7.9 线程互斥锁

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于 2022-04-21 19:34:28 发布

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分类专栏： Linux系统编程文章标签： linux

本文链接：https://blog.youkuaiyun.com/qq_38105227/article/details/124329666

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该实验代码展示了使用pthread库实现的生产者消费者问题。在一个多线程环境中，一个线程作为生产者增加蛋糕数量，另一个线程作为消费者减少蛋糕数量，通过互斥锁mutex保证了数据的一致性。生产者每3秒增加一个蛋糕，消费者每1秒尝试消费蛋糕，当没有蛋糕时，消费者会继续等待。

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1、实验代码

#include <stdio.h>
#include <pthread.h>
#include <unistd.h>

int cake_count = 0;
pthread_mutex_t mutex;

static void *produce_thread (void *arg)
{
    cake_count = *(int*)arg;
    while (1)
    {
        pthread_mutex_lock (&mutex);
        cake_count += 1;
        printf ("produce thread: cake_count = %d\n", cake_count);
        pthread_mutex_unlock (&mutex);
        sleep (3);
    }
    printf ("sub thread1: exit\n");
    pthread_exit (NULL);
}
static void *consume_thread (void *arg)
{
    int consume_unit = *(int*)arg;
    while (1)
    {
        pthread_mutex_lock (&mutex);
        if (cake_count == 0)
        {
            pthread_mutex_unlock (&mutex);
            continue;
        }
        cake_count -= consume_unit;
        printf ("consume thread consume 1: cake_count = %d\n", cake_count);
        pthread_mutex_unlock (&mutex);
        sleep (1);
    }
    printf ("sub thread2: exit\n");
    pthread_exit (NULL);
}

int main (void)
{
    pthread_t tid1, tid2;

    pthread_mutex_init (&mutex, NULL);

    int arg1 = 10;
    int arg2 = 1;
    pthread_create (&tid1, NULL, produce_thread, (void *) &arg1);
    pthread_create (&tid2, NULL, consume_thread, (void *) &arg2);
    pthread_join (tid1, NULL);
    pthread_join (tid2, NULL);
    
    pthread_mutex_destroy (&mutex);
    printf ("main thread: exit\n");
    return 0;
}