c++ - 为什么嵌套类型的基类不需要 `typename`？

Question

我非常惊讶地发现，当依赖类型作为基类出现时，没有必要添加 typename:

struct B {};

struct wr
{ typedef B type; };

template<class T>
struct A : T::type
{};

int main()
{
    A<wr> a;
    (void)a;
}

为什么T::type前面不需要typename？

Answer 1

Why isn't typename required in front of T::type?

因为你不能从值继承。您使用 typename 告诉编译器给定的嵌套标识符是一种类型，但对于继承来说，无论如何都必须是这种情况，因此您可以省略它 - 这就是为什么该语言为 提供了一个异常(exception)>typename - 基本说明符的规则。来自 cppreference (强调我的):

The typename disambiguator for dependent names

In a declaration or a definition of a template, including alias template, a name that is not a member of the current instantiation and is dependent on a template parameter is not considered to be a type unless the keyword typename is used or unless it was already established as a type name, e.g. with a typedef declaration or by being used to name a base class.

请注意，我们会得到更多可以省略 typename 的地方，请参阅 P0634 .