//本文的詳細講解內容請大家下載word文件：http://download.csdn.net/detail/chenqiai0/4611801
#include <stdio.h>
#include <pthread.h>//執行緒
#include <semaphore.h>//因為用到了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;
//訊號量的資料型別為結構sem_t，它本質上是一個長整型的數
sem_t full;//訊號量 
sem_t empty;
pthread_mutex_t mutex;//pthread_mutex_t是一個結構,是互斥鎖


void producer(void)
{ 
  int m;
  while(i<10)
  {
   pthread_mutex_lock(&mutex);//宣告開始用互斥鎖上鎖
   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函式的作用是給訊號量的值加上一個“1”，它是一個“原子操作"即同時對同一個訊號量做加“1”操作的兩個執行緒是不會衝突的
   pthread_mutex_unlock(&mutex);//宣告開始用互斥鎖解鎖
  }
}

void consumer(void)
{ 
  int m;
  while(j<10)
  {
   int x;
   pthread_mutex_lock(&mutex);
   sem_wait(&full);// sem_wait函式也是一個原子操作，它的作用是從訊號量的值減去一個“1”，但它永遠會先等待該訊號量為一個非零值才開始做減法。
   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() 初始化一個定位在 sem 的匿名訊號量。value 引數指定訊號量的初始值。 pshared 引數指明訊號量是由程序內執行緒共享，還是由程序之間共享。如果 pshared 的值為 0，那麼訊號量將被程序內的執行緒共享，並且應該放置在所有執行緒都可見的地址上
  sem_init(&empty,0,10);
  pthread_create(&thread1,NULL,(void *)producer,NULL);//函式pthread_create用來建立一個執行緒
  pthread_create(&thread2,NULL,(void *)consumer,NULL);//函式pthread_join用來等待一個執行緒的結束
  pthread_join(thread1,NULL);
  pthread_join(thread2,NULL);
}
 

當我們在終端輸入：gcc -o producer_consumer.out producer_consumer.c

就會出現錯誤：

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'

這個是因為pthread並非Linux系統的預設庫，編譯時注意加上-lpthread引數，以呼叫連結庫

我們再一次在終端輸入：gcc -o producer_consumer.out producer_consumer.c -lpthread

此時編譯正確

執行：在終端輸入：./producer_consumer.out

輸出為：

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