Re: problem with template friend partial specialization

From:

Victor Bazarov <v.bazarov@comcast.invalid>

Newsgroups:

comp.lang.c++

Date:

Tue, 03 Apr 2012 08:52:07 -0400

Message-ID:

<jlerpn$ctc$1@dont-email.me>

On 4/2/2012 5:35 AM, Andrey Dj wrote:

Hello!

I found some interesting case of non-compiling C++ code.
Look at first example:

--------------------------------
template<class T> void Foo(T*);
class Bar;
template<> void Foo(Bar*);

class Bar
{
friend void ::Foo<>(Bar*);
};
--------------------------------

Compiling... All fine.

Now add to Foo return type of Baz:

--------------------------------
class Baz {};

template<class T> Baz Foo(T*);
class Bar;
template<> Baz Foo(Bar*);

class Bar
{
friend Baz ::Foo<>(Bar*);
};
--------------------------------

Trying to compile:
MSVC: error C2039: 'Foo' : is not a member of 'Baz'
GCC: error: 'Foo' in class 'Baz' does not name a type

Is there any way to declare such friend partial specialization?

It's a tokenization problem, obviously. Your compiler treats the colons
before 'Foo' in the friend declaration as part of an elaborate name
specifier that starts with 'Baz'. It knows that 'Baz' is a class name...

You could try adding a 'using' declaration before the 'friend':

     class Bar
     {
        using ::Foo;
        friend Baz Foo<>(Bar*);
     };

(I've not tried it). Or you could put 'Baz' in parentheses:

     class Bar
     {
        friend (Baz) ::Foo<>(Bar*);
     };

(I've not tried it either). What you essentially need is tell the
compiler that the token "Baz" should stand on its own and not be treated
as the beginning of an elaborate name specifier even though there is a
name resolution operator after it.

V
--
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