Re: Templated friend classes
Ralf Goertz wrote:
Hi,
while I was redesigning some of my programs to have them resemble cool
C++ code my compiler gave me a hard time. For instance, I use a
templated base class and two derived classes. These need to know about
each others private internals so that I decided to make them friends of
each other. But in order to do that I need a forward declaration of one
of the classes. I couldn't find out how to do that correctly. The code
below doesn't compile, g++ says "error: template argument required for
'struct Derived2'". But if I do it like in (**) it says "error:
'Derived2' is not a template". I tried almost all word permutations of
friend template class Derived2<T>
for line (*) but I had no success. So I ended up making both classes
public which is bad. What is the correct way of doing this?
template <class ID_TYPE> class Base
{
ID_TYPE id;
};
template <class T> class Derived1: public Base<T>
{
friend class Derived2; // (*)
};
template <class T> class Derived2: public Base<T>
{
friend class Derived1<T>; (**)
};
int main()
{
Derived1<int> d1;
return 0;
}
It depends on what you want to make a friend of what. It is very easy
to make all Derived2 instantiations friends of any particular Derived1
instantiation and vice versa. The syntax is
template<class T> class Derived1: public Base<T>
{
...
template<class U> friend class Derived2;
};
but that would make Derived2<int> a friend of Derived1<double> and so
on, which is not necessarily what you want.
It is impossible (at this point anyway) to make a particular
instantiation of a class template a friend of another instantiation. It
would require to declare a specialisation of the friend class template,
and that's not allowed (don't ask me why, I don't know). I would hope
that the syntax
template<class T> class Derived1: public Base<T>
{
...
template<> friend class Derived2<T>;
};
or
template<class T> class Derived1: public Base<T>
{
...
template friend class Derived2<T>;
};
would work, but IIRC they don't.
V
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