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实际上，主要场景是：从主线程有两个线程在运行。通过使用条件变量，两个线程将运行并休眠，然后返回主线程。我的意思是我不想要不同的输出模式。只有一个模式：从main->thread1->thread2->main。我已经为 C 线程编写了一个代码。它有时会显示我想要的结果，有时会显示我想要的结果。例如，输出是：

I am in thread 1 
before conditional wait
I am in thread 2
before conditional release
i am again in thread 2
i am again in thread 1
main exits here

问题有时是“这里的主要出口”不执行。请帮助我。需要注意的是我不能使用 pthread_join()。我的代码如下

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

pthread_mutex_t gLock;
pthread_cond_t gCondition;

pthread_mutex_t mLock;
pthread_cond_t mCondition;

void initialize()
{
      pthread_mutex_init(&gLock, NULL);
      pthread_cond_init (&gCondition, NULL);
      pthread_mutex_init(&mLock, NULL);
      pthread_cond_init (&mCondition, NULL);

      return;
}

void * threadOne(void * msg)
{
    printf("%s \n",(char*) msg);
    printf("before conditional wait\n");

    pthread_mutex_lock(&gLock);
    pthread_cond_wait(&gCondition,&gLock);
    pthread_mutex_unlock(&gLock);

    printf("i am again in thread 1\n");

    pthread_mutex_lock(&mLock);
    pthread_cond_signal(&mCondition);
    pthread_mutex_unlock(&mLock);

}

void * threadTwo(void * msg)
{
    printf("%s\n",(char*)msg);
    printf("before conditional release\n");
    pthread_mutex_lock(&gLock);
    pthread_cond_signal(&gCondition);
    pthread_mutex_unlock(&gLock);
    printf("i am again in thread 2\n");

}

int main()
{
        pthread_t thread1;
        pthread_t thread2;

        char * msg1="I am in thread 1";
        char * msg2="I am in thread 2";
        initialize();

        pthread_create(&thread1,NULL,threadOne,(void*) msg1);
        pthread_create(&thread2,NULL,threadTwo,(void*) msg2);

        pthread_mutex_lock(&mLock);
        pthread_cond_wait(&mCondition,&mLock);
        pthread_mutex_unlock(&mLock);

        printf("main exits here");

        return 0;
}
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1 回答 1

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问题是您错误地使用了条件变量。条件变量只是一种通知机制，而不是标志。除了当前等待的线程列表之外，它没有任何内部状态。因此，如果在其他线程调用时还main()没有实际执行到调用，则信号丢失，并且将永远等待。pthread_cond_wait()pthread_cond_signal()main()

您需要使用与条件变量关联的单独标志。main()然后可以检查这个标志，只有在标志没有设置时才等待。此外，它必须在循环中检查此标志，以确保处理“虚假唤醒”，在没有相应信号的情况下pthread_cond_wait()返回。这同样适用于 和 之间的通知。threadOnethreadTwo

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

pthread_mutex_t gLock;
pthread_cond_t gCondition;
int gFlag=0;

pthread_mutex_t mLock;
pthread_cond_t mCondition;
int mFlag=0;

void initialize()
{
    pthread_mutex_init(&gLock, NULL);
    pthread_cond_init (&gCondition, NULL);
    pthread_mutex_init(&mLock, NULL);
    pthread_cond_init (&mCondition, NULL);
}

void * threadOne(void * msg)
{
    printf("%s \n",(char*) msg);
    printf("before conditional wait\n");

    pthread_mutex_lock(&gLock);
    while(!gFlag)
    {
        pthread_cond_wait(&gCondition,&gLock);
    }
    pthread_mutex_unlock(&gLock);

    printf("i am again in thread 1\n");

    pthread_mutex_lock(&mLock);
    mFlag=1;
    pthread_cond_signal(&mCondition);
    pthread_mutex_unlock(&mLock);

}

void * threadTwo(void * msg)
{
    printf("%s\n",(char*)msg);
    printf("before conditional release\n");
    pthread_mutex_lock(&gLock);
    gFlag=1;
    pthread_cond_signal(&gCondition);
    pthread_mutex_unlock(&gLock);
    printf("i am again in thread 2\n");

}

int main()
{
    pthread_t thread1;
    pthread_t thread2;

    char * msg1="I am in thread 1";
    char * msg2="I am in thread 2";
    initialize();

    pthread_create(&thread1,NULL,threadOne,(void*) msg1);
    pthread_create(&thread2,NULL,threadTwo,(void*) msg2);

    pthread_mutex_lock(&mLock);
    while(!mFlag)
    {
        pthread_cond_wait(&mCondition,&mLock);
    }
    pthread_mutex_unlock(&mLock);

    printf("main exits here");

    return 0;
}
于 2011-06-29T09:41:39.550 回答