std :: function的替代方法，用于将函数作为参数传递（回调等） [英] Alternative to std::function for passing function as argument (callbacks, etc.)

问题描述

在使用C ++ 11进行实验时，我偶然发现了这一点。我发现这是一个显而易见的解决方案，但我无法在野外找到任何其他示例，因此我担心我缺少一些东西。

I stumbled across this during my experiments with C++11. I find that it is an obvious solution, but I haven't been able to find any other examples of it in the wild, so I'm concerned that there's something I'm missing.

我指的是这种做法（在 addAsync函数中）：

The practice I'm referring to (in the "addAsync" function):

#include <thread>
#include <future>
#include <iostream>
#include <chrono>

int addTwoNumbers(int a, int b) {
    std::cout << "Thread ID: " << std::this_thread::get_id() << std::endl;

    return a + b;
}

void printNum(std::future<int> future) {
    std::cout << future.get() << std::endl;
}

void addAsync(int a, int b, auto callback(std::future<int>) -> void) { //<- the notation in question
    auto res = std::async(std::launch::async, addTwoNumbers, a, b);

    if (callback) //super straightforward nullptr handling
        return callback(std::move(res));
}

int main(int argc, char** argv) {
    addAsync(10, 10, [](std::future<int> number) { //lambda functions work great
        addAsync(number.get(), 20, [](std::future<int> number) {
            addAsync(893, 4387, printNum); //as do standard functions
            addAsync(2342, 342, nullptr); //executes, sans callback

            std::cout << number.get() << std::endl;
        });
    });

    std::cout << "main thread: " << std::this_thread::get_id() << std::endl;

    return 0;
}

这是不好的做法，还是不可携带（我已经仅在MSVC ++ 2015中尝试过）？另外，编译器如何处理呢？通过转换为std :: function？

Is it considered bad practice, or is it non-portable (I've only tried it in MSVC++ 2015)? Also, how does the compiler treat this; by conversion to std::function?

我很想在我的项目中继续使用它，因为它显然在签名中声明了必需的参数类型和返回类型，并接受nullptr作为可选项，并且似乎正当工作（我知道这些是C ++中著名的遗言）。

I would love to keep using this in my projects, as it obviously states the required argument types and return type in the "signature", accepts a nullptr for optionality, and seems to "just work" (I am aware that these are famous last words in C++).

推荐答案

您使用的是原始文件

与 std :: function 不同，这不适用于捕获的lambda或 std :: bind 的结果，或者具有实现 operator（）的通用类类型。

Unlike std::function, this will not work with a lambda that captures, or with a result of std::bind, or with a general class type that implements operator().

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