概述
//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:0,1,2,3,4,5,6,7,8,9n");
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
最后
以上就是俊逸白昼为你收集整理的linux下生产者与消费者问题代码,以及编译c代码时error:undefined reference to sem_wait 解决方法之一的全部内容,希望文章能够帮你解决linux下生产者与消费者问题代码,以及编译c代码时error:undefined reference to sem_wait 解决方法之一所遇到的程序开发问题。
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