Re: How to declare a friend function of a template class?

From:
"Bo Persson" <bop@gmb.dk>
Newsgroups:
comp.lang.c++
Date:
Tue, 13 Jul 2010 19:48:45 +0200
Message-ID:
<8a3n7kFk8iU1@mid.individual.net>
Francesco S. Carta wrote:

Francesco S. Carta <entuland@gmail.com>, on 12/07/2010 22:29:59,
wrote:

Rui Maciel <rui.maciel@gmail.com>, on 12/07/2010 21:08:03, wrote:

I have a class which has a set of friend functions. Now I wish to
convert that class to a template
class but I'm having trouble declaring the friend functions of
that template class. Can anyone
help? Any tip is more than welcomed.


It's a bit tricky, but possible:

//-------
template<class T> class foo {
public:
foo(T t): t(t) {}
protected:
T t;
template<class U> friend void inspect_foo(foo<U> f);
};

template<class T> void inspect_foo(foo<T> f) {
cout << f.t << endl;
}
//-------


Well, no, the word I meant to use wasn't exactly "tricky" but more
on the side of "verbose / tedious" - if you have lots of overloads
it's way easier to put them in a class and make that a friend, just
like suggested by Victor.

I wonder what's the rationale of not allowing to simply say:

friend function_name;

and set all overloads of "function_name" as friends, regardless of
their arguments or return types - I suppose the mechanism is not as
simple as I'd expect it to be, with these words of mine.


That would let anyone write additional unrelated overloads, and get
access to the private class members. Could as well make them public.

Bo Persson

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