来自 std::get 的完美转发元组值

Question

我有一个类似于 std::apply 的很好的实现，它将元组扩展为函数的参数。它工作得很好，除了 std::get 总是返回一个左值并且它无法匹配正确的重载。

可在此处找到 POC 代码：https://wandbox.org/permlink/OUYMQY2afL8vRMUu

这个想法是将 std::forward 添加到 apply_sequence 所以它打印 ONE TWO THREE

void printNumber(const int& x, const int& y)
{
    std::cout << "ONE" << std::endl;
}

void printNumber(const int& x, int&& y)
{
    std::cout << "TWO" << std::endl;
}

void printNumber(int&& x, const int& y)
{
    std::cout << "THREE" << std::endl;
}

template<typename... TTuple, std::size_t... Indices>
auto apply_sequence(const std::tuple<TTuple...>& tuple, std::index_sequence<Indices...>)
{
    // missing: forward value to proper type (currently is always lvalue)
    return printNumber(std::get<Indices>(tuple)...);
}

template<typename... TTuple>
auto apply_tuple(const std::tuple<TTuple...>& tuple)
{
    return apply_sequence(tuple, std::index_sequence_for<TTuple...>());
}

int main(int argc, char* argv[])
{
    std::tuple<int, int> one { 1, 2 };
    apply_tuple(one); // ONE

    std::tuple<int, int&&> two { 1, 2 };
    apply_tuple(two); // TWO

    std::tuple<int&&, int> three { 1, 2 };
    apply_tuple(three); // THREE

    return 0;
}

编辑：以防有人想要问题的解决方案https://wandbox.org/permlink/XkUjfypAMepJRPgZ

Answer 1

原因是如果元组参数是左值，std::get returns 是左值，即使元素是右值引用。你可以自己写my_get来解决：

// If the element is not an rvalue reference, behave the same as std::get
template< std::size_t I, class... Types >
constexpr std::enable_if_t <
              !std::is_rvalue_reference_v<std::tuple_element_t<I, std::tuple<Types...>>>, 
               std::tuple_element_t<I, std::tuple<Types...>>const&
          >
    my_get( const std::tuple<Types...>& t ) noexcept
{
    return std::get<I>(t);
}

// If the element is an rvalue reference, move the result of std::get
template< std::size_t I, class... Types >
constexpr std::enable_if_t <
              std::is_rvalue_reference_v<std::tuple_element_t<I, std::tuple<Types...>>>, 
              std::tuple_element_t<I, std::tuple<Types...>>
          >
    my_get( const std::tuple<Types...>& t ) noexcept
{
    return std::move(std::get<I>(t));
}

在代码中使用 my_get 而不是 std::get，那么一切顺利。

来自 std::get 的完美转发元组值

perfect forwading tuple values from std::get

c++

tuples

forward

perfect-forwarding