多线程第八篇:生产者消费者问题

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著名的生产者消费者问题,用到同步和互斥.

1.我们假设缓冲区大小为2,即最多只能放2个资源,并且肯定大于0.

2.生产者和消费者是可以同时进入缓冲区的.

代码设计:

1.由于window信号量在缓冲区满的时候无法阻塞,所以需要两个信号量,分别表示缓冲区剩余资源和可用资源分别用于阻塞消费者和生产者.

2.对于全局资源的存取,所有线程都要互斥.

3.对于消费者和生产者来说分别互斥自己的代码.

OK上代码:(相当easy)

#include <iostream>
#include <windows.h>
#include <process.h>
int g_count = 0;
CRITICAL_SECTION producer_critical, consumer_critical;
HANDLE producer_semaphore ,consumer_semaphore;
CRITICAL_SECTION all;
unsigned int __stdcall producer(void *)
{
int i =0;
while ( i<5 ){
WaitForSingleObject(consumer_semaphore ,INFINITE);
EnterCriticalSection(&producer_critical );
std::cout <<"生产者生产了."<<std ::endl;
++ i;
EnterCriticalSection(&all );
std::cout <<"生产者"<<++ g_count<<std ::endl;
LeaveCriticalSection(&all );
ReleaseSemaphore(producer_semaphore ,1,NULL);
LeaveCriticalSection(&producer_critical );
}
return 0;
}
unsigned int __stdcall consumer(void *)
{
int i =0;
while ( i<5 ){
WaitForSingleObject(producer_semaphore ,INFINITE);
EnterCriticalSection(&consumer_critical );
std::cout <<"消费者消费了."<<std ::endl;
++ i;
EnterCriticalSection(&all );
std::cout <<"消费者"<<-- g_count<<std ::endl;
LeaveCriticalSection(&all );
ReleaseSemaphore(consumer_semaphore ,1,NULL);
LeaveCriticalSection(&consumer_critical );
}
return 0;
}
int main ()
{
InitializeCriticalSection(& producer_critical);
InitializeCriticalSection(& consumer_critical);
InitializeCriticalSection(& all);
producer_semaphore = CreateSemaphore(NULL ,0,2,NULL);
consumer_semaphore = CreateSemaphore(NULL ,2,2,NULL);
HANDLE hproducer = (HANDLE )_beginthreadex( NULL,0,producer ,NULL,0, NULL);
HANDLE hconsumer = (HANDLE )_beginthreadex( NULL,0,consumer ,NULL,0, NULL);
WaitForSingleObject( hproducer,INFINITE );
WaitForSingleObject( hconsumer,INFINITE );
CloseHandle( hproducer);
CloseHandle( hconsumer);
return 0;
}