Inherit dependent typedef without using struct

Question

I have some code like this:

#include <string>
#include <map>

typedef std::less<std::string> Comparator; // simplified

typedef std::allocator<std::pair<const std::string, int>> Allocator; // simplified

template<class T>
struct Base
{
    typedef std::map<std::string, T, Comparator, Allocator> type; // VERY long declaration in the actual code
};

template<class T>
struct Container : public Base<T>::type
{
    Container(Allocator a) : Base<T>::type(Comparator(), a) {}
};

int main()
{
    Allocator a;
    Container<int> c(a);
}

Although the declarations are a bit more fancy in my actual code.

The Base struct is used so that I do not have to write the long map declaration multiple times.

I was wondering if there is a better way to inherit from the map without any Base struct?

No macros please. I hope in some way to hide the typedef in the Container class itself or something like that.

Thanks,

Answer 1

You can rely on templates having an injected class name. Inside the specialization of map<...> , the current specialization can be referred to simply by map . And that injected class name is also available to derived classes (and class templates). But since it's dependent, it requires a qualified name. It looks simpler than it sounds, here's how you roll the alias into Conatiner :

template<class T>
struct Container : public std::map<std::string, T, Comparator, Allocator>
{
    using Base = typename Container::map;
    Container(Allocator a) : Base(Comparator(), a) {}
};

Container::map is the injected class name. And the alias grabs it for convenient use.