how to implement is_pointer? - c++

Question

I want to implement is_pointer. I want something like this:
template <typename T >
bool is_pointer( T t )
{
// implementation
} // return true or false
int a;
char *c;
SomeClass sc;
someAnotherClass *sac;
is_pointer( a ); // return false
is_pointer( c ); // return true
is_pointer( sc ); // return false
is_pointer( sac ); // return true
How can I implement it?
Thanks

Answer 1

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 2

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 3

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.

Answer 4

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.