일치하는 함수 호출 오류가 없습니다. C++

Question

벡터 튜플을 튜플 벡터로 바꾸려고합니다 (그 반대의 경우도 마찬가지입니다). tuple_transpose 기능을 호출하는 데 문제가 있습니다. - 내가 할 수있는 잘못된 일을 어떻게하고 무엇 http://ideone.com/7AWiQQ#view_edit_box

_{prog.cpp: In function ‘int main()’:
prog.cpp:44:24: error: no matching function for call to ‘tuple_transpose(std::tuple >, std::vector > >&)’
prog.cpp:44:24: note: candidates are:
prog.cpp:30:6: note: template typename transpose::type tuple_transpose(std::tuple >...>&, seq)
prog.cpp:30:6: note: template argument deduction/substitution failed:
prog.cpp:44:24: note: candidate expects 2 arguments, 1 provided
prog.cpp:36:6: note: template typename transpose::type tuple_transpose(std::tuple >...>&)
prog.cpp:36:6: note: template argument deduction/substitution failed:
prog.cpp: In substitution of ‘template typename transpose::type tuple_transpose(std::tuple >...>&) [with T = {int, bool}]’:
prog.cpp:44:24: required from here
prog.cpp:36:6: error: no type named ‘type’ in ‘struct transpose >, std::vector > >&>’}

여기

#include <vector> 
#include <tuple> 
#include <type_traits> 

template <typename... T> 
struct transpose {}; 

template <typename... T> 
struct transpose<std::tuple<std::vector<T>...>> 
{ 
    using type = std::vector<std::tuple<T...>>; 
}; 

template <typename... T> 
struct transpose<std::vector<std::tuple<T...>>> 
{ 
    using type = std::tuple<std::vector<T>...>; 
}; 

// Indicies from Andy Prowl's answer 
template <int... Is> 
struct seq {}; 

template <int N, int... Is> 
struct gen_seq : gen_seq<N - 1, N - 1, Is...> {}; 

template <int... Is> 
struct gen_seq<0, Is...> : seq<Is...> {}; 

template <typename... T, int... Is> 
auto tuple_transpose(std::tuple<std::vector<T>...>& var, seq<Is...>) -> typename transpose<decltype(var)>::type 
{ 
    return { std::make_tuple(std::get<Is>(var)...) }; 
} 

template <typename... T> 
auto tuple_transpose(std::tuple<std::vector<T>...>& var) -> typename transpose<decltype(var)>::type 
{ 
    return tuple_transpose(var, gen_seq<sizeof...(T)>{}); 
} 

int main() 
{ 
    std::tuple<std::vector<int>, std::vector<bool>> var; 
    tuple_transpose(var); // error 
    ... 
} 

는 거기에 오류가있는 데모입니다 : 나는 하나 개의 인자로 호출하면 나는 일치 함수 호출 오류 고쳐? 감사. 당신은 동일한 크기의 벡터를 가정하는 경우

Answer 1

,이 작업을 수행해야합니다

template <int... Is> 
struct seq {}; 

template <int N, int... Is> 
struct gen_seq : gen_seq<N - 1, N - 1, Is...> {}; 

template <int... Is> 
struct gen_seq<0, Is...> : seq<Is...> {}; 

template <typename... T, int... Is> 
auto transpose(std::tuple<std::vector<T>...>& var, seq<Is...>) 
    -> std::vector<std::tuple<T...>> 
{ 
    std::vector<std::tuple<T...>> result; 
    for (std::size_t i = 0; i < std::get<0>(var).size(); i++) 
    { 
     std::tuple<T...> t = std::make_tuple(std::get<Is>(var)[i]...); 
     result.push_back(t); 
    } 

    return result; 
} 

template <typename... T, int... Is> 
auto transpose(std::tuple<std::vector<T>...>& var) 
    -> std::vector<std::tuple<T...>> 
{ 
    return transpose(var, gen_seq<sizeof...(T)>()); 
} 

을 그리고 여기 당신이 그것을 테스트 할 수있는 방법입니다 :

#include <iostream> 
#include <iomanip> 

int main() 
{ 
    std::vector<int> vi = {42, 1729, 6}; 
    std::vector<bool> vb = {true, false, false}; 
    std::vector<std::string> vs = {"Hi", "Hey", "Ho"}; 

    auto t = make_tuple(vi, vb, vs); 
    auto v = transpose(t); 

    std::cout << std::boolalpha; 
    for (auto const& t : v) 
    { 
     std::cout << "("; 
     std::cout << std::get<0>(t); 
     std::cout << ", " << std::get<1>(t); 
     std::cout << ", " << std::get<2>(t); 
     std::cout << ")" << std::endl; 
    } 
} 

마지막으로

하는 live example.