Finding names in templatized base classes in C++

Question

I'm reading Effective C++ 3rd Edition, item 43 "Know how to access names in templatized base classes".

template<typename T>
class B {
    T i;
};

template<typename T>
class D: public B<T> {
public:
    void Foo() {
        T a = B<T>::i;
    }
};

int main() {
    D<int> d;
}

For the above codes, I know if the B<T>:: is not added before i in D::Foo(), compilers will complain "i was not declared in this scope". (But it didn't complain i is private in B.)

However, if T i; is not declared in B, like the following, the compiling goes well.

template<typename T>
class B {
};

template<typename T>
class D: public B<T> {
public:
    void Foo() {
        T a = B<T>::i;
    }
};

int main() {
    D<int> d;
}

Compilers are defaulted not to find names in templatized base classes. But why they still don't do even I told them?

songyuanyao · Accepted Answer · 2017-08-19T13:19:16.837

But why they still don't do even I told them?

Because the member function Foo is not used, then it's not instantiated at all.

This applies to the members of the class template: unless the member is used in the program, it is not instantiated, and does not require a definition.

You might get an error if Foo is called, like

D<int> d;
d.Foo();

BTW

But it didn't complain i is private in B.

Because accessibility check is performed after name lookup. The name i is not found, then accessibility of nothing could be checked.

Finding names in templatized base classes in C++

1 Answers1