How to name a nested template in a templated base class?

base-class, c++, name-lookup, templates

Solution

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.

Problem

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 "`int`s", 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.

Original source