//本文的详细讲解内容请大家下载word文档:http://download.csdn.net/detail/chenqiai0/4611801 #include <stdio.h> #include <pthread.h>//线程 #include <semaphore.h>//因为用到了sem_init等 #include <unistd.h> #include <stdlib.h> #include <time.h> int buffer[10]={0,0,0,0,0,0,0,0,0,0}; int i=0; int j=0; //信号量的数据类型为结构sem_t,它本质上是一个长整型的数 sem_t full;//信号量 sem_t empty; pthread_mutex_t mutex;//pthread_mutex_t是一个结构,是互斥锁 void producer(void) { int m; while(i<10) { pthread_mutex_lock(&mutex);//声明开始用互斥锁上锁 buffer[i]=i+1; i=i+1; printf("\nthere are 10 blocks in the buffer:0,1,2,3,4,5,6,7,8,9\n"); printf("\nbuffer:"); for(m=0;m<10;m++) printf("%3d",buffer[m]); printf("\nthe number added by the producer is:"); printf("%d", buffer[i]); printf("\npointer is %d",i); sem_post(&full);// int sem_post(sem_t * sem);------sem_post函数的作用是给信号量的值加上一个“1”,它是一个“原子操作"即同时对同一个信号量做加“1”操作的两个线程是不会冲突的 pthread_mutex_unlock(&mutex);//声明开始用互斥锁解锁 } } void consumer(void) { int m; while(j<10) { int x; pthread_mutex_lock(&mutex); sem_wait(&full);// sem_wait函数也是一个原子操作,它的作用是从信号量的值减去一个“1”,但它永远会先等待该信号量为一个非零值才开始做减法。 if(buffer[j]==0) break; x=buffer[j]; buffer[j]=0; j=j+1; printf("\nbuffer:"); for(m=0;m<10;m++) printf("%3d",buffer[m]); printf("\n %d is removed form the buffer by consumer",x); printf("\nThe present pointer is %d",j); pthread_mutex_unlock(&mutex); sem_post(&empty); } } int main() { pthread_t thread1,thread2,thread3,thread4; sem_init(&full,0,0);//int sem_init(sem_t *sem, int pshared, unsigned int value)---------sem_init() 初始化一个定位在 sem 的匿名信号量。value 参数指定信号量的初始值。 pshared 参数指明信号量是由进程内线程共享,还是由进程之间共享。如果 pshared 的值为 0,那么信号量将被进程内的线程共享,并且应该放置在所有线程都可见的地址上 sem_init(&empty,0,10); pthread_create(&thread1,NULL,(void *)producer,NULL);//函数pthread_create用来创建一个线程 pthread_create(&thread2,NULL,(void *)consumer,NULL);//函数pthread_join用来等待一个线程的结束 pthread_join(thread1,NULL); pthread_join(thread2,NULL); }
当我们在终端输入:gcc -o producer_consumer.out producer_consumer.c
就会出现错误:
producer_consumer.c:(.text+0xc9): undefined reference to `sem_post'
/tmp/ccvFyPLL.o: In function `consumer':
producer_consumer.c:(.text+0x108): undefined reference to `sem_wait'
producer_consumer.c:(.text+0x1c6): undefined reference to `sem_post'
/tmp/ccvFyPLL.o: In function `main':
producer_consumer.c:(.text+0x1fe): undefined reference to `sem_init'
producer_consumer.c:(.text+0x21a): undefined reference to `sem_init'
producer_consumer.c:(.text+0x23f): undefined reference to `pthread_create'
producer_consumer.c:(.text+0x264): undefined reference to `pthread_create'
producer_consumer.c:(.text+0x278): undefined reference to `pthread_join'
producer_consumer.c:(.text+0x28c): undefined reference to `pthread_join'
这个是因为pthread并非Linux系统的默认库,编译时注意加上-lpthread参数,以调用链接库
我们再一次在终端输入:gcc -o producer_consumer.out producer_consumer.c -lpthread
此时编译正确
执行:在终端输入:./producer_consumer.out
输出为:
there are 10 blocks in the buffer:0,1,2,3,4,5,6,7,8,9
buffer: 1 0 0 0 0 0 0 0 0 0
the number added by the producer is:0
pointer is 1
there are 10 blocks in the buffer:0,1,2,3,4,5,6,7,8,9
buffer: 1 2 0 0 0 0 0 0 0 0
the number added by the producer is:0
pointer is 2
there are 10 blocks in the buffer:0,1,2,3,4,5,6,7,8,9
buffer: 1 2 3 0 0 0 0 0 0 0
the number added by the producer is:0
pointer is 3
there are 10 blocks in the buffer:0,1,2,3,4,5,6,7,8,9
buffer: 1 2 3 4 0 0 0 0 0 0
the number added by the producer is:0
pointer is 4
there are 10 blocks in the buffer:0,1,2,3,4,5,6,7,8,9
buffer: 1 2 3 4 5 0 0 0 0 0
the number added by the producer is:0
pointer is 5
there are 10 blocks in the buffer:0,1,2,3,4,5,6,7,8,9
buffer: 1 2 3 4 5 6 0 0 0 0
the number added by the producer is:0
pointer is 6
there are 10 blocks in the buffer:0,1,2,3,4,5,6,7,8,9
buffer: 1 2 3 4 5 6 7 0 0 0
the number added by the producer is:0
pointer is 7
there are 10 blocks in the buffer:0,1,2,3,4,5,6,7,8,9
buffer: 1 2 3 4 5 6 7 8 0 0
the number added by the producer is:0
pointer is 8
there are 10 blocks in the buffer:0,1,2,3,4,5,6,7,8,9
buffer: 1 2 3 4 5 6 7 8 9 0
the number added by the producer is:0
pointer is 9
there are 10 blocks in the buffer:0,1,2,3,4,5,6,7,8,9
buffer: 1 2 3 4 5 6 7 8 9 10
the number added by the producer is:10
pointer is 10
buffer: 0 2 3 4 5 6 7 8 9 10
1 is removed form the buffer by consumer
The present pointer is 1
buffer: 0 0 3 4 5 6 7 8 9 10
2 is removed form the buffer by consumer
The present pointer is 2
buffer: 0 0 0 4 5 6 7 8 9 10
3 is removed form the buffer by consumer
The present pointer is 3
buffer: 0 0 0 0 5 6 7 8 9 10
4 is removed form the buffer by consumer
The present pointer is 4
buffer: 0 0 0 0 0 6 7 8 9 10
5 is removed form the buffer by consumer
The present pointer is 5
buffer: 0 0 0 0 0 0 7 8 9 10
6 is removed form the buffer by consumer
The present pointer is 6
buffer: 0 0 0 0 0 0 0 8 9 10
7 is removed form the buffer by consumer
The present pointer is 7
buffer: 0 0 0 0 0 0 0 0 9 10
8 is removed form the buffer by consumer
The present pointer is 8
buffer: 0 0 0 0 0 0 0 0 0 10
9 is removed form the buffer by consumer
The present pointer is 9
buffer: 0 0 0 0 0 0 0 0 0 0
10 is removed form the buffer by consumer