如何从元组中获取元组中元素的引用元组？

Question

我有一个C++ 11元组，我想得到元组的相同元素std::reference_wrapper。有没有简单的方法来做到这一点？

感谢

Answer 1

映射一个元组很容易给出a pack of indices，如：

#include <tuple> 
#include <functional> 
#include <iostream> 

template <int...> 
struct Seq {}; 

template <int n, int... s> 
struct Gens : Gens<n-1, n-1, s...> {}; 

template <int... s> 
struct Gens<0, s...> { 
    typedef Seq<s...> Type; 
}; 

// The above are taken from https://stackoverflow.com/q/7858817 

使用指标，我们可以使用std::get适用std::ref的元组的每一个元素：

template <int... s, typename Tuple> 
auto ref_tuple_impl(Seq<s...> seq, Tuple& tup) 
-> std::tuple< 
    std::reference_wrapper< 
     typename std::tuple_element<s, Tuple>::type 
    >... 
> 
{ 
    return std::make_tuple(std::ref(std::get<s>(tup))...); 
} 

template <typename Tuple> 
auto ref_tuple(Tuple& tup) 
    -> decltype(ref_tuple_impl(typename Gens<std::tuple_size<Tuple>::value>::Type>(), tup)) 
{ 
    return ref_tuple_impl(typename Gens<std::tuple_size<tuple>::value>::Type(), tup); 
} 

用法示范：

int main() { 
    auto t = std::make_tuple(1, 4.5, "66"); 
    auto rt = ref_tuple(t); 

    std::get<0>(rt).get() = 123; 
    std::get<1>(rt).get() = 5.67; 
    std::get<2>(rt).get() = "34"; 

    std::cout << std::get<0>(t) << std::endl; 
    std::cout << std::get<1>(t) << std::endl; 
    std::cout << std::get<2>(t) << std::endl; 
}