Question

In the following setup, how can I make it so that I can refer to the name Bar inside the derived class Derived<T>?

template <typename T> struct Foo
{
    template <typename U> struct Bar { };
};

template <typename T> struct Derived : Foo<T>
{
    // what goes here?

    Bar<int> x;  // Error: 'Bar' does not name a type
};

I’ve tried using Foo<T>::Bar;, but that doesn’t help. Is there any sort of using declaration that can make the name of a nested base template known to the derived class, so that I can keep the simple de­cla­ration Bar<int> x?

I know that I can say typename Foo<T>::template Bar<int> x;, but I have a lot of those cases and I don’t want to burden the code needlessly with so much verbosity. I also have a lot of distinct “ints”, so a typedef for each nested-template instance is not feasible, either.

Also, I cannot use GCC 4.7 at this point nor C++11, and would thus like a “traditional” solution without template aliases.

Answer 1

In C++11 you can use an alias template:

template <typename T> struct Derived : Foo<T>
{
  template<typename X> using Bar = typename Foo<T>::template Bar<X>;
  Bar<int> x;
};

Edit

The traditional solution is what you’re already said, typename Foo<T>:template Bar<int>, or to emulate a “template typedef”

template <typename T> struct Derived : Foo<T>
{
  template<typename X>
    struct Bar
    { typedef typename Foo<T>::template Bar<X> type; };
  typename Bar<int>::type x;
};

One of the reasons for adding alias templates to the language is they support things which can’t be expressed easily in C++03.