在C ++中推导继承方法的父类 [英] Deduce parent class of inherited method in C++

问题描述

采取以下类层次结构：

  template< typename T& 
 class Foo {
 public：
 T fooMethod（）{...} 
}; 
 class Moo：public Foo< bool> {
 ... 
}; 
  

如果我现在在某处写 Moo :: fooMethod 编译器会推导出 Foo< bool> :: fooMethod 。在编译时之前，如何推导 Foo 为 fooMethod 的父类？

动机：编译器不会允许 bool（）的 Foo< bool> :: fooMethod Moo :: *）（），因为它在上下文中将是 bool（Foo< bool> :: *）（）但是因为我有多个继承，我不知道什么父 fooMethod 将在，它必须被推导。

如果我正确理解问题，可以使用下面的trait来推导出成员函数定义的类：

  template< typename> 
 struct member_class_t; 
 
 template< typename R，typename C，typename ... A> 
 struct member_class_t< R（C :: *）（A ...）> {using type = C; }; 
 
 template< typename R，typename C，typename ... A> 
 struct member_class_t< R（C :: *）（A ...）const> {using type = C const; }; 
 
 // ...其他限定词特殊化
 
 template< typename M> 
 using member_class = typename member_class_t< M> :: type; 
  

之后您可以写

  member_class< decltype（& Moo :: fooMethod）> 
  

给予 Foo< bool> p>



要更一般地定义 member_class ，您应该考虑 volatile 和 ref - 限定符，总共产生大约12个专业。完整的定义是此处 a>。

Take the following class hierarchy:

template<typename T>
class Foo {
public:
    T fooMethod() { ... }
};
class Moo : public Foo<bool> {
    ...
};

If I now somewhere write Moo::fooMethod the compiler will deduce Foo<bool>::fooMethod. How can I deduce Foo<bool> as parent of fooMethod myself before compile time?

Motivation: the compiler will not allow Foo<bool>::fooMethod to be passed as template parameter for bool (Moo::*)() since it will be of type bool (Foo<bool>::*)() in that context. But since I have multiple inheritance I dont know what parent fooMethod will be in, it must be deduced.

解决方案

If I understand the problem correctly, it is possible to deduce the class a member function is defined in, using the following trait:

template<typename>
struct member_class_t;

template<typename R, typename C, typename... A>
struct member_class_t <R(C::*)(A...)> { using type = C; };

template<typename R, typename C, typename... A>
struct member_class_t <R(C::*)(A...) const> { using type = C const; };

// ...other qualifier specializations

template<typename M>
using member_class = typename member_class_t <M>::type;

after which you can write

member_class<decltype(&Moo::fooMethod)>

giving Foo<bool>.

To define member_class more generally, you should take into account volatile and ref-qualifiers as well, yielding a total of about 12 specializations. A complete definition is here.

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