正确使用C ++模板模板参数

Question

I've some trouble to make use of template template parameters. Here is a very simplified example:

template <typename T> 
struct Foo {
  T t;
};

template <template <class X> class T>
struct Bar {
  T<X> data;
  X x;
};

int main()
{
  Bar<Foo<int>> a;
}

The compiler (g++ (Ubuntu 4.8.2-19ubuntu1) 4.8.2) reports the following error:

main.cpp:8:5: error: ‘X’ was not declared in this scope
   T<X> data;
     ^

main.cpp:8:6: error: template argument 1 is invalid
   T<X> data;
      ^

Any idea what's wrong?

Answer 1

So I would like so make use of something like Bar<Foo<>>

template <typename T = int> 
struct Foo {
  T t;
};

template <typename T>
struct Baz {
  T t;
};

template <typename T>
struct Bar;

template <template <typename> class T, typename X>
struct Bar<T<X>> {
  T<X> data;
  X x;
};

int main()
{
  Bar<Foo<>> a;
  Bar<Baz<float>> b;
}

Answer 2

template <typename T> 
struct Foo {
  T t;
};

template <template <class> class T, class X>
struct Bar {
  T<X> data;
  X x;
};

int main()
{
  Bar<Foo, int> a;
}

Answer 3

In

template <template <class X> class T>

The template type parameter X is not a template parameter to the outermost template: it is a template parameter to the innermost template. It's rather similar to

int foo(int (*bar)(int x))
{
    int y = x;  // compiler error
}

which doesn't work since the function takes a single argument, bar : there is no argument x .

Depending upon what you are truly trying to do, you could add the second template parameter, with something like

template <typename X, template <typename> class T >
struct Bar
{
    // ...
};

you can keep the declaration with a single type parameter, but pattern match to give a partial specialization that would define the class in the example context

template <typename T>
struct Bar;

template <typename X, template <typename> class T >
struct Bar<T<X>>
{
    // ...
};

you could modify Foo to have a useful nested type, and grab it that way

template <typename T>
struct Bar
{
    using X = T::value_type;
};

or you could define a metafunction that extracts a template parameter from a template type, and get it that way:

template <typename T>
struct Bar
{
    using X = get_parameter<T>;
};

The most flexible is the last version, except rather than a metafunction that extracts template arguments, you would declare a get_bar_parameter function, and define a partial specialization that extracts the template parameter from a Foo<X> (or a T<X> ). That way, if you ever decide in the future to use Bar with classes where the right value of X isn't computed that way, you can do so by giving an appropriate specialization for get_bar_parameter .

Answer 4

// method 1
template <typename T> 
struct Foo {
  typedef T Type;
  T t;
};

template <typename T>
struct Bar {
  T data;
  typename T::Type x;
};

// method 2
template <typename T>
struct Hack
{
  T t;
};

template <typename T>
struct TypeOf
{
  typedef struct UnknownType Type;
};

template<>
struct TypeOf< Hack<int> >
{
  typedef int Type;
};

template <typename T>
struct Baz {
  T data;
  typename TypeOf<T>::Type X;
};

int main()
{
  Bar< Foo<int> > a;
  Baz< Hack<int> > b;
  return 0;
}

In method 1 information is provided with nested type. This requires changing original class.

In method 2 this information is provided with specialization of another template.