我们可以使用可变参数模板函数过滤特定类型的参数，然后将其余参数传递给另一个函数吗? [英] Can we use variadic template function to filter parameters of specific type, then pass the rest to another function?

问题描述

例如

// we have a variadic function
void print(...);    

// I need such a function to filter parameters of specific type
template<typename... Args>
void print_filter(const Args&... args)
{
    // filter non-integral type
    print(integral args);
}

// my goal, all non-integral type can be ignored:
print_filter(1.0, 2, "abc", 3) == print(2, 3)

我已经用尽我的知识来做到这一点...您能帮忙吗?或只是证明这是不可能的，这也非常有帮助.谢谢

I have used up my knowledge to do that... can you help? or just to prove it's impossible, which also be very helpful. Thanks

推荐答案

一个巧妙的技巧是将所需的参数转换为1元素转发元组，将不需要的参数转换为空元组，tuple_cat 结果，然后将结果元组 apply (C ++ 17)应用于要调用的函数:

A neat trick is to convert the arguments you want into a 1-element forwarding tuple, the arguments you don't want into an empty tuple, tuple_cat the results, then apply (C++17) the resulting tuple to the function you want to invoke:

template<typename... Args>
void print_filter(Args&&... args) {
    std::apply(
        [](auto&&... args) { return print(std::forward<decltype(args)>(args)...); },
        std::tuple_cat(
            std::get<std::is_integral<typename std::decay<Args>::type>::value ? 0 : 1>(
                std::make_tuple(
                    [](Args&& arg) { return std::tuple<Args&&>{std::forward<Args>(arg)}; },
                    [](Args&&) { return std::tuple<>{}; }))(
                std::forward<Args>(args))...));
}

请注意，这采用了另一个技巧，即使用 get 有条件地将两个函数之一应用于参数.

Note that this employs another trick, which is to use get to conditionally apply one of two functions to an argument.

示例.

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