将任何函数作为模板参数传递？ [英] Passing any function as a template parameter?

问题描述

我正在寻找一种方法来传递一个通用（constexpr，明显）函数到模板。它必须能够采取任何数量的参数，而不使用lambda。这是我到目前为止：

I am looking for a way to pass a generic (constexpr, obviously) function to a template. It has to be able to take any amount of parameters, without using a lambda. This is what I have so far:

template<typename T, T(*FUNC)()> struct CALL
{
    static inline constexpr decltype(FUNC()) EXEC()
    {
        return FUNC();
    }
};

但是，这只有在传递的函数不带参数时才有效。有没有办法使模板接受任何constexpr函数？传递std ::函数似乎不工作。
我想关键是可变参数模板参数，但我不知道如何利用它们在这种情况下的优势。

This however only works if the passed function takes no parameters. Is there a way to make the template accept ANY constexpr function? Passing a std::function does not seem to work. I suppose the key is variadic template parameters, but I have no idea how to take advantage of them in this situation.

推荐答案

如果我正确理解你想要实现什么，你可以使用模板函数，函数：

If I understand correctly what you are trying to achieve, you can use a template function rather than a template class with a static function:

#include <iostream>

template<typename T, typename... Ts>
constexpr auto CALL(T (*FUNC)(Ts...), Ts&&... args) -> decltype(FUNC(args...))
{
    return FUNC(std::forward<Ts>(args)...);
}

constexpr double sum(double x, double y)
{
    return (x + y);
}

int main()
{
    constexpr double result = CALL(sum, 3.0, 4.0);
    static_assert((result == 7.0), "Error!");
    return 0;
}

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