推入并弹出std :: tuple的第一个元素 [英] Pushing and popping the first element of a std::tuple

问题描述

我正在用这种方式用可变数量的参数（和不同类型）用C ++编写一个函数

I am writing a function in C++ with a variable number of arguments (and different types) in this way

template<typename ...Ts>
void myFunction(Ts ...args)
{
    //create std::tuple to access and manipulate single elements of the pack
    auto myTuple = std::make_tuple(args...);    

    //do stuff

    return;
}

我想做什么，但是我不知道怎么做从元组中推送和弹出元素，尤其是第一个元素...

What i would like to do, but I don't know how, is to push and pop elements from the tuple, in particular the first element... something like

//remove the first element of the tuple thereby decreasing its size by one
myTuple.pop_front()

//add addThis as the first element of the tuple thereby increasing its size by one
myTuple.push_front(addThis)

可能吗？

推荐答案

您可以执行以下操作

template <typename T, typename Tuple>
auto push_front(const T& t, const Tuple& tuple)
{
    return std::tuple_cat(std::make_tuple(t), tuple);
}

template <typename Tuple, std::size_t ... Is>
auto pop_front_impl(const Tuple& tuple, std::index_sequence<Is...>)
{
    return std::make_tuple(std::get<1 + Is>(tuple)...);
}

template <typename Tuple>
auto pop_front(const Tuple& tuple)
{
    return pop_front_impl(tuple,
                          std::make_index_sequence<std::tuple_size<Tuple>::value - 1>());
}

演示

请注意，它确实很基本，不处理引用元组或const限定类型的元组，但它可能足够。

Note that it is really basic and doesn't handle tuple of reference, or tuple of const qualified type, but it might be sufficient.

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