如何使用顶级const解决此问题？

Question

I have a class template that contains two similar member functions:

template<class T>
class MyTemplate {
   // other stuff, then
   static T* member( T& ); //line 100
   static const T* member( const T& ); //line 101

};

which I instantiate like this:

MyTemplate<int* const>

and Visual C++ 9 complains:

mytemplate.h(101) : error C2535: 'int* MyTemplate<T>::member(T &)' :
   member function already defined or declared
with
[
    T=int *const 
]
mytemplate.h(100) : see declaration of 'MyTemplate::member'
with
[
    T=int *const 
]
somefile.cpp(line) : see reference to class template instantiation
    'MyTemplate<T>' being compiled
with
[
    T=int *const 
]

I certainly need the two versions of member() - one for const reference and one for non-const reference. I guess the problem has something with top-level const qualifiers, but can't deduce how to solve it.

How do I resolve this problem so that I still have two versions of member() and the template compiles?

Answer 1

When T is int * const , T is already const , so T& and const T& are both int * const .

Or do you mean in this case, you need your class to look like:

class MyTemplate_int_p_const{
     static int * member (int *&);
     static int * const member (int * const &);
};

You can add this to your main template to achieve this:

template<class T>
class MyTemplate<const T>
{
    static T * member(T&);
    static const T* member(const T&);
};

As a responds to the OP's comment, if you don't want to use partial specialization, you'll need type_traits. It is supported by C++0x, and for VC++9, you can use boost .

In the following code, the non_const version of member will take a dummy_type ( a pointer to member function) if T is already const. So the non_const overload would not exist.

#include <type_traits>
template<class T>
class MyTemplate {
   // other stuff, then

   //void dummy(void);
   typedef void (*dummy_type)(void);
   typedef typename std::conditional<std::is_const<T>::value, dummy_type, T>::type T_no_const;
   typedef typename std::remove_const<T>::type T_remove_const;
   static T_no_const* member( T_no_const& t ) //line 100
   {
       if (std::is_same<T, T_no_const>::value)
       {
           return member_portal(t);
       }
       else
           return NULL;
   }
   static T_no_const* member_portal(dummy_type&){return NULL;};
   static T_remove_const* member_portal(T_remove_const&);
   static const T* member( const T& ); //line 101

};


int main()
{
    MyTemplate<int * const> mt;
    MyTemplate<int *> mtt;
    return 0;
}

This is the first time that I play with type_traits . It can pass compilation under g++ 4.5.2 with C++0x enabled. But I've never run it.

Main idea is, when T is const, the non_const version of member takes a argument of an arbitrary type ( a type that is not likely to be used any where else, and to not likely to be implicitly converted to), thus the non_const version disappears. But in the way, the logic breaks in the implementation of member ( as the argument type is to be used, but is not expected). So the main logic of member is move another function of member_portal .

Answer 2

The explanation given by fefe is correct. Foo const& and Foo const const& will simply evaluate to the same type, hence your function overloading does not work. In case your template argument is const, I'd suggest specialisation.

Version A:

template<class T>
class MyTemplate {
   static T* member( T& );
   static const T* member( const T& );
};
template<class T>
class MyTemplate<T const> {
   static const T* member( const T& );
};

Version B:

template<class T>
class MyTemplate_mutableImpl {
   static T* member( T& );
};
template<class T>
class MyTemplate_constImpl {
   static const T* member( const T& );
};
template<class T>
class MyTemplate : public MyTemplate_mutableImpl<T>, public MyTemplate_constImpl<T> {
};
template<class T>
class MyTemplate<T const> : public MyTemplate_constImpl<T const> {
};

Answer 3

A simple solution would be to disable the function if T is const :

#include <boost/mpl/if.hpp>
#include <boost/type_traits/is_const.hpp>

template<class T>
class MyTemplate {
   static T* member( T& );
   struct Disabled {};
   static const T* member(typename boost::mpl::if_<boost::is_const<T>, Disabled, const T&>::type);
};