使用引用基类的成员模板的声明.

Question

任何人都可以解释这个

 "A using-declaration in a derived class cannot refer to a specialization 
 of a template conversion function in a base class."

它来自ISO C++标准.14.5.2,第7点

Answer 1

这意味着这是不正确的:

struct A { template<typename T> operator T(); };
struct B : A { using A::operator int; }; // ill-formed: refers to specialization

同样适用于其他功能模板专业化(不仅仅是转换功能)

struct A { template<typename T> void f(); };
struct B : A { using A::f<int>; }; // ill-formed: refers to specialization