可变模板，函数作为参数 [英] Variadic template, function as argument

问题描述

我想在可变参数模板中使用函数作为参数，为什么以下不起作用?我如何使它工作?

I would like to use a function as argument in a variadic template, why does the following not work? How do I make it work?

template<typename F, typename... Args>
F test(F f, const Args&&... args) {
return f(std::forward<Args>(args)...);
}

int simple(int i) { 
    return i; 
}


int main()
{
    std::cout << test(simple, 2); // error, 'std::forward': none of the 2 overloads could convert all the argument types
}

推荐答案

您的代码存在一些问题.

There are a couple of problems with your code.

首先，您应该使用转发引用，因此您需要将const Args&&... 更改为Args&&....

First of all, you should use forwarding references, so you need to change const Args&&... to Args&&....

那么，test 不必返回 F.所以这里使用decltype(auto)是合理的.

Then, test does not have to return F. So it is reasonable to use decltype(auto) here.

除此之外，转发 f 也是有意义的.

In addition to that, it makes sense to forward f too.

固定版本可能如下所示:

The fixed version might look like this:

template<typename F, typename... Args>
decltype(auto) test(F&& f, Args&&... args) {
    return std::forward<F>(f)(std::forward<Args>(args)...);
}

WANDBOX 示例

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