friend of template method in template class, within a template class

Question

The description can be quite mind boggling so I get straight to the example:

#include <iostream>
#include <typeinfo>
using namespace std;

template<typename T> class fr{
    static void privatestuff(){
        cout<<"private of "<<typeid(T).name()<<endl;
    }
public:
    template<typename TT> void callsomeone(){
        fr<TT>::privatestuff();
    }
    //template<typename TT> friend void fr<TT>::callsomeone<T>();
    //template<> template<typename TT> friend void fr<TT>::callsomeone<T>();
    //template<typename TT> template<> friend void fr<TT>::callsomeone<T>();
    //no other combinations... how to get it?
};

int main(){
    fr<bool> obj;
    obj.callsomeone<int>();
}

Basically, I want fr to be able to call fr<int>::privatestuff. But I'd like also to not expose more than what is needed, so make fr<int> friend of only fr<bool>::callsomeone<int>, not fr<bool>::callsomeone<char> or others.

I can count on c++11 if that's needed.

The question is in a comment in the code. What it means? No idea... — scientiaesthete, Apr 09 '12 at 20:13
If I understand you correct - you need to declare callsomeone() as friend function of class fr to be able to call private fr's functions. Is it correct? — SVGreg, Apr 09 '12 at 20:20
But `callsomeone()` is a *member function* template, it doesn't need to be a friend in the first place. Or am I missing something? — jrok, Apr 09 '12 at 20:34

Johannes Schaub - litb · Accepted Answer · 2012-04-09T20:19:21.043

1

template<class x> template<class y> friend void fr<x>::callsomeone();

You shall not pass any template arguments to callsomeone, because you want to befriend a function template, and not a specialization of it (in other words, you want to befriend all specializations of it).

edited Apr 09 '12 at 20:19

answered Apr 09 '12 at 20:14

Johannes Schaub - litb

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Should that be fd? And where does y fit in? – SirGuy Apr 09 '12 at 20:19
doesn't this make declare friendness to any TT? I want to make it friend only when TT=T – Lorenzo Pistone Apr 09 '12 at 20:21
no, his class template is called fr, not fd. I have no idea why he used fd in the friend declaration. – Johannes Schaub - litb Apr 09 '12 at 20:22
i believe i see.your intention now. i will look up in the spec the conditions of befriending a single function template secialization of a class template later, but i seem to remember that is not possible. – Johannes Schaub - litb Apr 09 '12 at 20:32
Could you link me these specs by the way? – Lorenzo Pistone Apr 09 '12 at 20:36
the problem is that a.friend.declaration must refer.to somethong that can be.declared. but an instantiation of a.member.template first has its.enclosing template be instantiated. you are trying to reverse the instantiation of the enclosing and enclosed template, but there will never be.a declaration of fr::f – Johannes Schaub - litb Apr 09 '12 at 20:36
sorry for the typos, im writing.on my phone – Johannes Schaub - litb Apr 09 '12 at 20:36
@Lorenzo : The 'spec' in this case is the C++ standard, which is not freely available, however an extremely close draft is freely available [here](http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2012/n3337.pdf). – ildjarn Apr 09 '12 at 23:23

friend of template method in template class, within a template class

1 Answers1