Const and non const template specialization - c++

Question

I have a class template which I use to get the size of a variable:
template <class T>
class Size
{
unsigned int operator() (T) {return sizeof(T);}
};
This works fine but for strings I want to use strlen instead of sizeof:
template <>
class Size<char *>
{
unsigned int operator() (char *str) {return strlen(str);}
};
The problem is when I create an instance of size with const char * it goes to the unspecialized version. I was wondering if there is a way to capture both the const and non-const versions of char * in on template specialization? Thanks.

Answer 1

Use this technique:
#include <type_traits>
template< typename T, typename = void >
class Size
{
unsigned int operator() (T) {return sizeof(T);}
};
template< typename T >
class Size< T, typename std::enable_if<
std::is_same< T, char* >::value ||
std::is_same< T, const char* >::value
>::type >
{
unsigned int operator() ( T str ) { /* your code here */ }
};
EDIT: Example of how to define the methods outside of the class definition.
EDIT2: Added helper to avoid repeating the possibly long and complex condition.
EDIT3: Simplified helper.
#include <type_traits>
#include <iostream>
template< typename T >
struct my_condition
: std::enable_if< std::is_same< T, char* >::value ||
std::is_same< T, const char* >::value >
{};
template< typename T, typename = void >
struct Size
{
unsigned int operator() (T);
};
template< typename T >
struct Size< T, typename my_condition< T >::type >
{
unsigned int operator() (T);
};
template< typename T, typename Dummy >
unsigned int Size< T, Dummy >::operator() (T)
{
return 1;
}
template< typename T >
unsigned int Size< T, typename my_condition< T >::type >::operator() (T)
{
return 2;
}
int main()
{
std::cout << Size< int >()(0) << std::endl;
std::cout << Size< char* >()(0) << std::endl;
std::cout << Size< const char* >()(0) << std::endl;
}
which prints
1
2
2

Answer 2

And you should also be able to write, of course:
template <>
class Size<const char *>
{
unsigned int operator() (const char *str) {return strlen(str);}
};
template <>
class Size<char *> : public Size<const char *>
{ };
...and, should you need to:
template <size_t size>
class Size<char[size]> : public Size<const char *>
{ };
template <size_t size>
class Size<const char[size]> : public Size<const char *>
{ };

Answer 3

One way is to use a helper function in order to determine if the template type is a char * or a char const *. You can do this with a simple struct. Then you can use SFINAE to select the proper specialization of your Size struct.
#include <cstring>
#include <iostream>
#include <boost/utility/enable_if.hpp>
template<typename T>
struct is_cstring {
enum { value = false };
};
template<>
struct is_cstring<char *> {
enum { value = true };
};
template<>
struct is_cstring<char const *> {
enum { value = true };
};
template<typename T, typename = void>
struct Size;
template<typename T>
struct Size<T, typename boost::disable_if<is_cstring<T> >::type> {
unsigned int operator ()(T const &) const {
return sizeof(T);
}
};
template<typename T>
struct Size<T, typename boost::enable_if<is_cstring<T> >::type> {
unsigned int operator ()(T const &str) const {
return strlen(str);
}
};
int main() {
std::string blah = "afafasa";
char *x = "asdfsadsad";
std::cout << Size<int>()(4) << std::endl;
std::cout << Size<char const *>()("blahblah") << std::endl;
std::cout << Size<char *>()(x) << std::endl;
}
The printed result is:
4
8
10