I'm trying to get access to members from another templated class and it doesn't let me. The issue is I don't know why.

#include <iostream>
    
template <typename T0, typename T1, typename T2, typename T3>
struct Obj{};
    
template <typename T0, typename T1, typename T2, typename T3>
struct Obj2
{
    template <typename U0, typename U1, typename U2, typename U3> 
    friend Obj<U0, U1, U2, U3>;
    
    template <typename T> 
    Obj2(Obj<T, T0, T1, T2> const&) {}
};
    
int main() {}

Output:

<source>:10:31: error: expected unqualified-id before ';' token [-Wtemplate-body]
   10 |     friend Obj<U0, U1, U2, U3>;
      |                               ^
Compiler returned: 1

Godbolt link: code

Is this even doable?

I'm trying to get access to members from another templated class and it doesn't let me. The issue is I don't know why.

#include <iostream>
    
template <typename T0, typename T1, typename T2, typename T3>
struct Obj{};
    
template <typename T0, typename T1, typename T2, typename T3>
struct Obj2
{
    template <typename U0, typename U1, typename U2, typename U3> 
    friend Obj<U0, U1, U2, U3>;
    
    template <typename T> 
    Obj2(Obj<T, T0, T1, T2> const&) {}
};
    
int main() {}

Output:

<source>:10:31: error: expected unqualified-id before ';' token [-Wtemplate-body]
   10 |     friend Obj<U0, U1, U2, U3>;
      |                               ^
Compiler returned: 1

Godbolt link: code

Is this even doable?

• c++
• templates
• friend

• @wohlstad yes, any template types. – Saitama10000 Commented Nov 27, 2024 at 14:14
• You mean like in cppreference - Template friends examples? – pptaszni Commented Nov 27, 2024 at 14:16
• @Saitama10000 I posted an answer with both options. – wohlstad Commented Nov 27, 2024 at 14:17
• @wohlstad can you add the third option of only some parameters? thank you :) – Saitama10000 Commented Nov 27, 2024 at 14:19

1 Answer 1

I assume you want Obj with any template types to be a friend.
This means that any instantiation of Obj will be a friend of Obj2.

In this case the proper syntax is:

template <typename U0, typename U1, typename U2, typename U3> 
friend struct Obj;

Live demo 1

If on the other hand you wish only Obj with template types that match those of Obj2 to be a friend, the syntax for that is:

friend struct Obj<T0,T1,T2,T3>;

Live demo 2