C++：调用 tuple_transpose 函数时没有匹配的函数调用

Question

我正在尝试将向量元组更改为元组向量（反之亦然）。我无法调用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: 没有匹配函数调用‘tuple_transpose(std::tuple >, std::vector > >&)’
prog.cpp:44:24：注意：候选人是：
prog.cpp:30:6: 注意：模板类型名 transpose::type tuple_transpose(std::tuple >...>&, seq)
prog.cpp:30:6：注意：模板参数扣除/替换失败：
prog.cpp:44:24: 注意：候选人需要 2 个参数，提供 1 个
prog.cpp:36:6: 注意：模板类型名 transpose::type tuple_transpose(std::tuple >...>&)
prog.cpp:36:6：注意：模板参数扣除/替换失败：
prog.cpp: 代替'模板类型名转置::type tuple_transpose(std::tuple >...>&) [with T = {int, bool}]':
prog.cpp:44:24: 从这里需要
prog.cpp:36:6: 错误：在‘struct transpose>，std::vector >>&>’中没有名为‘type’的类型}

#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
    ...
}

这是一个包含错误的演示：http://ideone.com/7AWiQQ#view_edit_box

我做错了什么，我该如何解决？谢谢。

Answer 1

如果您假设向量的大小相同，这应该可以完成工作：

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)>());
}

您可以通过以下方式对其进行测试：

#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;
    }
}

最后是live example。