template specialization for a const pointer to a const type

Question

I was reading http://bartoszmilewski.wordpress.com/2009/10/21/what-does-haskell-have-to-do-with-c/ and came across this code to check if a type is a pointer or not:

template<class T> struct
isPtr {
 static const bool value = false;
};

template<class U> struct
isPtr<U*> {
 static const bool value = true;
};

template<class U> struct
isPtr<U * const> {
 static const bool value = true;
};

How do i specialize the general template to handle case for const pointer to a const type? If i do this:

std::cout << isPtr <int const * const>::value << '\n';

i get a true when i am expecting false. Can someone explain?

EDIT: using VC++ 2010 compiler (express :-)

score 5 · Accepted Answer · answered Jul 19 '11 at 02:43

The result is coming correct only; your third specialization is invoked when you call isPtr <int const * const>; which you are setting to true.

You can choose enum over bool in this case as you have 3 states:

enum TYPE
{
  NOT_POINTER,
  IS_POINTER,
  IS_CONST_POINTER
};

template<class T> struct
isPtr {
 static const TYPE value = NOT_POINTER;
};

template<class U> struct
isPtr<U*> {
 static const TYPE value = IS_POINTER;
};

template<class U> struct
isPtr<U * const> {
 static const TYPE value = IS_CONST_POINTER;
};

Here is the demo.

template specialization for a const pointer to a const type

1 Answers1