'What are dependent names in C++?

When is a C++ name "dependent"? Apparently it is not dependent when it uses a type-definition in the same scope. It is however dependent, when it uses a type definition in a nested scope of this scope. Consider the following examples:

Example A:

template <typename T>
struct some_struct
{
    struct B
    {
        int b_;
    };
    T some_member_func();
    T a_;
};

template <typename T>
T some_struct<T>::some_member_func() 
{
    using hello = B;
    return a_;
}

.. compiles fine.

Example B (Nested):

template <typename T>
struct some_struct
{
    struct B
    {
        struct C {
            int c_;
        };
    };
    T some_member_func();
    T a_;
};

template <typename T>
T some_struct<T>::some_member_func() 
{
    using hello = B::C;
    return a_;
}

..issues the following error:

<source>: In member function 'T some_struct<T>::some_member_func()':
<source>:365:19: error: need 'typename' before 'some_struct<T>::B::C' because 'some_struct<T>::B' is a dependent scope
  365 |     using hello = B::C;
      |                   ^
      |                   typename

Why is some_struct::B a dependent scope but not some_struct (example A)?

c++c++11typesscope


Solution 1:[1]

If I'm reading https://en.cppreference.com/w/cpp/language/dependent_name correctly, both B and B::C are dependent.

But B also "refers to the current instantiation", which allows it to be used without typename. It seems to be a fancy way of saying that "whether it's a type can't be affected by specializations".

You can specialize B to make B::C not a type:

template <>
struct some_struct<int>::B
{
    static void C() {}
};

(interestingly, a partial specialization didn't compile)

But you can't prevent B from being a type via specializations.

Sources

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Source: Stack Overflow

Solution Source
Solution 1 HolyBlackCat

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