【发布时间】:2020-10-22 22:35:43
【问题描述】:
我正在尝试编写一个代码,该代码使用从一组可变元组中提取的输入参数调用 lambda。但是,我的尝试没有编译:
#include <iostream>
#include <tuple>
#include <utility>
#include <type_traits>
template <typename ...>
struct first_of;
template <typename T, typename ... Args>
struct first_of<T, Args...> {
using type = std::decay_t<T>;
};
template <typename T>
struct first_of<T> {
using type = std::decay_t<T>;
};
template <typename ... T>
using first_of_t = typename first_of<T...>::type;
template <typename Fn, typename... Tuples, std::size_t... Idxs>
void run_impl(Fn&& fn, std::index_sequence<Idxs...>, Tuples... t) {
auto temp = {(fn(std::get<Idxs>(t)...), true)...};
(void)temp;
}
template <typename Fn, typename... Tuples>
void run(Fn&& fn, Tuples&&... tuples) {
run_impl(std::forward<Fn>(fn), std::make_index_sequence<std::tuple_size<first_of_t<Tuples...>>::value>{}, std::forward<Tuples>(tuples)...);
}
int main() {
auto a = std::make_tuple(1, 2.34, "one");
auto b = std::make_tuple(32, 5.34, "two");
auto print = [](auto& f, auto& g) { std::cout << f << ", " << g << std::endl; };
run(print, a, b);
}
我期待以下输出:
1、32
2.34, 5.34
一,二
我使用的是 c++14,所以很遗憾,没有折叠表达式。 这是代码的上帝螺栓链接:https://godbolt.org/z/G19n5z
【问题讨论】:
标签: c++ tuples c++14 variadic-templates template-meta-programming