生产者消费者问题代码示例

生产者消费者问题是一个经典的并发实例，因为项目需要代码优化实现程序并行，因此直接学习这一部分知识拿来用作代码的优化，也是边学习边理解。程序菜鸟一只，只希望尽快入门，早点得道。

这段代码是从别人处偷来的，但是别人的源代码无法执行，我算是改正了之后的，经过自己亲自实验是保证可以运行的。

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

#define M 32             /*缓冲数目*/

#define P(x)     sem_wait(&x)//使信号量-1
#define V(x)    sem_post(&x)//使信号量+1

int in = 0;         /*生产者放置产品的位置*/
int out = 0;             /*消费者取产品的位置*/

int buff[M] = {0};     /*缓冲初始化为0， 开始时没有产品*/

sem_t empty_sem;         /*同步信号量，当满了时阻止生产者放产品*/
sem_t full_sem;         /*同步信号量，当没产品时阻止消费者消费*/
pthread_mutex_t mutex; /*互斥信号量， 一次只有一个线程访问缓冲*/

/*
*output the buffer
*/
void print()
{
int i;
for(i = 0; i < M; i++)
printf("%d ", buff[i]);
printf("\n");
}

/*
*producer
*/
void *producer(void *arg)
{
for(;;)
{
sleep(1);

P(empty_sem);
pthread_mutex_lock(&mutex);

in = in % M;
printf("(+)produce a product. buffer:");

buff[in] = 1;
print();
++in;

pthread_mutex_unlock(&mutex);
V(full_sem);
}
}

/*
*consumer
*/
void *consumer(void *arg)
{
for(;;)
{
sleep(2);

P(full_sem);
pthread_mutex_lock(&mutex);

out = out % M;
printf("(-)consume a product. buffer:");

buff[out] = 0;
print();
++out;

pthread_mutex_unlock(&mutex);
V(empty_sem);
}
}

void sem_mutex_init()
{
/*
*semaphore initialize
*/
int init1 = sem_init(&empty_sem, 0, M);//sem_init()函数用作进行信号量的初始化，第三个参数分别代表初始化之后的值，M就代表空余的位置，可以
//使生产者生产这么多产品，每次执行P操作，其值减一直到为0，只要大于零就可以继续执行该线程。
int init2 = sem_init(&full_sem, 0, 0);//full_sem信号被初始化为0，代表此时仓库中的产品个数。
if( (init1 != 0) && (init2 != 0))
{
printf("sem init failed \n");
exit(1);
}
/*
*mutex initialize
*/
int init3 = pthread_mutex_init(&mutex, NULL);
if(init3 != 0)
{
printf("mutex init failed \n");
exit(1);
}

}
int main()
{
pthread_t id1;
pthread_t id2;
int i;
int ret;

sem_mutex_init();

/*create the producer thread*/
ret = pthread_create(&id1, NULL, producer, NULL);
if(ret != 0)
{
printf("producer creation failed \n");
exit(1);
}

/*create the consumer thread*/
ret = pthread_create(&id2, NULL, consumer, NULL);
if(ret != 0)
{
printf("consumer creation failed \n");
exit(1);
}

pthread_join(id1,NULL);
pthread_join(id2,NULL);

exit(0);
}

该程序编译的时候注意在编译命令后加 -lpthread，

运行之后，结果如下：

(+)produce a product. buffer:1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
(-)consume a product. buffer:0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
(+)produce a product. buffer:0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
(+)produce a product. buffer:0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
(-)consume a product. buffer:0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
(+)produce a product. buffer:0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
(+)produce a product. buffer:0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
(-)consume a product. buffer:0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
(+)produce a product. buffer:0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
(+)produce a product. buffer:0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
(-)consume a product. buffer:0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
(+)produce a product. buffer:0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
(+)produce a product. buffer:0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
(-)consume a product. buffer:0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
(+)produce a product. buffer:0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
(+)produce a product. buffer:0 0 0 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
(-)consume a product. buffer:0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
(+)produce a product. buffer:0 0 0 0 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
(+)produce a product. buffer:0 0 0 0 0 0 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
(-)consume a product. buffer:0 0 0 0 0 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
(+)produce a product. buffer:0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
(+)produce a product. buffer:0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
(-)consume a product. buffer:0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

因为生产和消费均为无限循环，可以执行下去查看结果。