how to implement is_pointer?

Question

I want to implement is_pointer. I want something like this:

template 
bool is_pointer( T t )
{
   // implementation
} // return true or fa         


        

Answer 1

From Dr. Dobbs.

template <typename T> 
struct is_pointer 
{ static const bool value = false; };

template <typename T> 
struct is_pointer<T*> 
{ static const bool value = true; };

You can't do exactly what you want to do. You'll have to use this like:

is_pointer<int*>::value

It's not possible to determine this at run time.

Answer 2

template <typename T>
struct is_pointer_type
{
    enum { value = false };
};

template <typename T>
struct is_pointer_type<T*>
{
    enum { value = true };
};

template <typename T>
bool is_pointer(const T&)
{
    return is_pointer_type<T>::value;
}

Johannes noted:

This is actually missing specializations for T *const, T *volatile and T * const volatile i think.

Solution:

template <typename T>
struct remove_const
{
    typedef T type;
};

template <typename T>
struct remove_const<const T>
{
    typedef T type;
};

template <typename T>
struct remove_volatile
{
    typedef T type;
};

template <typename T>
struct remove_volatile<volatile T>
{
    typedef T type;
};

template <typename T>
struct remove_cv : remove_const<typename remove_volatile<T>::type> {};

template <typename T>
struct is_unqualified_pointer
{
    enum { value = false };
};

template <typename T>
struct is_unqualified_pointer<T*>
{
    enum { value = true };
};

template <typename T>
struct is_pointer_type : is_unqualified_pointer<typename remove_cv<T>::type> {};

template <typename T>
bool is_pointer(const T&)
{
    return is_pointer_type<T>::value;
}

...but of course this is just reinventing the std::type_traits wheel, more or less :)

Answer 3

You can use "typeid" operator defined in typeinfo.h for this. check this link : http://en.wikipedia.org/wiki/Typeid

The typeid operator will give an object of std::type_info class, which has a name() function returning char *. Once you get the type in string form, you can identify the pointer easily.

Hope it helps.

Romil.

Answer 4

template <typename T>
bool is_pointer(T const &t) // edited: was "T t"; see the comments
{
   return false;
}

template <typename T>
bool is_pointer(T *t)
{
   return true;
}

You might not believe it, but it works. The reason is that the most specific template implementation will be chosen, which is the one which takes the pointer type.