Ive got a problem that if I have a template class, which in turn has a template method that takes a parameter of another instance of the class (with different template argum
They are different types: templates construct new types from a template.
You have to make other instantiations of your class friends:
template <typename T>class MyClass
{
T v;
public:
MyClass(T v):v(v){}
template<typename T2>void foo(MyClass<T2> obj)
{
std::cout << v << " ";
std::cout << obj.v << " ";
std::cout << v + obj.v << std::endl;
}
// Any other type of MyClass is a friend.
template <typename U>
friend class MyClass;
// You can also specialize the above:
friend class MyClass<int>; // only if this is a MyClass<int> will the
// other class let us access its privates
// (that is, when you try to access v in another
// object, only if you are a MyClass<int> will
// this friend apply)
};
Add MyClass as friend class:
template<typename T> class MyClass
{
template<typename TX>
friend class MyClass;
...
According to C++ Standard 14.5.3/3:
A friend template may be declared within a class or class template. A friend function template may be defined within a class or class template, but a friend class template may not be defined in a class or class template. In these cases, all specializations of the friend class or friend function template are friends of the class or class template granting friendship. [Example:
class A { template<class T> friend class B; // OK template<class T> friend void f(T){ /* ... */ } // OK };—end example]
NOTE: You should know that the code above still could lead to an error with some compilers due to Core Issue #602 which is still open. Despite this, the code above compiles on GCC, Visual C++ and Comeau.
To make only function foo a friend you could write the following:
template<typename T> class MyClass
{
template<typename TY> template<typename TX>
friend void MyClass<TY>::foo(MyClass<TX>);
...