增加计数器时避免竞争条件

Question

我以前就这个话题发过帖子，但到目前为止我还没有运气。我把它归结为我的一个坏问题。这次我做了一个简短的可编译示例，显示了我试图避免的不良行为。我希望这会受到赞赏。

问题是两个（或更多）线程被设置为运行同一个进程，它们的“id”决定了它们操作变量数据的哪一部分。目前两个线程都会更新计数器。

当前的输出是这样的，

tid = 0, var[tid] = 0
tid = 0, var[tid] = 1
tid = 0, var[tid] = 2
tid = 0, var[tid] = 3
tid = 0, var[tid] = 4
tid = 0, var[tid] = 5
tid = 0, var[tid] = 6
tid = 0, var[tid] = 7
tid = 0, var[tid] = 8
tid = 0, var[tid] = 9
tid = 1, var[tid] = 0
Press any key to continue . . .

想要的输出应该是这样的……

tid = 0, var[tid] = 0
tid = 1, var[tid] = 0
tid = 0, var[tid] = 1
tid = 1, var[tid] = 1
tid = 0, var[tid] = 2
tid = 1, var[tid] = 2
tid = 0, var[tid] = 3
tid = 1, var[tid] = 3 etc.

我们将不胜感激任何指导。

编辑：我已经使用按预期工作的代码更新了答案。

[注意这里效率很重要，我想尽快完成流程]

#include <iostream>  
#include <boost/thread.hpp>

int var[2];
int mT;
int mTotalSamples;
boost::mutex mCountMutex;
boost::thread *threadMap[2];

using namespace std;

void process()
{
    int tid = 1;

    // sleep for 1 seconds - just to make sure threadMap 
    // has been assigned (only ncessary for this demo).
    boost::this_thread::sleep(boost::posix_time::seconds(1));

    if (threadMap[0]->get_id() == boost::this_thread::get_id()){ tid = 0;}

    while ( mT < mTotalSamples ) 
    {
        // perform processing
        var[tid] = mT; 
        // processing complete

        mCountMutex.lock(); // (a thread waits to aquire mutex)
        cout << "tid = " << tid << ", var[tid] = " << var[tid] << endl;
        mT++;           // How to stop both threads incrementing this?      
        mCountMutex.unlock();       
    }   
}

int main()
{
    boost::thread_group threads;

    mT = 0;
    mTotalSamples = 10;

    threadMap[0] = threads.create_thread( boost::bind(&process) );
    threadMap[1] = threads.create_thread( boost::bind(&process) );

    threads.join_all();

    return 0;
}

Answer 1

从您的预期输出来看，您希望线程在每次更新后同步。 boost 库提供了 boost::barrier，如果你在 process 的 while 循环的开头或结尾放置一个 wait ，应该可以解决问题。

#include <iostream>  
#include <boost/thread.hpp>

int var[2];
int mT;
int mTotalSamples;
boost::mutex mCountMutex;
boost::thread *threadMap[2];
boost::barrier bar(2);

using namespace std;

void process()
{
    int tid = 1;        

    // sleep for 2 seconds - just to make sure threadMap 
    // has been assigned (only ncessary for this demo).
    boost::this_thread::sleep(boost::posix_time::seconds(2));

    if (threadMap[0]->get_id() == boost::this_thread::get_id()){ tid = 0;}

    while ( mT < mTotalSamples ) 
    {
        // perform processing
        var[tid] = mT; 
        // processing complete


        bar.wait();
        if (threadMap[0]->get_id() == boost::this_thread::get_id())
        {
            mT++;               
            cout << "var[0] = " << var[0] << endl;
            cout << "var[1] = " << var[1] << endl;                      
        }           
        bar.wait();
    }   
}

int main()
{
    boost::thread_group threads;

    mT = 0;
    mTotalSamples = 10;

    threadMap[0] = threads.create_thread( boost::bind(&process) );
    threadMap[1] = threads.create_thread( boost::bind(&process) );

    threads.join_all();

    return 0;
}

Answer 2

在 process() 中将此 int mT; 设为本地 - 不是全局的。 或者您需要int mT[2]; 那么你就不需要互斥锁了。