Create a std::function type with limited arguments

问题

Given the type of a callable function C, I want to get at compile time a std::function; the type of which:

• has the same return type of function C
• the argument types are the first N argument types of function C

This means that, for a given type void(int, char, double) and a given N, the type of the function is:

• N = 1 => result type: std::function<void(int)>
• N = 2 => result type: std::function<void(int, char)>
• N = 3 => result type: std::function<void(int, char, double)>
• N > 3 => compile time error

Example:

template<std::size_t N, typename R, typename... A>
constexpr auto get() {
    return /*(magically somehow)*/ std::function<R(FirstNFromA...)>
}

template<std::size_t N, typename R, typename... A>
struct S {
    using func = decltype(get<N, R, A...>());
};

回答1:

It follows a possible solution:

#include <tuple>
#include <utility>
#include <functional>
#include <type_traits>

template<
    typename R,
    typename... A,
    std::size_t... I,
    std::enable_if_t<(sizeof...(I)<=sizeof...(A))>* = nullptr
> constexpr auto
get(std::integer_sequence<std::size_t, I...>) {
    return std::function<R(std::tuple_element_t<I, std::tuple<A...>>...)>{};
}

template<std::size_t, typename>
struct FuncType;

template<std::size_t N, typename R, typename... A>
struct FuncType<N, R(A...)> {
    using Type = decltype(get<R, A...>(std::make_index_sequence<N>{}));
};

int main() {
    static_assert(
        std::is_same<
            FuncType<2, void(int, char, double, int)>::Type,
            std::function<void(int, char)>
        >::value,
        "!"
    );
}

The basic idea is to use a tuple and those utilities that are part of the Standard template Library (as an example, std::tuple_element), mix them with a pack expansion placed in the right place and that's all.
To do that, I used a constexpr function that returns an empty std::function object of the given type. Then the type of the function is taken by means of a decltype and exported as a type of the FuncType object using an alias.
Everything takes place at compile time.
In order to deduce the right type for that function, I used a well known pattern that involves a std::integer_sequence and actually unpack the types of a tuple by expanding the indexes.

回答2:

Another solution could be:

#include <tuple>
#include <utility>
#include <functional>
#include <type_traits>

template <size_t N, class R, class Pack, class ResultPack, class Voider>
struct FuncTypeImpl;

template <size_t N, class R, template <class...> class Pack, class First, class... Args, class... ResultArgs>
struct FuncTypeImpl<N, R, Pack<First, Args...>, Pack<ResultArgs...>,  std::enable_if_t<(N > 0)>>: FuncTypeImpl<N-1, R, Pack<Args...>, Pack<ResultArgs..., First>, void> {
   using typename FuncTypeImpl<N-1, R, Pack<Args...>, Pack<ResultArgs..., First>, void>::Type;
};

template <size_t N, class R, template <class...> class Pack, class... Args, class... ResultArgs>
struct FuncTypeImpl<N, R, Pack<Args...>, Pack<ResultArgs...>, std::enable_if_t<(N == 0)>> {
   using Type = std::function<R(ResultArgs...)>;
};

template<std::size_t, typename>
struct FuncType;

template<std::size_t N, typename R, typename... A>
struct FuncType<N, R(A...)> {
    using Type = typename FuncTypeImpl<N, R, std::tuple<A...>, std::tuple<>, void>::Type;
};

int main() {
    static_assert(
        std::is_same<
            FuncType<3, void(int, char, double, int)>::Type,
            std::function<void(int, char, double)>
        >::value,
        "!"
    );
}

Edit: Yet another maybe a little bit simpler (that does not need a std::tuple) solution:

#include <utility>
#include <functional>
#include <type_traits>

template <class T>
struct ResultOf;

template <class R, class... Args>
struct ResultOf<R(Args...)> {
   using Type = R;
};

template<std::size_t N, class Foo, class ResultFoo = typename ResultOf<Foo>::Type() , class Voider = void>
struct FuncType;

template<std::size_t N, class R, class First, class... Args, class... ResultArgs >
struct FuncType<N, R(First, Args...), R(ResultArgs...), std::enable_if_t<(N > 0)>>: FuncType<N-1, R(Args...), R(ResultArgs..., First), void> {
};

template<std::size_t N, class R, class First, class... Args, class... ResultArgs >
struct FuncType<N, R(First, Args...), R(ResultArgs...), std::enable_if_t<(N == 0)>> {
   using Type = std::function<R(ResultArgs...)>;
};

int main() {
    static_assert(
        std::is_same<
            FuncType<3, void(int, char, double*, int)>::Type,
            std::function<void(int, char, double*)>
        >::value,
        "!"
    );
}

回答3:

Another tuple based solution.

Should work with C++11 too.

I suppose it can be simplified but I don't know how to avoid the use of the bool template parameter (I'm just learning C++11).

#include <tuple>
#include <utility>
#include <functional>
#include <type_traits>


template<std::size_t N, bool Z, typename R, typename...>
struct FTH1;

template <typename R, typename... A, typename... B>
struct FTH1<0U, true, R, std::tuple<A...>, B...>
 { using type = decltype(std::function<R(A...)>{}); };

template <std::size_t N, typename R, typename... A, typename B0, typename... B>
struct FTH1<N, false, R, std::tuple<A...>, B0, B...>
 { using type = typename FTH1<N-1U, (N-1U == 0U), R, std::tuple<A..., B0>, B...>::type; };


template <std::size_t N, typename>
struct FuncType;

template<std::size_t N, typename R, typename... A>
struct FuncType<N, R(A...)>
 { using Type = typename FTH1<N, (N == 0), R, std::tuple<>, A...>::type; };



int main() {
    static_assert(
        std::is_same<
            FuncType<2, void(int, char, double, int)>::Type,
            std::function<void(int, char)>
        >::value,
        "!"
    );
}

p.s.: sorry for my bad English.

回答4:

As I mentioned in the comments, if such problem arises then my first approach would be to change the problem itself! Sometimes, instead of finding a "tough solution" for a "tough problem", it's better to make the problem itself "simpler"!
There should never be a need, where one has to write FuncType<2, R(X,Y,Z)>::type instead of simple std::function<R(X,Y)>.

Above is my real answer. To solve your problem as a coding delight, I am putting a simple macro based answer. Instead of compile time, it will give you the required type at preprocessing time.

#define F(R, ...) std::function<R(__VA_ARGS__)>  // shorthand for std::function...
#define FuncType(N, R, ...) FUNC_##N(R, __VA_ARGS__)  // Main type
#define FUNC_0(R, ...) F(R)  // overload for 0 arg
#define FUNC_1(R, _1, ...) F(R, _1)  // overload for 1 arg
#define FUNC_2(R, _1, _2, ...) F(R, _1, _2)  // overload for 2 args
#define FUNC_3(R, _1, _2, _3, ...) F(R, _1, _2, _3)  // overload for 3 args

Usage:

int main() {
  static_assert(std::is_same<
    FuncType(2, void, int, char, double, int),  // <--- see usage
    std::function<void(int, char)>
    >::value, "!");  
}

As you can see there is a little change in usage. Instead of FuncType<2, void(int, char, double, int)>::type, I am using FuncType(2, void, int, char, double, int). Here is the demo.

---

As of now, the FUNC_N macros are overloaded upto 3 arguments. For more arguments, if we want to avoid copy pastes, then we can generate a header file with a simple program:

  std::ofstream funcN("FUNC_N.h"); // TODO: error check for argv & full path for file
  for(size_t i = 0, N = stoi(argv[1]); i < N; ++i)
  {
    funcN << "#define FUNC_" << i << "(R";  // FUNC_N
    for(size_t j = 1; j <= i; ++j)
      funcN << ", _" << j;  // picking up required args
    funcN << ", ...) ";  // remaining args

    funcN << "F(R";  // std::function
    for(size_t j = 1; j <= i; ++j)
      funcN << ", _" << j;  // passing only relevant args
    funcN <<")\n";
  }

And simply #include it:

#define F(R, ...) std::function<R(__VA_ARGS__)>
#define FuncType(N, R, ...) FUNC_##N(R, __VA_ARGS__)
#include"FUNC_N.h"

Here is the demo for how to generate the header file.

来源：https://stackoverflow.com/questions/37541787/create-a-stdfunction-type-with-limited-arguments