将模板化基类转换运算符引入派生作用域 [英] Bring templated base class conversion operator into derived scope

问题描述

我有一个基类，它定义了一个受约束的模板化转换运算符

I have a base class that defines a constrained templated conversion operator

struct base {
    template <typename C, std::enable_if_t<some_constraints<C>, int> = 0>
    operator C() const;
};

我还有一个派生类，它实现了另一个具有不同约束的转换运算符:

I also have a derived class that implements another conversion operator with different constraints:

struct derived : base {
    template <typename P, std::enable_if_t<different_constraints<P>, int> = 0>
    operator P() const;
};

不幸的是，派生类中的声明隐藏了基类中的运算符.我想将基本运算符带入派生范围，但明显"语法

Unfortunately, the declaration in the derived class hides the operator in the base class. I would like to bring the base operator into the derived scope, but the "obvious" syntax

template <typename C>
using base::operator C;

不起作用(编译器似乎试图将其解析为别名模板声明).

doesn't work (the compiler seems to try to parse it as an alias template declaration).

有谁知道实现此目的的正确语法吗?

Does anyone know of the correct syntax to achieve this?

推荐答案

我会说这是不可能的.即使是这样，根据 namespace.udecl#15.sentence-1:

I would say it is not possible. and even if it was, your derived operator would hide base one as template argument is not part according to namespace.udecl#15.sentence-1:

当使用声明器将基类的声明引入派生类时，派生类中的成员函数和成员函数模板将覆盖和/或隐藏同名的成员函数和成员函数模板，参数类型列表基类中的 、cv 限定符和 ref 限定符(如果有)(而不是冲突)

When a using-declarator brings declarations from a base class into a derived class, member functions and member function templates in the derived class override and/or hide member functions and member function templates with the same name, parameter-type-list, cv-qualification, and ref-qualifier (if any) in a base class (rather than conflicting)

不幸的是，模板参数不计算在内，转换 operator 的参数类型列表为空，const 没有引用限定符.

Unfortunately, template parameter doesn't count and both conversion operator has empty parameter-type-list, are const and no ref-qualifier.

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