1. C/C++

多线程之生产者消费者问题

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使用Linux平台的pthread库进行多线程编程实现生产者消费者问题。

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

#define P sem_wait
#define V sem_post

sem_t empty;
sem_t full;

pthread_mutex_t mutex;
int val = 0;

void *provider(){
    while (1)
    {
        P(&empty);
        pthread_mutex_lock(&mutex);
        val++;
        printf("provider: provide one good!/now have %d\n",val);
        pthread_mutex_unlock(&mutex);
        V(&full);
        sleep(3);
    }
}

void *consumer(){
    while (1)
    {
        P(&full);
        pthread_mutex_lock(&mutex);
        val--;
        printf("consumer: consume one good!/now have %d\n",val);
        pthread_mutex_unlock(&mutex);
        V(&empty);
        sleep(1);
    }
}

int main(int argc, char const *argv[])
{
    pthread_mutex_init(&mutex,NULL);
    sem_init(&empty,0,20);
    sem_init(&full,0,0);
    pthread_t t1;
    pthread_t t2;
    pthread_create(&t1,NULL,provider,NULL);
    pthread_create(&t1,NULL,consumer,NULL);
    pthread_join(t1,NULL);  // wiat t1 end
    pthread_join(t2,NULL);  // wait t2 end
    pthread_mutex_destroy(&mutex);
    return 0;
}

定义了两个信号量full，empty和一个互斥变量mutex。其中full和empty用于实现同步，mutex用于实现互斥访问val进行存取（即改变val的值）。

由main方法中的**sem_init(&empty,0,20)**可以看出val的取值范围为[0-20],程序开始时初始化val值为0，full为0，empty为20。然后起t1，t2两个线程分别为生产者和消费者线程。

  • 生产者线程主要是先申请（P)一个empty的空间然后互斥访问val进行生产（即val++）生产完成后释放（V）full表示val中已经有一个生产完成的资源了。
  • 消费者线程主要是先申请（P)一个full的空间然后互斥访问val进行消费（即val–）消费完成后释放（V）empty表示val中已经有一个资源被消费了。

由于消费者线程中sleep(3);而生产者线程中为slee(1);

所以执行结果为：

michael@DESKTOP-6N96H5I:/mnt/e/test$ gcc threadTest.c -lpthread && ./a.out
provider: provide one good!/now have 1
consumer: consume one good!/now have 0
provider: provide one good!/now have 1
consumer: consume one good!/now have 0
provider: provide one good!/now have 1
consumer: consume one good!/now have 0
provider: provide one good!/now have 1
consumer: consume one good!/now have 0
provider: provide one good!/now have 1
consumer: consume one good!/now have 0
provider: provide one good!/now have 1
consumer: consume one good!/now have 0

如果将其对调即消费者线程中sleep(1);而生产者线程中为slee(3);则执行结果为：

michael@DESKTOP-6N96H5I:/mnt/e/test$ gcc threadTest.c -lpthread && ./a.out
provider: provide one good!/now have 1
consumer: consume one good!/now have 0
provider: provide one good!/now have 1
provider: provide one good!/now have 2
consumer: consume one good!/now have 1
provider: provide one good!/now have 2
provider: provide one good!/now have 3
provider: provide one good!/now have 4
consumer: consume one good!/now have 3
provider: provide one good!/now have 4
provider: provide one good!/now have 5
provider: provide one good!/now have 6
consumer: consume one good!/now have 5
provider: provide one good!/now have 6
provider: provide one good!/now have 7
provider: provide one good!/now have 8
consumer: consume one good!/now have 7
provider: provide one good!/now have 8
provider: provide one good!/now have 9
provider: provide one good!/now have 10
consumer: consume one good!/now have 9
provider: provide one good!/now have 10
provider: provide one good!/now have 11
provider: provide one good!/now have 12
consumer: consume one good!/now have 11
provider: provide one good!/now have 12
provider: provide one good!/now have 13
provider: provide one good!/now have 14
consumer: consume one good!/now have 13
provider: provide one good!/now have 14
provider: provide one good!/now have 15
provider: provide one good!/now have 16
consumer: consume one good!/now have 15
provider: provide one good!/now have 16
provider: provide one good!/now have 17
provider: provide one good!/now have 18
consumer: consume one good!/now have 17
provider: provide one good!/now have 18
provider: provide one good!/now have 19
provider: provide one good!/now have 20
consumer: consume one good!/now have 19
provider: provide one good!/now have 20
consumer: consume one good!/now have 19
provider: provide one good!/now have 20

P.s.其实mutex也可以用semphore来定义即

sem_t mutex;

sem_init(&mutex,0,1);

这样PV操作更加统一！

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