c++ - clang++ - 将模板类名视为类范围内的模板

Question

似乎 clang++(我试过 clang 3.2)将模板类的名称视为实例化类，而不是类范围内任何事件的模板。比如下面的代码

template <template <class> class T>
class A {};

template <typename T>
class B {
    A<B> member;
   // ^---- clang++ treats B as an instantiated class
         // but I want it to be a template here
         // this code could compile in g++
};

int main()
{
    B<int> b;
    return 0;
}

我应该怎么做才能编译它？

Answer 1

C++03

以这种方式解析 B(称为 injected-class-name，每个类的隐式声明成员，包括模板实例化)旨在提供便利。我从来没有见过它像那样妨碍!

要变通，通过在名称前添加 :: 来限定名称(如果需要，还可以添加命名空间的名称)。

template <typename T>
class B {
    A< ::B> member; // whitespace required to prevent digraph; see comments
};

C++11

C++11 §14.6.1/1 指定(强调我的)

Like normal (non-template) classes, class templates have an injected-class-name (Clause 9). The injected- class-name can be used as a template-name or a type-name. When it is used with a template-argument-list, as a template-argument for a template template-parameter, or as the final identifier in the elaborated-type- specifier of a friend class template declaration, it refers to the class template itself. Otherwise, it is equivalent to the template-name followed by the template-parameters of the class template enclosed in <>.

因此，如果在 C++11 下出现此问题，则为编译器错误。如上所述的解决方法。

注——为了对比，C++03中对应的段落是

Like normal (non-template) classes, class templates have an injected-class-name (clause 9). The injected- class-name can be used with or without a template-argument-list. When it is used without a template- argument-list, it is equivalent to the injected-class-name followed by the template-parameters of the class template enclosed in <>. When it is used with a template-argument-list, it refers to the specified class template specialization, which could be the current specialization or another specialization.

如您所见，已经有一种特殊情况允许标识符是类或模板，具体取决于它是否出现在模板名称中。他们只是增加了几个案例。