tags:

views:

417

answers:

4

I have code which works in VC9 (Microsoft Visual C++ 2008 SP1) but not in GCC 4.2 (on Mac):

struct tag {};

template< typename T >
struct C
{   
    template< typename Tag >
    void f( T );                 // declaration only

    template<>
    inline void f< tag >( T ) {} // ERROR: explicit specialization in
};                               // non-namespace scope 'structC<T>'

I understand that GCC would like me to move my explicit specialization outside the class but I can't figure out the syntax. Any ideas?

// the following is not correct syntax, what is?
template< typename T >
template<>
inline void C< T >::f< tag >( T ) {}
+1  A: 

Try this:

template <> template<typename T> inline void C<T> :: foo<tag2>(T) {}
Alexander Gessler
nope, that doesn't work
jwfearn
sorry - I think I mixed up the order of thw two template qualifiers. Try placing the one with the empty brackets on the very left.
Alexander Gessler
nope again (error: template partial specialization not allowed) 'foo<tag2' is not allowed). Thanks for the suggestion though.
jwfearn
A: 

I know this may not satisfy you, but I do not believe you may not have a specialization enclosed within a non-explicitly-specialized structure.

template<>
template<>
inline void C< tag1 >::foo< tag2 >( t_type ) {}
ephemient
+4  A: 

GCC is in the clear, here. MSVC has a non-standard extension that allows in-class specialization. The standard, however, says:

14.7.3.2:
2. An explicit specialization shall be declared in the namespace of which the template is a member, or, for member templates, in the namespace of which the enclosing class or enclosing class template is a member. An explicit specialization of a member function, member class or static data member of a class template shall be declared in the namespace of which the class template is a member.

Additionally, you can't partially specialize a function. (Though I'm unsure about the details in your case, that would be the final blow.)

You could do this:

#include <iostream>

struct true_type {};
struct false_type {};

template <typename T, typename U>
struct is_same : false_type
{
    static const bool value = false;
};

template <typename T>
struct is_same<T, T> : true_type
{
    static const bool value = true;
};

struct tag1 {};
struct tag2 {};

template< typename T >
struct C
{
    typedef T t_type;

    template< typename Tag >
    void foo( t_type pX)
    {
        foo_detail( pX, is_same<Tag, tag1>() );
    }

private:
    void foo_detail( t_type, const true_type& )
    {
        std::cout << "In tag1 version." << std::endl;
    }
    void foo_detail( t_type, const false_type& )
    {
        std::cout << "In not tag1 version." << std::endl;
    }
};

int main(void)
{
    C<int> c;
    c.foo<tag1>(int());
    c.foo<tag2>(int());
    c.foo<double>(int());
}

Though this is somewhat ugly.

GMan
+2  A: 

You can't specialize a member function without explicitly specializing the containing class.
What you can do however is forward calls to a member function of a partially specialized type:

template<class T, class Tag>
struct helper {
    static void f(T);   
};

template<class T>
struct helper<T, tag1> {
    static void f(T) {}
};

template<class T>
struct C {
    // ...
    template<class Tag>
    void foo(T t) {
        helper<T, Tag>::f(t);
    }
};
Georg Fritzsche
This is similar to the solution I'm using, thanks! I wonder if there's a way to do this less verbosely using Boost?
jwfearn
Not really, it doesn't get easier in general. You need a helper function to avoid the issue of *"enclosing class has to be explicitly specialized"* and you need to move that one into a class to do partial specialization.
Georg Fritzsche