c++ - 如何编写连接两个函数的函数？

Question

我正在尝试编写一个通用函数，该函数连接两个可以使用同一组参数调用的函数，但我遇到了一些麻烦。这是我目前所拥有的(无法编译)

//A functor to store the input functions and call them
template <typename LEFT, typename RIGHT>
struct combine_functions {
  combine_functions(const LEFT &left, const RIGHT &right)
   : left(left), right(right) {}

  template <typename ...ARGS>
  std::enable_if_t<
    //My compiler doesn't have support for C++17 std library so I 
    //found an implementation of callable on SO
    is_callable_v<LEFT, std::decay_t<ARGS>...> &&
    is_callable_v<RIGHT, std::decay_t<ARGS>...>
  > operator()(ARGS... args) const {
    //the return value doesn't matter in my situation and can be 
    //completely discarded
    left(std::forward<ARGS>(args)...);
    right(std::forward<ARGS>(args)...);
  }

private:
  mutable LEFT left;
  mutable RIGHT right;
};

//I should probably have an enable if that checks the arguments 
//are function pointers or functors
template <typename LEFT, typename RIGHT>
combine_functions<
  std::decay_t<LEFT>,
  std::decay_t<RIGHT>
>
operator+(
  const LEFT &left,
  const RIGHT &right
) {
  return {left, right};
}

如果不清楚我要实现的目标，那么这里是一个测试。

#include <iostream>
#include "combine functions.hpp"    

struct A {
  void operator()(float &f, int i) {
    std::cout << "running A with float " << f << " and int " << i << '\n';
    f++;
  }
};

struct B {
  void operator()(float &f, int i) {
    std::cout << "running B with float " << f << " and int " << i << '\n';
    f++;
  }
};

struct C {
  void operator()(float &f, int i) {
    std::cout << "running C with float " << f << " and int " << i << '\n';
    f++;
  }
};

int main(int, const char**) {
  A a;
  B b;
  C c;
  auto abc = concat(concat(a, b), c);
  //or
  //auto abc = a + b + c;
  std::function<void(float &, int)> abcFunc = abc;
  float f = 5.0f;
  int i = 9;
  abcFunc(f, i);

  return EXIT_SUCCESS;
}

这是预期的输出

running A with float 5 and int 9
running B with float 6 and int 9
running C with float 7 and int 9    

• 如何在 C++ 中实现它？
• 在这种情况下使用重载运算符是否不明智？
• “串联”是表示此操作的最佳术语吗？

Answer 1

我认为这是一个合理的起点。通过完美转发支持任意数量的连接和任意数量的参数:

#include <tuple>
#include <utility>
#include <iostream>

namespace detail 
{
    template<class Tuple, std::size_t...Is, class...Args>
    void exec(Tuple&& tuple, std::index_sequence<Is...>, Args&&...args)
    {
        using expand = int[];
        void(expand{
            0,
            (std::get<Is>(tuple)(std::forward<Args>(args)...),0)...
        });

    }
}

template<class...Funcs>
auto concat(Funcs&&...funcs)
{
    constexpr auto nof_funcs = sizeof...(funcs);
    return [funcs = std::make_tuple(std::forward<Funcs>(funcs)...)](auto&&...args) mutable
    {
        detail::exec(funcs, 
                     std::make_index_sequence<nof_funcs>(), 
                     std::forward<decltype(args)>(args)...);
    };
};

int main()
{
    auto f1 = [](auto&& arg) { std::cout << arg << std::endl; };
    auto f2 = [](auto&& arg) { std::cerr << arg << std::endl; };

    concat(f1, f2)("Hello, World");
}