具有递归继承和使用声明的可变参数模板

Question

I want to do something similar to the code below, except that I don't want to implement func() twice because it will be the same implementation. Do you have any proposal how to accomplish this?

template <typename First, typename... Rest>
class Base : public Base<Rest...> {
public:
    using Base<Rest...>::func;
    void func(First object) {
        // implementation
    }
};

template <typename First>
class Base<First> {
public:
    void func(First object) {
        // implementation
    }
};

struct X {};
struct Y {};
struct Z {};

class Derived : public Base<X, Y, Z> {};

// ...

Derived d;
d.func(X{});
d.func(Y{});
d.func(Z{});

Answer 1

With multiple inheritance and an additional using directive:

template <typename First, typename... Rest>
struct Base : Base<First>,  Base<Rest...> {
    using Base<First>::func;
    using Base<Rest...>::func;
};

template <typename First>
struct Base<First> {
    void func(First object) {/*...*/}
};

Live Demo (C++11)

Answer 2

Could be this a solution? (Move definition of func to a common base?)

template <class First>
class Func_impl
{
public:
     void func(First obj) {
     // implementation
     }
}

template <typename First, typename... Rest>
class Base : public Base<Rest...>, Func_impl<First> {
public:
    // no need to declare func here
};


template <typename First>
class Base<First> : public Func_impl<First> {
public:
   // no need to declare func here
};

Answer 3

I propose to (1) declare Base as receiving zero or more types

template <typename...>
class Base;

(2) transform the recursive version in a partial specialization that receive one or more types

template <typename First, typename... Rest>
class Base<First, Rest...> : public Base<Rest...> {
public:
    using Base<Rest...>::func;
    void func(First object) {
        // implementation
    }
};

(3) delete the only-one-parameter partial specialization and (4) add a ground case zero parameter specialization

template <>
class Base<>
 { public: void func () {} };

Observe the dummy (without implementation) func() in the ground specialization: it's required (something with the name func is required) by using Base<Rest...>::func; .

The following is a full working example.

#include <iostream>

template <typename...>
class Base;

template <typename First, typename... Rest>
class Base<First, Rest...> : public Base<Rest...> {
public:
    using Base<Rest...>::func;
    void func(First object)
     { std::cout << "func() recursive" << std::endl; }
};

template <>
class Base<>
 { public: void func () {} };

struct X {};
struct Y {};
struct Z {};

class Derived : public Base<X, Y, Z> {};

int main()
 {
   Derived d;
   d.func(X{});
   d.func(Y{});
   d.func(Z{});
 }

Answer 4

In C++17 there is a simple non-recursive solution based on "variadic" multiple inheritance and "variadic using" (pack expansion in using-declaration). Although this is not what you are asking for, I consider it worth an answer. As a bonus, the implementations for each type know their index in the original sequence.

http://coliru.stacked-crooked.com/a/f67d00da68226d5f

#include <iostream>
#include <utility>

template<std::size_t i, class T>
struct IndexedNode {
  void func(T) {
    std::cout << __PRETTY_FUNCTION__ << std::endl;
  }
};

template<class Indices, class... Ts>
struct IndexedBase;

template<std::size_t... is, class... Ts>
struct IndexedBase<
  std::index_sequence<is...>,
  Ts...
>
  : IndexedNode<is, Ts>...
{
  using IndexedNode<is, Ts>::func...;
};

template<class... Ts>
struct Base
  : IndexedBase<std::index_sequence_for<Ts...>, Ts...>
{
  using Is = std::index_sequence_for<Ts...>;
  using IndexedBase<Is, Ts...>::func;
};