c++ - 为每次 c++ 创建一个障碍

Question

如何组织下面的程序，使其按如下方式工作:对于每次，只要所有其他线程还没有达到那个时间，每个线程就必须等待。在给定时间内所有线程都“执行”时，应该打印出结果值。

#include <iostream>
#include <thread>
#include <mutex>

const int numThreads = 4;

typedef double Time;
double resultForGivenTime = 0;

class Printer
{
public:
    void print(Time time, double result)
    {
        mtx.lock();
        std::cout << "Time:"  << time << " -> Result:" << result << std::endl;
        resultForGivenTime = 0;
        mtx.unlock();
    }

private:
    std::mutex mtx;
};

Printer p;

void doIt (Printer& p, Time& t, int& id)
{
    //Is it possible to create here a barier so that
    //program output will look like this:
    //Time: 0 -> Result 6             # one or four time
    //Time: 1 -> Result 6            
    //Time: 2 -> Result 6
    //Time: 3 -> Result 6
    //Time: 4 -> Result 6
    resultForGivenTime += id;
    p.print(t, resultForGivenTime);
}

void handler(int id)
{
    for (Time time = 0.0; time < 5.0; ++time)
    {
        doIt(p, time, id);
    }
}

int main()
{
    std::thread threads[numThreads];

    for (int i = 0; i < numThreads; ++i)
        threads[i] = std::thread(handler, i);

     for (auto& th : threads) th.join();

    return 0;
}

Answer 1

您可以结合使用条件变量和计数器。您可以在此处找到一个很好的用法示例:

http://www.cplusplus.com/reference/condition_variable/condition_variable/

或者，如果您有可用的 Boost 库，您可以使用 barrier类，它提供了一个很好的包装器:

#include <boost/thread/barrier.hpp>

class barrier
{
public:
    barrier(barrier const&) = delete;
    barrier& operator=(barrier const&) = delete;

    barrier(unsigned int count);
    template <typename F>
    barrier(unsigned int count, F&&);

    ~barrier();

    bool wait();
    void count_down_and_wait();
};