c++ - 重新声明类名 class-key

Question

以下引用来自 14.5.1/4 [temp.class]:

In a redeclaration, partial specialization, explicit specialization or explicit instantiation of a class template, the class-key shall agree in kind with the original class template declaration

我认为这意味着我们不能使用另一个类键声明显式特化，例如:

template <class T, class W>
struct A
{
    void foo();
};

template <class T, class W>
class A<T*, W>                 // Should have printed an error
{
    void foo();
};

<强> DEMO

但是效果很好。那么这条规则有什么意义呢？

Answer 1

在引用的句子之后是对 [dcl.type.elab] 的引用。 [dcl.type.elab]/p3 描述了“同意”的含义:

The class-key or enum keyword present in the elaborated-type-specifier shall agree in kind with the declaration to which the name in the elaborated-type-specifier refers. [...] Thus, in any elaborated-type-specifier, the enum keyword shall be used to refer to an enumeration (7.2), the union class-key shall be used to refer to a union (Clause 9), and either the class or struct class-key shall be used to refer to a class (Clause 9) declared using the class or struct class-key.

换句话说，如果主模板是 union ，那么“重新声明、部分特化、显式特化或显式实例化”必须使用union；否则它可以使用class或struct，但不能使用union。