Have you tried to put "inline" in the function definition?

"Anthony Williams" <anthony_w.geo@yahoo.com> wrote in message
news:8pi4bd$d807e$1@ID-49767.news.cis.dfn.de...
> I posted this about a week ago, and it didn't show up on Deja, so I'm
going
> to
> try again.
> - ---
> I have a class template, and I want to declare a friend function that
> performs
> some operation on this class. The problem is, I want this friend
> function to be
> a template
>
> template<typename T>
> class MyClass
> {
>   template<typename U>
>   friend void func(MyClass<T> mc,U data);
> };
>
> How can I define this function outside the class template?
>
> I tried:
>
> template<typename T>
> template<typename U>
> void func(MyClass<T> mc,U data)
> {}
>
> and
>
> template<typename T,typename U>
> void func(MyClass<T> mc,U data)
> {}
>
> both of which compiled OK, but didn't link, as the compiler didn't
> match these
> template definitions to that declared in the friend declaration.

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-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1

I posted this about a week ago, and it didn't show up on Deja, so I'm going
to
try again.
- ---
I have a class template, and I want to declare a friend function that
performs
some operation on this class. The problem is, I want this friend
function to be
a template

template<typename T>
class MyClass
{
  template<typename U>
  friend void func(MyClass<T> mc,U data);
};

How can I define this function outside the class template?

I tried:

template<typename T>
template<typename U>
void func(MyClass<T> mc,U data)
{}

and

template<typename T,typename U>
void func(MyClass<T> mc,U data)
{}

both of which compiled OK, but didn't link, as the compiler didn't
match these
template definitions to that declared in the friend declaration.

Help is much appreciated.

Anthony

- --
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