Question

我需要使用互斥锁（仅限）在2个线程之间实现屏障同步。屏障同步是指2个线程在继续之前会在预定义的步骤中等待彼此满足。

我能够使用seamaphore，但我怎样才能使用互斥锁来实现这一点。我得到了一个提示，我需要2个互斥量而不是1个。

使用Seamaphore：

#include <pthread.h>
#include <semaphore.h>
using namespace std;

sem_t s1;
sem_t s2;


void* fun1(void* i)
{
    cout << "fun1 stage 1" << endl;
    cout << "fun1 stage 2" << endl;
    cout << "fun1 stage 3" << endl;
    sem_post (&s1);
    sem_wait (&s2);
    cout << "fun1 stage 4" << endl;
}

void* fun2(void* i)
{
    cout << "fun2 stage 1" << endl;
    cout << "fun2 stage 2" << endl;
//    sleep(5);
    sem_post (&s2);
    sem_wait (&s1);
    cout << "fun2 stage 3" << endl;
}

main()
{
    sem_init(&s1, 0, 0);
    sem_init(&s2, 0, 0);
    int value; 
    sem_getvalue(&s2, &value);
    cout << "s2 = " << value << endl;

    pthread_t iThreadId;

    cout << pthread_create(&iThreadId, NULL, &fun2, NULL) << endl;
//    cout << pthread_create(&iThreadId, NULL, &fun2, NULL) << endl;
    pthread_create(&iThreadId, NULL, &fun1, NULL);
    sleep(10);
}

将上述代码编译为“g ++ barrier.cc -lpthread”

Answer 1

没有MUTEXES 怎么样没有锁？仅使用原子操作：

#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#include <signal.h>

static sigset_t _fSigSet;
static volatile int _cMax=20, _cWait = 0;
static pthread_t    _aThread[1000];

void * thread(void *idIn)
{
int nSig, iThread, cWait, id = (int)idIn;

    printf("Start %d\n", id, cWait, _cMax);
    // do some fake weork
    nanosleep(&(struct timespec){0, 500000000}, NULL);
    // barrier
    cWait = __sync_add_and_fetch(&_cWait, 1);
    printf("Middle %d, %d/%d Waiting\n", id, cWait, _cMax);
    if (cWait < _cMax)
    {        
        // if we are not the last thread, sleep on signal
        sigwait(&_fSigSet, &nSig); // sleepytime
    }
    else
    {
        // if we are the last thread, don't sleep and wake everyone else up
        for (iThread = 0; iThread < _cMax; ++iThread)
            if (iThread != id)
                pthread_kill(_aThread[iThread], SIGUSR1);
    }

    // watch em wake up    
    cWait = __sync_add_and_fetch(&_cWait, -1);
    printf("End %d, %d/%d Active\n", id, cWait, _cMax);

    return 0;
}

int main(int argc, char** argv)
{
    pthread_attr_t attr;
    int i, err;

    sigemptyset(&_fSigSet);
    sigaddset(&_fSigSet, SIGUSR1);
    sigaddset(&_fSigSet, SIGSEGV);

    printf("Start\n");
    pthread_attr_init(&attr);
    if ((err = pthread_attr_setstacksize(&attr, 16384)) != 0)
    {
        printf("pthread_attr_setstacksize failed: err: %d %s\n", err, strerror(err));
        exit(0);
    }

    for (i = 0; i < _cMax; i++)
    {
        if ((err = pthread_create(&_aThread[i], &attr, thread, (void*)i)) != 0)
        {
            printf("pthread_create failed on thread %d, error code: %d %s\n", i, err, strerror(err));
            exit(0);
        }
    }

    for (i = 0; i < _cMax; ++i)
        pthread_join(_aThread[i], NULL);

    printf("\nDone.\n");
    return 0;
}

Answer 2

我不确定你需要两个互斥锁，一个互斥锁和一个条件变量，一个额外的标志就足够了。这个想法是你通过获取互斥锁进入临界区，然后你检查你是否是第一个来的线程，如果是，你等待条件。如果你是第二个线程，那么你就会唤醒等待线程并离开。

使用互斥锁的2个进程之间的屏障同步

2 个答案: