Question

A simple example:

template<typename _X> // this template parameter should be usable outside!
struct Small {
   typedef _X X; // this is tedious!
   X foo;
};

template<typename SomeSmall>
struct Big {
   typedef typename SomeSmall::X X; // want to use X here!
   SomeSmall bar;
   X toe;
};

Is there a way to access the template parameter X of Small without using a typedef in the Small class?

Answer 1

Depending on what you’re doing, template template parameters might be a better option:

// "typename X" is a template type parameter. It accepts a type.
// "template <typename> class SomeSmall" is a template template parameter.
// It accepts a template that accepts a single type parameter.
template<typename X, template <typename> class SomeSmall> 
struct Big {
   SomeSmall<X> bar; // We pass X to the SomeSmall template.
   X toe; // X is available to this template.
}; 

// Usage example:

template<typename X>
struct Small { 
   X foo; 
};

struct MyType {};

// The foo member in Small will be of type MyType.
Big<MyType, Small> big;