• "Steve Strand" <snstrand comcast.net> Sep 05 2003
• mjs NOSPAM.gmx.de (Mark Junker) Sep 07 2003
• "Steve Strand" <snstrand comcast.net> Sep 07 2003
• mjs NOSPAM.gmx.de (Mark Junker) Sep 07 2003

The program below compiles fine, as long as the definition of fun2 uses the
same "template dummy variable" used in the declaration.
Member function fun1 has no such restriction. Why is this? I ask because
sometimes inquiring into mysteries pays off with unexpected
valuable knowledge.


#include <iostream.h>

template <int kk>                   //dummy variable is kk
class test {
    int val;
public:
    test() {val=0;};
    int fun1();
    template <int rr>               //dummy variable is rr
    friend int fun2(test<rr> a);
};

template <int ss>                   //dummy variable is ss
int test<ss>::fun1() {return val+ss;};

template <int rr>                   //dummy variable has to be rr?!
int fun2(test<rr> a) {return a.val+rr;};

int main()
{
    test<5> a;
    cout << a.fun1() << '\n';    //test member function
    cout << fun2(a)  << '\n';    //test friend function
}

Try using a reference instead.

     template <int rr>               //dummy variable is rr
     friend int fun2(test<rr> a);



template<int rr>
friend int fun2(test<rr> &a);

 template <int rr>                   //dummy variable has to be rr?!
 int fun2(test<rr> a) {return a.val+rr;};



template<int rr>
int fun2(test<rr> &a) { return a.val+rr; }

Should compile without warning when you don't use const objects.

Regards,
Mark Junker

Experiment shows that passing the argument to fun2 by value or
reference makes no difference - the "template dummy variable"
still has to match.

 Experiment shows that passing the argument to fun2 by value or
 reference makes no difference - the "template dummy variable"
 still has to match.


Yes, it seems that you're right. It should only take care of the type not of  
the name.

Regards,
Mark Junker