C++ 将嵌套的基模板类实例声明为派生类的友元

假设我有：

class A {};

template <typename T> class B {};

template <typename T> class C {};

class D : public C<B<A>> {
    //friend ... ??
};

---

恐怕每个班你都得单独交朋友。你不能继承友谊，所以我认为用一个朋友声明不可能做到这一点。

看起来你在写混音。如果使用多重继承，可能会更干净。但这并不能消除多个好友声明。

#include <iostream>

template <typename T>
struct static_base {
  T& self() { return static_cast<T&>(*this); }
  T const& self() const { return static_cast<T const&>(*this); }
};

template <typename Derived>
class InterfaceBase : public static_base<Derived> {};

template <typename Derived>
class Interface1 : public Derived {
 public:
  void foo() { this->self().foo_(); }
};

template <typename Derived>
class Interface2 : public Derived {
 public:
  void bar() { this->self().bar_(); }
};

template <typename Derived>
class Interface3 : public Derived {
 public:
  void baz() { this->self().baz_(); }
};

class Impl : public Interface3<Interface2<Interface1<InterfaceBase<Impl>>>> {
    friend Interface3<Interface2<Interface1<InterfaceBase<Impl>>>>;
    friend Interface2<Interface1<InterfaceBase<Impl>>>;
    friend Interface1<InterfaceBase<Impl>>;

    protected:
        void foo_() { std::cout << "foo" << "\n"; }
        void bar_() { std::cout << "bar" << "\n"; }
        void baz_() { std::cout << "baz" << "\n"; }
};

int main() {
    auto impl = Impl();
    impl.foo();
    impl.bar();
    impl.baz();
}