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I'm trying to call the class function A<F>::f() from within class S , but I'm getting the following errors when I instantiate an S object ( S<int>s ) and call it's f member function ( s.f() ) : source.cpp: In instantiation of ' int S<F>::f() [with F = int] ': source.cpp:30:21: required from here source.cpp:22:25: error: ' A<int> ' is not an accessible base of ' S<int> ' Note that this works when I replace return A<F>::f(); inside the declaration of class S with return C<A, F>::f(); . But I'm wondering why I can't do it the other way... #include <iostream> template <typename T> class A { public: int f(); }; template <typename T> int A<T>::f() { return sizeof(T); } template <template <typename> class E, typename D> class C : E<D> { public: int f() { return E<D>::f(); } }; template <typename F> class S : C<A, F> { public: int f() { return A<F>::f(); } }; int main() { S<int>s; std::cout << s.f(); } Any help is appreciated and if you require further clarification please feel free to comment. Update Since this questions is resolved I guess I should post the code that actually worked: #include <iostream> template <typename T> class A { public: int f(); }; template <typename T> int A<T>::f() { return sizeof(T); } template <template <typename> class E, typename D> class C : public E<D> { public: int f() { return E<D>::f(); } }; class S : public C<A, int> {}; int main() { S s; std::cout << s.f

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