I have the following code, which doesn't compile:
class A {
public:
typedef int A_t;
};
template<class T>
class IB {
public:
typedef typename T::A_t B_t;
virtual B_t get() = 0;
protected:
B_t v;
IB(B_t in) : v(in) { }
};
template<class T>
class B : public IB<T> {
public:
B(B_t in) : IB<T>(in) { } // error: expected ')' before 'in'
B_t get() { return B_t(); } // error: B_t does not name a type
};
int main() {
B<A> a(3); // Errors here since int can't be converted to const B& or const B&&
// Also because A is virtual, since A::get() doesn't compile
return 0;
}
As far as I can understand the error messages, the main problem is that the typedef in IB for some reason isn't available in B.
If I add a line typedef IB<T>::B_t B_t; above the B constructor, the code compiles OK, but I don't understand why I need this. Shouldn't B_t be available through inheritance?
Also, the following doesn't work either:
B(IB<T>::B_t in) : IB<T>(in) { }
// etc, with qualified references to B_t
yields the same error messages.
What is the mechanism here that prevents me from using B_t it in B without the extra typedef?
