继承模板化转换运算符

Question

考虑以下代码：

template <class R, class... Args>
using function_type = R(*)(Args...);

struct base {
    template <class R, class... Args>
    constexpr operator function_type<R, Args...>() const noexcept {
        return nullptr;
    }
};

struct derived: private base {
    template <class R, class... Args>
    using base::operator function_type<R, Args...>; // ERROR
};

在 C++20 中是否有可行的替代方法来继承和公开模板化转换函数？

Answer 1

GCC 支持这个：[demo]

template <class R, class... Args>
using function_type = R(*)(Args...);

struct base {
    template <class R, class... Args>
    constexpr operator function_type<R, Args...>() const noexcept {
        return nullptr;
    }
};

struct derived: private base {
  
    using base::operator function_type<auto, auto...>; // No error!
};


int main (){
  derived d;
  static_cast <int(*)(int)>(d);
}

但我认为这是对可能来自概念-TS的语言的扩展。

Answer 2

Is there a working alternative in C++20 to inherit and expose a templated conversion function?

我不知道直接公开它的方法，通过 using。

但您可以将其包装在派生运算符中

struct derived: private base {

    template <typename R, typename... Args>
    constexpr operator function_type<R, Args...>() const noexcept {
       return base::operator function_type<R, Args...>();
    }

};

继承模板化转换运算符

Inheriting a templated conversion operator

c++

inheritance

conversion-operator

template-meta-programming

c++20