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如果我想使用但不将一个函数应用于整个事物来遍历一个元组std::apply，我该如何分离元组，即将一个函数应用于第一个n值，另一个应用于它之后的所有值？

some_values将是一个元组，它可以具有任何长度和类型，并且length_of_first_part（可能命名得不那么冗长）在编译时就知道了。

std::tuple<char, long long, double, long double, float> some_values(33, 2, 3.4, 5.6, 7.8);
const size_t length_of_first_part = 2;
std::apply(
    [](auto&&... current_val) {
        ((std::cout << "(Should be first part) " << current_val << "\n"), ...); //Would obviously do a litle more than cout, but this is just a minimal example
    }, some_values
);

std::apply(
    [](auto&&... current_val) {
        ((std::cout << "(Should be second part) " << current_val << "\n"), ...);
    }, some_values
);
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1 回答 1

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试一试。在这里查看“可能的实现”： https ://en.cppreference.com/w/cpp/utility/apply

它将一个传递std::index_sequence给一个辅助函数来调用std::invoke。我将其更改为通过std::make_sequence<N>（N 是拆分索引）而不是元组大小。然后我通过逆（元组大小 - N）来调用std::invoke第二个函数：

template <int N, typename F1, typename F2, typename Tuple, size_t... I1s, size_t... I2s>
decltype(auto) split_apply_impl(F1&& f1, F2&& f2, Tuple&& t, std::index_sequence<I1s...>, std::index_sequence<I2s...>) {
    std::invoke(std::forward<F1>(f1), std::get<I1s>(std::forward<Tuple>(t))...);
    return std::invoke(std::forward<F2>(f2), std::get<(I2s+N)>(std::forward<Tuple>(t))...);
}

template <size_t N, typename F1, typename F2, typename Tuple>
decltype(auto) split_apply(F1&& f1, F2&& f2, Tuple&& t)
{
    return split_apply_impl<N>(
        std::forward<F1>(f1), 
        std::forward<F2>(f2),
        std::forward<Tuple>(t),
        std::make_index_sequence<N>{},
        std::make_index_sequence<(std::tuple_size_v<std::remove_reference_t<Tuple>>-N)>{});
}

用法：

std::tuple<char, long long, double, long double, float> some_values(33, 2, 3.4, 5.6, 7.8);
constexpr size_t length_of_first_part = 2;
split_apply<length_of_first_part>(
    [](auto&&... current_val) { 
        ((std::cout << "(Should be first part) " << current_val << "\n"), ...); 
    },
    [](auto&&... current_val) {
        ((std::cout << "(Should be second part) " << current_val << "\n"), ...);
    },
    some_values);

概念证明： https ://godbolt.org/z/6dvMre

于 2020-09-25T07:19:19.340 回答