-declaration?

c++

typename

using-declaration

c++11