templates - 通过模板别名显式实例化类

Question

是否可以通过模板别名显式实例化模板类？

如果是这样，如何？否则，有人可以指出讨论并决定反对的 ISO 文件吗？

template<class T>
struct A { };

/// Explicit instantiate A for int:
template struct A<int>;

/// Alias
template<class T>
using B = A<T>;

/// Explicitly instantiate A for double via alias B:
template struct B<double>;
/// error: elaborated type refers to a non-tag type

这不应该实例化 A<double>自 B<T>只是 A<T> 的不同名称?

Answer 1

这是间接禁止的，因为:

7/3 禁止在没有类键的情况下编写显式特化( class 、 struct 或 union ):

In a simple-declaration, the optional init-declarator-list can be omitted only when declaring a class (Clause 9) or enumeration (7.2), that is, when the decl-specifier-seq contains either a class-specifier, an elaborated-type-specifier with a class-key (9.1), or an enum-specifier.

7.1.6.3/2 禁止将类键与别名模板特化结合:

3.4.4 describes how name lookup proceeds for the identifier in an elaborated-type-specifier. ... If the identifier resolves to a typedef-name or the simple-template-id resolves to an alias template specialization, the elaborated-type-specifier is ill-formed.