std::async 使用相同的线程，我的代码没有实现并行性

Question

我在Mac OS Xcode 4.3.2上使用C++11std：：async 使用相同的线程，我的代码没有实现并行性。在下面的示例代码中，我想创建 10 个新线程。在每个线程中，我想计算输入变量的平方根并将结果设置为 promise。在主函数中，我想显示从线程计算的结果。我正在调用 std：：async 与策略启动：：async，所以我希望它创建一个新线程（10 次）。

    #include <mutex>
    #include <future>
    #include <thread>
    #include <vector>
    #include <cmath>
    #include <iostream>
    using namespace std;
    mutex iomutex;
    void foo(int i, promise<double> &&prms)
    {
        this_thread::sleep_for(chrono::seconds(2));
        prms.set_value(sqrt(i));
        {
            lock_guard<mutex> lg(iomutex);
            cout << endl << "thread index=> " << i << ", id=> "<< this_thread::get_id();
        }
    }
    int main() 
    {   
        {
            lock_guard<mutex> lg(iomutex);
            cout << endl << "main thread id=>"<< this_thread::get_id();
        }
        vector<future<double>> futureVec;
        vector<promise<double>> prmsVec;
        for (int i = 0; i < 10; ++i) {
            promise<double> prms;
            future<double> ftr = prms.get_future();
            futureVec.push_back(move(ftr));
            prmsVec.push_back(move(prms));
            async(launch::async, foo, i, move(prmsVec[i]));
        }
        for (auto iter = futureVec.begin(); iter != futureVec.end(); ++iter) {
            cout << endl << iter->get();
        }
        cout << endl << "done";
        return 0;
    }

但是，如果我使用 std：：thread，那么我可以实现并行性。

    #include <mutex>
    #include <future>
    #include <thread>
    #include <vector>
    #include <cmath>
    #include <iostream>
    using namespace std;
    mutex iomutex;
    void foo(int i, promise<double> &&prms)
    {
        this_thread::sleep_for(chrono::seconds(2));
        prms.set_value(sqrt(i));
        {
            lock_guard<mutex> lg(iomutex);
            cout << endl << "thread index=> " << i << ", id=> "<< this_thread::get_id();
        }
    }
    int main() 
    {   
        {
            lock_guard<mutex> lg(iomutex);
            cout << endl << "main thread id=>"<< this_thread::get_id();
        }
        vector<future<double>> futureVec;
        vector<promise<double>> prmsVec;
        vector<thread> thrdVec;
        for (int i = 0; i < 10; ++i) {
            promise<double> prms;
            future<double> ftr = prms.get_future();
            futureVec.push_back(move(ftr));
            prmsVec.push_back(move(prms));
            thread th(foo, i, move(prmsVec[i]));
            thrdVec.push_back(move(th));
        }
        for (auto iter = futureVec.begin(); iter != futureVec.end(); ++iter) {
            cout << endl << iter->get();
        }
        for (int i = 0; i < 10; ++i) {
            thrdVec[i].join();
        }
        cout << endl << "done";
        return 0;
    }

Answer 1

            async(launch::async, foo, i, move(prmsVec[i]));

此行返回一个future但由于您没有将其分配给任何内容，因此 future 的析构函数在语句末尾运行，该语句通过调用 std::future::wait() 来阻止并等待结果

为什么你手动调用std::async一个承诺，当它返回未来时？ 异步的要点是你不需要手动使用承诺，这是在内部为你完成的。

重写您的foo()以返回double然后用async调用它

#include <mutex>
#include <future>
#include <thread>
#include <vector>
#include <cmath>
#include <iostream>
using namespace std;
mutex iomutex;
double foo(int i)
{
    this_thread::sleep_for(chrono::seconds(2));
    lock_guard<mutex> lg(iomutex);
    cout << "nthread index=> " << i << ", id=> "<< this_thread::get_id();
    return sqrt(i);
}
int main()
{
    cout << "nmain thread id=>" << this_thread::get_id();
    vector<future<double>> futureVec;
    for (int i = 0; i < 10; ++i)
        futureVec.push_back(async(launch::async, foo, i));
    for (auto& fut : futureVec)
    {
        auto x = fut.get();
        lock_guard<mutex> lg(iomutex);
        cout << endl << x;
    }
    cout << "ndonen";
}