Member templates

[Guysl]

Say I have a template class, with a member template as the following:

template <typename T>
class CLASS{

template <typename T2>
    void f(T2 const&);
};

Is it possible to have a specialization for f(T2 const&) withot specifying T?
I mean,can I define a special treatment for T2=int no matter what T stands for?

[Graham]

Have you tried

template <typename T>
void CLASS<T>::f2<int>(int const&)
{
	// ...
}

?

[Guysl]

The truth is that member is an operator , and by looking at the compiler's
error I assume the syntax is different.

I get the following compiler errors:
"`operator+' used as template "
"use `CLASS<T>::template operator+' to indicate that it is a template "
"expected init-declarator at end of input "

I've tried to use the following :

CLASS<T> CLASS<T>::operator+<double>(const double &d){
                std::cout << "2" << std::endl;
                return *this;}

OR

CLASS<T> CLASS<T>::template operator+<double>(const double &d){
                std::cout << "2" << std::endl;
                return *this;}

but its not working.

[MrViggy]

This works for me, in Dev Studio 7.x:

template<typename _T>
class TempStruct
{
    TempStruct<_T> operator+(const double &d1);

    int m_nI;
};

template<typename _T>
TempStruct<_T> TempStruct<_T>::operator+(const double &d1)
    {
        std::cout << "2" << std::endl;
        return *this;
    }

As an aside, don't you want to return a reference to the class?

Viggy

[Guysl]

I wish it was that simple. My first general code shows two levels
for template parameters,
T for the class, and T2 for a function member:

template <typename T>
class CLASS{

template <typename T2>
    void f(T2 const&);
};

when it comes to operator, I can't get it work.
(gcc 3.4.2 on Dev-C++ 4.9.9.2)

=============
Edit:
Indeed the return type should be a reference,
but my concern is on templates practice

[MrViggy]

Ahh, I missed that. Yeah, that's an interesting one.

Hmmm, Stroustrup doesn't say anything about specializing a template member function of a template class.

Viggy

[exterminator]

I suggest something similar to what Graham has given. Here is something I think should work but not on VC++ 6.0 -, may be on VS 2003 (I shall provide a link that tells this). I was not able to test this because I have VC++ 6.0 and I am a bit confused about what Comeau documentation has to say about this. First the code that I think should work.

#include <iostream>

template <typename T>
class CLASS{
public:
	template<typename T2>
	void f(const T2&);
        void f(const int&);
};

template<typename T>
template<typename T2>
void CLASS<T>::f(const T2&){std::cout << "template function called\n";}
	
template<typename T>
void CLASS<T>::f(const int&){std::cout << "overload/specialization called\n";}

int main(){
	int i = 10;
	double d = 10.001;
	CLASS<double> obj;
	obj.f(i);
	obj.f<double>(d);
	return 0;
}

This code compiles fine with Comeau. Let's see if this works on the compiler that you are using. Now here are some links that might help you:
1. The comeau documentation - here - Comeau Template FAQ - How to define a member function outside of its template?
2. The MSDN documentation - Member function Templates

See if this helps - and if not please mention the errors and provide the exact compilable code that you are using. You gave us a small example and then you said it is more than that so that keeps people just guessing.

//EDIT: I tested the above quoted code on gcc and it works fine for me. Hope this helps (provided this is what you were looking for).

[marten_range]

One thing to note is that template specializations for methods don't overload (http://www.gotw.ca/gotw/049.htm).

Also, I agree with exterminator. You should provide the forums with source code that encapsulates the whole problem. Whether it's your original source code or an example illustrating the problem. It gets frustrating to invest time in an answer to the original question only to get the response: "Yeah, but...".

Anyway, I think exterminators suggestion by using regular overload is probably closer to what you want (esp. considering that specializations don't overload). Provide a regular non-template member function that overloads the template method. The compiler will choose the non-template function if it's an exact match.

Hope this helps

[Guysl]

Thanks alot guys,
I've used this information to suggest a solution for :
http://www.codeguru.com/forum/showth...06#post1241906

[NMTop40]

VC7 got all of them right according to Sutter. I got 11/12. If I reproduce it here with my code, ask people what they think the output will be.

#include <iostream>
#include <complex>
template<typename T> void writeln( const T& t ) { std::cout << t << std::endl; }

template<typename T1, typename T2>
void f( T1, T2 ) { writeln( 1 ); }                       // 1
template<typename T> void f( T ) { writeln( 2 ); }       // 2
template<typename T> void f( T, T ) { writeln( 3 ); }    // 3
template<typename T> void f( T* ){ writeln( 4 ); }      // 4
template<typename T> void f( T*, T ) { writeln( 5 ); }   // 5
template<typename T> void f( T, T* ) { writeln( 6 ); }   // 6
template<typename T> void f( int, T* ) { writeln( 7 ); } // 7
template<> void f<int>( int ) { writeln( 8 ); }          // 8
void f( int, double ) { writeln( 9 ); }                  // 9
void f( int ) { writeln( 10 ); }                          // 10
    
int main()
{
   int             i=0;
    double          d=0.0;
    float           ff=0.0f;
    std::complex<double> c;

    f( i );         // a
    f<int>( i );    // b
    f( i, i );      // c
    f( c );         // d
    f( i, ff );     // e
    f( i, d );      // f
    f( c, &c );     // g
    f( i, &d );     // h
    f( &d, d );     // i
    f( &d );        // j
    f( d, &i );     // k
    f( &i, &i );    // l
}

[Improving]

a) 10
b) 8
c) 3
d) 2
e) 1
f) 9
g) 6
h) 7
i) 5
j) 4
k) 1
l) 3

I think.

[exterminator]

One thing to note is that template specializations for methods don't overload (http://www.gotw.ca/gotw/049.htm).

I had deferred reading this article because I was quite busy these days and still am, now I am through with it though a bit late. The things there fit in quite well and logically. But I still was not able to figure out what you meant when you said that template specializations for methods don't overload? I saw in that article that the correct functions (or template instantiations) were called depending upon the best match (and the rules set up by the standard) and hence I would conclude that overloading is taking place. I would be needing an explanation for the words you used because I am finding them a bit confusing to me. Hoping for a response. Regards.

[exterminator]

Were you referring to something like - the rules for calling the overloads of function templates (their instantiations), their specializations and normal overload forms (for those functions themselves) are different from that of normal function overloading?