TITLE: forward declaring friends

(Newsgroups: comp.lang.c++.moderated, 20 Oct 99)

CHAMBERLAND: Eric Chamberland

>> [snip]
>> > template <class T>
>> > class A
>> > {
>> > friend void foo<T>();
>> > };
>> >
>> > template <class T>
>> > void foo(){
>> > }
>> > --------------------------
>> >
>> > Whith my compiler (KAI --strict -D_KAI_STRICT v3.4d) the second
>> > doesn't pass, but the first does.


BUSHUYEV: Gene Bushuyev

>> The syntax of your friend declaration is wrong.


SHIVA: Shiva <jshiva@bigfoot.com>

> I can't see anything wrong with his friend declaration
>
> template<class T>
> class A
> {
> friend void foo<T>();
> };
>
> seems to be perfectly correct.
> It means foo<int> is a friend of A<int>
> foo<string> is a friend of A<string>

BUSHUYEV:

>> And the definition of
>> the friend should precede the declaration in a template class that it
>> may be found during the lookup.

SHIVA:

> I don't know about this but another post in the same thread by Biju
> Thomas indicates that
> friend decl is taken as a forward declaration.


POTTER: John Potter <jpotter@falcon.lhup.edu>

I think that if you read that again, you will see that Biju said "yes"
a forward declaration is required.  The rules have changed on this
along the road to the standard.  C++PL3e has a recent correction to
an example.  The state of compilers has also changed and g++ for
example accepts the old and new rules in such a mix that it is of
no value in deciding what is correct.

Play it safe.  Forward declare the friends prior to the class that
declares them as friends.

I think that I have this example right.

void f();

class C {
   friend void f(); // The f above is a friend
   friend void g(); // Some unknown g is a friend if it gets declared.
   };
void h1 () {
   g();  // error, there is no g in scope
   }
void g(); // Here is that g and it is a friend
void h2 () {
   g();  // ok, calls the friend
   }

When parameters get involved, there are some other rules.  Templates
just make it messier.  Practical advice.  Assume that a friend
declaration does not declare the function.  Put the function
declaration ahead of the class that makes it a friend.


TRIBBLE: David R Tribble [david@tribble.com] (private correspondence)

This is also a requirement if you're declaring a friend function having
extern C linkage:
 
    extern "C"
    {
        int  cCall(int a);
    }
 
    class Foo
    {
        friend int  cCall(int a);
        ...
    };
 
The declaration of the extern C function must precede its declaration as
a friend.