用C语言实现--生产者与消费者的问题(PV操作)

用C语言实现--生产者与消费者的问题(PV操作)
2011-06-28 16:48
转载自 智达高远lee
最终编辑 智达高远lee

#include   <windows.h>  
#include   <stdio.h>  
#include   <stdlib.h>  
   
typedef   HANDLE   Semaphore;   //   信号量的Windows原型  
#define   P(S)   WaitForSingleObject(S,   INFINITE)   //   定义Windows下的P操作  
#define   V(S)   ReleaseSemaphore(S,   1,   NULL)     //   定义Windows下的V操作  
   
#define   rate   1000  
   
#define   CONSUMER_NUM   10   /*   消费者个数   */  
#define   PRODUCER_NUM   10   /*   生产者个数   */  
#define   BUFFER_NUM       4     /*   缓冲区个数   */  
   
char   *thing[10]   =   {"猪耳朵",   "牛肝",   "羊蹄",   "驴肉",   "马肉",   "狗肉",   "猴脑",   "老虎屁股",   "大象肚",   "河马大肠"};  
   
struct   Buffer  
{  
int   product[BUFFER_NUM];   //   缓冲区  
int   start,   end;                   //   两个指针  
}   g_buf;  
   
Semaphore   g_semBuffer,   g_semProduct,   g_mutex;    
   
   
//   消费者线程  
DWORD   WINAPI   Consumer(LPVOID   para)  
{  
//   i表示第i个消费者  
int   i   =   *(int   *)para;  
int   ptr;   //   待消费的内容的指针  
   
printf("     猪头-%03d:   猪头我来啦！\n",   i);  
   
Sleep(300);  
while   (1)  
{  
printf("     猪头-%03d:   我要吃.........!\n",   i);  
//   等待产品  
P(g_semProduct);    
   
//   有产品，先锁住缓冲区   g_buf  
P(g_mutex);  
   
//   记录消费的物品  
ptr   =   g_buf.start;  
   
//   再移动缓冲区指针  
g_buf.start   =   (g_buf.start+1)%BUFFER_NUM;  
   
//   让其他消费者或生产者使用   g_buf  
V(g_mutex);  
   
printf("     猪头-%03d:   我吃   buf[%d]   =   %s\n",   i,   ptr,   thing[g_buf.product[ptr]]);  
   
Sleep(rate*rand()%10+110);  
   
//   消费完毕，并释放一个缓冲  
printf("     猪头-%03d:   太好了！   buf[%d]   =   %s\n",   i,   ptr,   thing[g_buf.product[ptr]]);  
V(g_semBuffer);  
}  
return   0;  
}  
   
//   生产者线程  
DWORD   WINAPI   Producer(LPVOID   para)  
{  
int   i   =   *(int   *)para   -   CONSUMER_NUM;  
int   ptr;  
int   data;   //   产品  
   
printf("工作狂-%03d:   我来啦！\n",   i);  
   
Sleep(300);  
while   (1)  
{  
printf("工作狂-%03d:   努力！…………\n",   i);  
Sleep(rate*rand()%10+110);  
   
data   =   rand()%10;  
printf("工作狂-%03d:   发现一个东西   data   =   %s!\n",   i,   thing[data]);  
//   等待存放空间  
P(g_semBuffer);    
   
//   有地方，先锁住缓冲区   g_buf  
P(g_mutex);  
   
//   记录消费的物品  
ptr   =   g_buf.end;  
   
//   再移动缓冲区指针  
g_buf.end   =   (g_buf.end+1)%BUFFER_NUM;  
   
//   让其他消费者或生产者使用   g_buf  
V(g_mutex);  
   
printf("工作狂-%03d:   搁到   buf[%d]   =   %s\n",   i,   ptr,   thing[data]);  
g_buf.product[ptr]   =   data;  
   
Sleep(rate/2*rand()%10+110);  
   
//   放好了完毕，释放一个产品  
printf("工作狂-%03d:   buf[%d]   =   %s   放好了，大家吃！\n",   i,   ptr,   thing[g_buf.product[ptr]]);  
V(g_semProduct);  
}  
return   0;  
}  
   
int   main(int   argc,   char   *argv[])  
{  
//   线程技术，前面为消费者线程，后面为生产者线程  
HANDLE     hThread[CONSUMER_NUM+PRODUCER_NUM];     //   线程计数  
   
//srand(time());  
DWORD   tid;  
int   i=0;  
   
//   初始化信号量  
g_mutex   =   CreateSemaphore(NULL,   BUFFER_NUM,   BUFFER_NUM,   "mutexOfConsumerAndProducer");  
g_semBuffer   =   CreateSemaphore(NULL,   BUFFER_NUM,   BUFFER_NUM,   "BufferSemaphone");  
g_semProduct   =   CreateSemaphore(NULL,   0,   BUFFER_NUM,   "ProductSemaphone");  
   
if   (   !g_semBuffer   ||   !g_semProduct   ||   !g_mutex)  
{  
printf("Create   Semaphone   Error!\n");  
return   -1;  
}  
   
int   totalThreads   =   CONSUMER_NUM+PRODUCER_NUM;  
//   开启消费者线程  
printf("先请猪头们上席！\n");  
for   (i=0;   i<CONSUMER_NUM;   i++)  
{  
hThread[i]   =   CreateThread(NULL,   0,   Consumer,   &i,   0,   &tid);  
if   (   hThread[i]   )   WaitForSingleObject(hThread[i],   10);  
}  
   
printf("厨子们就位！\n");  
for   (;   i<totalThreads;   i++)  
{  
hThread[i]   =   CreateThread(NULL,   0,   Producer,   &i,   0,   &tid);  
if   (   hThread[i]   )   WaitForSingleObject(hThread[i],   10);  
}  
   
//   生产者和消费者的执行  
WaitForMultipleObjects(totalThreads,   hThread,   TRUE,   INFINITE);  
return   0;  
}

原文链接：http://hi.baidu.com/125580956/blog/item/6708fbc9def91a51f31fe72c.html