如何检查类中是否存在成员名称(变量或函数)，无论是否指定类型? [英] How to check if a member name (variable or function) exists in a class, with or without specifying type?

问题描述

这个Q是以下内容的扩展:
模板化检查类成员函数是否存在?

This Q is an extension of:
Templated check for the existence of a class member function?

是否有任何实用程序可以帮助您找到:

Is there any utility which will help to find:

• 成员名是否存在于类中?该成员可以是变量或方法.
• 指定成员的类型应该是可选的

推荐答案

C++03

#define HasMember(NAME) 
  template<class Class, typename Type = void> 
  struct HasMember_##NAME 
  { 
    typedef char (&yes)[2]; 
    template<unsigned long> struct exists; 
    template<typename V> static yes Check (exists<sizeof(static_cast<Type>(&V::NAME))>*); 
    template<typename> static char Check (...); 
    static const bool value = (sizeof(Check<Class>(0)) == sizeof(yes)); 
  }; 
  template<class Class> 
  struct HasMember_##NAME<Class, void> 
  { 
    typedef char (&yes)[2]; 
    template<unsigned long> struct exists; 
    template<typename V> static yes Check (exists<sizeof(&V::NAME)>*); 
    template<typename> static char Check (...); 
    static const bool value = (sizeof(Check<Class>(0)) == sizeof(yes)); 
  }

用法:只需使用您想要查找的任何成员调用宏:

Usage: Simply invoke the macro with whatever member you want to find:

HasMember(Foo);  // Creates a SFINAE `class HasMember_Foo`
HasMember(i);    // Creates a SFINAE `class HasMember_i`

现在我们可以使用HasMember_X来检查任何class中的X，如下所示:

Now we can utilize HasMember_X to check X in ANY class as below:

#include<iostream>
struct S
{
  void Foo () const {}
//  void Foo () {}  // If uncommented then type should be mentioned in `HasMember_Foo`    
  int i;
};
int main ()
{
  std::cout << HasMember_Foo<S, void (S::*) () const>::value << "
";
  std::cout << HasMember_Foo<S>::value << "
";
  std::cout << HasMember_i<S, int (S::*)>::value << "
";
  std::cout << HasMember_i<S>::value << "
";
}

捕获:

1. 在方法的情况下，如果我们不提及类型，那么class不得有重载方法.如果有，那么这个技巧就失败了.即，即使命名成员出现不止一次，结果也会是 false.
2. 如果成员是基类的一部分，那么这个技巧就失败了；例如如果 B 是 S 的基础 &void B::Bar() 存在，则 HasMember_Bar::value 或 HasMember_Bar::value 或 HasMember_Bar::value 将给出 false

1. In case of methods, if we don't mention the type then the class must not have overloaded methods. If it has then this trick fails. i.e. even though the named member is present more than once, the result will be false.
2. If the member is part of base class, then this trick fails; e.g. if B is base of S & void B::Bar () is present, then HasMember_Bar<S, void (B::*)()>::value or HasMember_Bar<S, void (S::*)()>::value or HasMember_Bar<S>::value will give false

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