Can one access the template parameter outside of a template without a typedef?

问题

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?

回答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;

回答2:

Yes, define a second "getter" template with partial specialization.

template< typename >
struct get_Small_X; // base template is incomplete, invalid

template< typename X > // only specializations exist
struct get_Small_X< Small< X > > {
    typedef X type;
};

Now instead of Small<X>::X you have typename get_Small_X< Small<X> >::type.

By the way, _X is a reserved identifier, so you shouldn't use it for anything. X_ is a better choice.

Advanced topic: template introspection.

While I'm thinking about it, you don't need to define this separately for every template. A single master template should do it.

This compiles in Comeau, I know there are rules about matching template template arguments but I think it's OK… template template arguments are forbidden from the master template in partial specialization.

template< typename >
struct get_first_type_argument;

template< template< typename > class T, typename X >
struct get_first_type_argument< T< X > > {
    typedef X type;
};

template< typename X >
struct simple;

get_first_type_argument< simple< int > >::type q = 5;

This only works with "unary" templates but could be adapted in C++0x for the general case.

来源：https://stackoverflow.com/questions/3696286/can-one-access-the-template-parameter-outside-of-a-template-without-a-typedef

标签

c++

templates

metaprogramming

introspection