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我正在尝试将向量元组更改为元组向量（反之亦然）。我无法调用该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 > >&)”<br> prog.cpp:44:24: 注意: 候选是:
prog.cpp:30:6: 注意: 模板类型名 transpose::type tuple_transpose(std::tuple >...>&, seq)
prog.cpp:30: 6：注意：模板参数推导/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：模板参数推导/替换失败：
prog.cpp: 替换'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: 'struct transpose >, std::vector > >&>' 中没有名为'type'的类型</sub>

#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。

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1 回答 1

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如果您假设向量的大小相同，那么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)>());
}

这里是你如何测试它：

#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，一个活生生的例子。

于 2013-05-15T23:22:55.070 回答