Question

我有一个类A的对象，可以用不同的类型调用，并在每次调用时返回self。出于此问题的目的，A将执行

struct A {
    A call(const int&) {
    }
    A call(const string& s) {
    }
    ////
} a;

所以我有一个未知类型的元组：

std::tuple<Types...> t;

我想用每个元组元素调用a，所以我希望得到类似的内容：

b = a;
b = b.call(get<0>(t));
b = b.call(get<1>(t));
b = b.call(get<2>(t));
//...

或

b = a.call(get<0>(t)).call(get<1>(t)).call(get<2>(t)...)

订单并不重要（我的意思是如果呼叫顺序被颠倒了甚至改组它就行了）。

我确实理解它可能与递归有关，但它很难看。是否有可能在没有递归的情况下实现？

Answer 1

您可以使用std::index_sequence<Is...>，例如：

namespace detail
{

    template <std::size_t...Is, typename T>
    void a_call(A& a, std::index_sequence<Is...>, const T& t)
    {
        int dummy[] = {0, ((a = a.call(std::get<Is>(t))), void(), 0)...};
        static_cast<void>(dummy); // Avoid warning for unused variable.
    }

}

template <typename ... Ts>
void a_call(A& a, const std::tuple<Ts...>& t)
{
    detail::a_call(a, std::index_sequence_for<Ts...>{}, t);
}

在C ++ 17中，折叠表达式允许：

    template <std::size_t...Is, typename T>
    void a_call(A& a, std::index_sequence<Is...>, const T& t)
    {
        (static_cast<void>(a = a.call(std::get<Is>(t))), ...);
    }

甚至是std::apply：

template <typename ... Ts>
void a_call(A& a, const std::tuple<Ts...>& t)
{
    std::apply([&](const auto&... args){ (static_cast<void>(a = a.call(args)), ...); }, t);
}

Answer 2

您可以使用boost.fusion来迭代元组

http://theboostcpplibraries.com/boost.fusion

Answer 3

在＆＃34; get part of std::tuple＆＃34;：

的帮助下

#include <iostream>
#include <string>
#include <algorithm>
#include <tuple>

//SEE https://stackoverflow.com/questions/8569567/get-part-of-stdtuple 
//vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

template <size_t... n>
struct ct_integers_list {
    template <size_t m>
    struct push_back
    {
        typedef ct_integers_list<n..., m> type;
    };
};

template <size_t max>
struct ct_iota_1
{
    typedef typename ct_iota_1<max - 1>::type::template push_back<max>::type type;
};

template <>
struct ct_iota_1<0>
{
    typedef ct_integers_list<> type;
};

template <size_t... indices, typename Tuple>
auto tuple_subset(const Tuple& tpl, ct_integers_list<indices...>)
-> decltype(std::make_tuple(std::get<indices>(tpl)...))
{
    return std::make_tuple(std::get<indices>(tpl)...);
    // this means:
    //   make_tuple(get<indices[0]>(tpl), get<indices[1]>(tpl), ...)
}

template <typename Head, typename... Tail>
std::tuple<Tail...> tuple_tail(const std::tuple<Head, Tail...>& tpl)
{
    return tuple_subset(tpl, typename ct_iota_1<sizeof...(Tail)>::type());
    // this means:
    //   tuple_subset<1, 2, 3, ..., sizeof...(Tail)-1>(tpl, ..)
}

//^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
//SEE https://stackoverflow.com/questions/8569567/get-part-of-stdtuple 

struct A
{

    A call(const int&) {
        std::cout << "int" << std::endl;
        return *this;
    }

    A call(const std::string& s) {
        std::cout << "string" << std::endl;
        return *this;
    }

    template <typename T>
    A call(std::tuple<T> tpl)
    {
        return call(std::get<0>(tpl));
    }

    template <typename T, typename...Types>
    A call(std::tuple<T, Types...> tpl)
    {       
        return call(std::get<0>(tpl)).call(tuple_tail(tpl));
    }
} a;

int main()
{
    std::tuple<int, std::string, int, int, std::string> t(0, "1", 2, 3, "4");   
    A b = a.call(t);
}

输出：

int
string
int
int
string

Answer 4

使用std :: integer_sequence和std :: make_integer_sequence，如下所示：

struct A {
    A call(const int&) {
        std::cout << "Calling A::call(int)" << std::endl;
        return A{};
    }
    A call(const std::string& s) {
        std::cout << "Calling A::call(std::string)" << std::endl;
        return A{};
    }
    ////
} a;

template<typename Sequence>
struct call_helper;

template<std::size_t Current, std::size_t... Rest>
struct call_helper<std::integer_sequence<std::size_t, Current, Rest...>>
{
    template<typename Tuple, typename Result>
    static auto call(Tuple const& tup, Result&& res)
    {
        // Call the next helper with the rest sequence
        return call_helper<
            std::integer_sequence<std::size_t, Rest...>
        >::call(tup, res.call(std::get<Current>(tup)));
    }
};

template<>
struct call_helper<std::integer_sequence<std::size_t>>
{
    // End reached, just return the value
    template<typename Tuple, typename Result>
    static auto call(Tuple const&, Result&& res)
        -> std::decay_t<Result>
    {
        return std::forward<Result>(res);
    }
};

template<typename Tuple, typename Result>
auto call_all(Tuple const& tup, Result res)
{
    return call_helper<std::make_integer_sequence<
        std::size_t, std::tuple_size<Tuple>::value>
    >::call(tup, std::forward<Result>(res));
}

// Test call:
std::tuple<int, std::string, int, std::string, std::string> mytup;
call_all(mytup, a);

<强> Demo

在没有递归的情况下为一个对象上的每个元组元素调用函数

4 个答案: