c++ - 如何迭代打印 std::tuple

Question

大多数 C++ 书籍示例都利用递归机制来打印 std::tuple。是否可以通过利用 sizeof...(Typename) 迭代打印 std::tuples？

例如，函数签名如下:

template<typename... Ts>
constexpr void PrintTuple(std::tuple<Ts...>& tuple)

然后我可以使用 sizeof...(Ts) 来知道元组中有多少元素，然后我可以使用 std::get< i >(tuple) 来检索单个元素吗？

Answer 1

这是可能的解决方案之一:

#include <cstddef>
#include <iostream>
#include <tuple>
#include <type_traits>
#include <utility>

template <typename T, std::size_t ...I, typename F>
void tuple_foreach_impl(T &&tuple, std::index_sequence<I...>, F &&func)
{
    // In C++17 we would use a fold expression here, but in C++14 we have to resort to this.
    using dummy_array = int[];
    dummy_array{(void(func(std::get<I>(tuple))), 0)..., 0};
}

template <typename T, typename F> void tuple_foreach(T &&tuple, F &&func)
{
    constexpr int size = std::tuple_size<std::remove_reference_t<T>>::value;
    tuple_foreach_impl(std::forward<T>(tuple), std::make_index_sequence<size>{},
                       std::forward<F>(func));
}

int main()
{
    auto x = std::make_tuple("Meow", 1, 2.3);

    tuple_foreach(x, [](auto &&value)
    {
        std::cout << value << ' ';
    });
    // Prints:
    // Meow
    // 1
    // 2.3
}

---

使用 tuple_foreach 制作合适的打印机应该很简单。

template <typename T> void print_tuple(const T &tuple)
{
    std::cout << '{';
    tuple_foreach(tuple, [first = true](auto &value) mutable
    {
        if (!first)
            std::cout << "; ";
        else
            first = 0;
        std::cout << value;
    });
    std::cout << '}';
}

// ...

print_tuple(std::make_tuple("Meow", 1, 2.3)); // Prints `{Meow; 1; 2.3}`