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?

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:

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

#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 << "
";
}