Using dependent name in base class name without "typename"

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I have a template class for which the base class is also a template parameterized with a type member of one of the outer template type parameters. Example:

template <typename X>
class Adapter : public Generator<typename X::generated_type>{
    using G = typename X::generated_type;
};

Here X is some type that has a member generated_type and we want to subclass a Generator for X::generated_type. Because X::generated_type is a dependent name, we have to prefix it with typename when we use it in the base class name.

In the rest of the definition of Adapter, I can use G instead of typename X::generated_type, which is obviously a lot more convenient.

Is there any way to use G in the name of the base class too?

In the actual code it's not this simple; I've tried to distill the example down to the core issue.

The only thing I could think of was to have G as a template parameter and create a wrapper that passes X::generated_type as G.

typename c++
1 Answers

One possible solution is to cheat, something along these lines:

template <typename X, typename G=typename X::generated_type>
class Adapter : public Generator<G>{
   // ...
};

You were almost there, heading into an additional template parameter territory, but you don't need to explicitly pass it in, just default it.

Complete example:

template<typename> class Generator {};

template <typename X, typename G=typename X::generated_type>
    class Adapter : public Generator<G>{
       // ...
    };

struct Y {
    typedef int generated_type;
};

int main()
{
    Adapter<Y> a;

    return 0;
}
user_39433120
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