Suppose I define a template T that uses a nested class of the template parameter P, as follows:
template<class P> class T
{
public:
    T(P& p) : p(p) {}
    P& p;
    typename P::Nested& get_nested()    { return p.nested; }
};
If I declare a class A that includes a nested class named Nested, I can define a variable of type T<A> with no problem:
class A
{
public:
    class Nested
    {
    public:
        int i;
    };
    Nested nested;
};
void test2a()
{
    A a;
    a.nested.i = 1;
    T<A> t_a(a);
    t_a.get_nested().i = 2;
}
Now, I want to declare a class B that, in the same way, includes a nested class named Nested and that inherits from T<B>, as follows:
class B : public T<B>
{
public:
    class Nested
    {
    public:
        int i;
    };
    Nested nested;
};
Compilation of the above code fails with error: "Nested is not a member of B"
I think I understand what's happening: at the time the template is entered, class B is incompletely defined because of inheritance.
However, I am wondering if there is any way to do such a thing...
Thanks for help.
You need to defer the resolution of the return type of get_nested until it is called.
One way to do this is to make the return type dependent on a template parameter:
  template<typename unused = void>
    typename std::conditional<false, unused, P>::type::Nested&
    get_nested()    { return p.nested; }
Another way (since C++14) is to use return type deduction:
  auto& get_nested()    { return p.nested; }
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