我_153434__无法调用 tuple_transpose
函数 . 当我用一个参数调用它时,我得到一个 no matching function call
错误:
prog.cpp:在函数'int main()'中:prog.cpp:44:24:错误:没有用于调用'tuple_transpose(std :: tuple>,std :: vector >> &&)'prog的匹配函数 . cpp:44:24:注意:候选人是:prog.cpp:30:6:注意:模板typename transpose :: type tuple_transpose(std :: tuple> ...>&,seq)prog.cpp:30:6:注意:模板参数推断/替换失败:prog.cpp:44:24:注意:候选者需要2个参数,1个提供prog.cpp:36:6:注意:模板类型名称transpose :: type tuple_transpose(std :: tuple> . ..> && prog.cpp:36:6:注意:模板参数演绎/替换失败:prog.cpp:代替'template typename transpose :: type tuple_transpose(std :: tuple> ...>&)[与T = {int,bool}]':prog.cpp:44:24:从这里需要prog.cpp:36:6:错误:'struct transpose>中没有名为'type'的类型,std :: vector >> &>”
#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
...
}
这是一个带有错误的演示:http://ideone.com/7AWiQQ#view_edit_box
我做错了什么,我该如何解决?谢谢 .
1 回答
如果你假设相同大小的向量,这应该做的工作:
这是你如何测试它:
最后,live example .