How to check if a member name (variable or function) exists in a class, with or without specifying type?

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)); \
  }

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`

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 << "\n";
  std::cout << HasMember_Foo<S>::value << "\n";
  std::cout << HasMember_i<S, int (S::*)>::value << "\n";
  std::cout << HasMember_i<S>::value << "\n";
}

Catches:

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