通过 lambda 表达式将函数转发给 std::thread

Question

我正在阅读注释 here 以了解示例异步函数实现。

正在尝试编译以下代码

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


int f(int x) {
    auto start = std::chrono::system_clock::now();
    std::time_t start_time = std::chrono::system_clock::to_time_t(start);
    std::cout << "returning " << x << " at " << std::ctime(&start_time) << std::endl;
    return x;
}


template<typename Function, typename... Args>
auto async(Function&& function, Args&&... args) {

    std::promise<typename std::result_of<Function(Args...)>::type> outer_promise;
    auto future = outer_promise.get_future();
    auto lambda = [function, promise = std::move(outer_promise)](Args&&... args) mutable {
        try {
            promise.set_value(function(args...));
        } catch (...) {
            promise.set_exception(std::current_exception());
        }
    };

    std::thread t0(std::move(lambda), std::forward<Args>(args)...);
    t0.detach();
    return future;
}

int main() {

    auto result1 = async(f, 1); 
    auto result2 = async(f, 2); 

}

我遇到以下编译器错误，我很难理解这些错误。

我想要一些指导，说明为什么这个 function arg 没有根据编译器正确声明。

In instantiation of 'async(Function&&, Args&& ...)::<lambda(Args&& ...)> mutable [with Function = int (&)(int); Args = {int}]':
22:61:   required from 'struct async(Function&&, Args&& ...) [with Function = int (&)(int); Args = {int}]::<lambda(int&&)>'
28:3:   required from 'auto async(Function&&, Args&& ...) [with Function = int (&)(int); Args = {int}]'
37:27:   required from here
22:88: error: variable 'function' has function type
22:88: error: variable 'function' has function type
 In instantiation of 'struct async(Function&&, Args&& ...) [with Function = int (&)(int); Args = {int}]::<lambda(int&&)>':
28:3:   required from 'auto async(Function&&, Args&& ...) [with Function = int (&)(int); Args = {int}]'
37:27:   required from here
22:18: error: field 'async(Function&&, Args&& ...) [with Function = int (&)(int); Args = {int}]::<lambda(int&&)>::<function capture>' invalidly declared function type

Answer 1

要获得 result_of 的结果类型，您必须访问 type:

std::promise< typename std::result_of<Function(Args...)>::type > outer_promise;

通过 lambda 表达式将函数转发给 std::thread

Forwarding a function to std::thread via a lambda expression

c++

c++17