I want to declare member function of template class as friend. But I got warning message such as warning: dependent nested name specifier 'Schedule<T>::' for friend class declaration is not supported; turning off access control for 'Manager' [-Wunsupported-friend]
My code is like below
template<typename T>
class Task{};
template<typename T>
class Schedule {
public:
void dispatch(Task<T>*) {}
};
class Manager {
template<typename T>
friend class Task;
template<typename T>
friend void Schedule<T>::dispatch(Task<T>*); // <== Warning here!!!!
template<typename T>
friend int ticket() {
return ++Manager::count;
}
static int count;
};
What causes this warning and how can I fix it?
This is a warning you get when using a feature that is not fully implemented in LLVM.
From the LLVM source code:
/// True if this 'friend' declaration is unsupported. Eventually we
/// will support every possible friend declaration, but for now we
/// silently ignore some and set this flag to authorize all access.
unsigned UnsupportedFriend : 1;
You can safely ignore the warning if you are not worried about the "all access" granted to all Schedule<T>s member functions. I'm assuming you got the warning when compiling with -Weverything, so just add -Wno-unsupported-friend afterwards - or be pragmatic and make the whole class a friend:
template<typename T>
friend class Schedule;
The effect of the authorization of all access to all Schedule<T>s member functions can be illustrated like this:
class Manager;
template <typename T>
class Schedule {
public:
void dispatch() {}
void other(Manager&); // not befriended
};
class Manager {
public:
template <typename T>
friend void Schedule<T>::dispatch(); // becomes `friend class Schedule<T>;`
private:
int x = 0;
};
template <typename T>
void Schedule<T>::other(Manager& m) {
++m.x; // should not compile, but does
}