PV操作哲学家问题和生产者-消费者问题剖析

1、PV操作（哲学家问题）给每个哲学家编号，规定奇数号的哲学家先拿他的左筷子，然后再去拿他的右筷子；而偶 数号的哲学家则相反。这样总可以保证至少有一个哲学家可以进餐。include include include processhinclude include using namespace std;DWORD WINAPI philosopher(LPVOID IpParameter);void thinking(int):void eating(int);void waiting(int);void print(int ,const char *);全局变量CRITICAL_SECTION c

2、rout;/这个变量用来保证输出时不会竟7CRITICAL_SECTION fork5;/定义五个临界变量，代表五更筷J： int main(int argc, char *argv)HANDLE hthread5;int i;int arg5;int count = 5;long a=0;unsigned long retval:InitializeCriticalSection(&crout);初始化临界变量for(i=0;i5;i+)InitializeCriticalSection(fork + i):/创建五个哲学家for(i = 0; i5;i+)argi = i;hthreadi

3、 = CreateThread(NULL, 0, philosopher, (void*) (arg-i), 0, NULL); for(a=0;a30000000;a+);if( hthreadCi = INVALID_HANDLE_VALl：E)/如果线程创建失败返回Tcerr error while create thread i endl;cerr ，zerror code : GetLastError 0 endl;等待所有线程结束retval = WaitForMultipleObjects (5, hthread, true, INFINITE); /等待多个线程 for(a=0

4、;a30000000:a+);if(retval = WAIT.FAILED)cerr wait error, error code: z，GetLastError0endl;for(i = 0; i5;i+)for(a=0;a30000000;a+);i f (C loseHandl e (ht hread i ) = false) /关闭句柄cerr error while close thread iendl;cerr ，zerror code: /zGetLastError0endl;return 0;DWORD WINAPI ph订osopher(LPVOID IpParameter

5、) long a=0;int n = (int *)IpParameter)0;for(a=0;a30000000:a+):print(n, is in!);/srand(time(NULL);while(true)thinking(n); waiting(n); eating(n);print(n, is out!);return n;void thinking(int k)long a=0;for(a=0;a30000000;a+):print (k, is thinking);Sleep (1);/Sleep(rand0 %100) *5);void eating(int k)long

6、a=0;for(a=0;a30000000;a+):print (k, is eating);/Sleep(rand 0%100) *5);Sleep (1);LeaveCriticalSection(fork + (k+l)%5); / /放下右边的筷子/print(k, give left);LeaveCriticalSection(fork + k); /放下左边的筷子/./print (k, give right);void waiting(int k)long a=0;for(a=0;a30000000;a+):print(k, is waiting);Sleep (1);Enter

7、CriticalSection(fork + k) ;/获得左边的筷子/print(k, take left);EnterCriticalSection(fork + (k + 1)%5);/获得右边的筷子/print(k, take right);void print(int who, const char *str)tvpedefint semaphore; /*信号量是一种特殊的整型变量*/semaphore mutex=l; /* 互斥访问 */semaphore emptv=N;/*记录缓冲区中空的槽数*/semaphore fiill=0; /*记录缓冲区中满的槽数*/semaph

8、ore buffN; /*有N个槽数的缓冲区bufN,并实现循环缓冲队列*/ semaphore front=0, rear=0;void p(seniaphore *x) /* p 操作 */ *x=(*x)-l;void v(semaphore *v)/* v 操作 */*y=(*y)+i;void produce_item(int *item_ptr)/*pnntf(Mpioduce ail itemnn);*/*item_ptr=,m,; /* nf is Mmaii 满；*/ void enter_item(mt x)fiont=(fiont+1 )%N;buffront=x;pnn

9、tf(Mentei_item %c to buf%dn, buffront, front);void remove_item(int *yy)reai(reai+l)%N;printfCemovetem %c fiom buff%d, buffieai, rear);*yy=bufrear;buflreai; /* k is nkong 空” */ pnntf(M so the buff%d changed to empty%ciiH, fear, bufrear);void consume_item(mt y)printf(Mcosume tlie item :tlie screem pri

10、nt %cn9 y)； void producer(void); void consumer(void);/*生产者*/void producer(void) mt item;wlule(l)produce_item(&item);p(&emptv);/*递减空槽数*/p(&mutex);/*进入临界区*/enter_item(item); /*将一个新的数据项放入缓冲区*/ V(&mutex);/*离开临界区*/v(&hill);/*递增满槽数/if(fiill=N)/*若缓冲区满的话，唤醒消费者进程*/consumerQ;/*消费者*/void consumer(void) mt get_

11、item;wlule(l)p(&fbll);/*递减满槽数*/p(&mutex);/*进入临界区/remove_item(&g亡/*从缓冲区中取走一个数据项*/ v(&mutex);/*离开临界区*/v(&empty);/*递增空槽数*/consume_item(get_item);/*对数据项进行操作(消费)*/ if(empty=N)/*若缓冲区全空的话，唤生产者进程*/pioducerO;/*调用生产者一消费者进程实现进程间同步*/ niaui()producer();return 0;EEE:0Ssample1Debugsample1.execosune the item :the s

12、creen tt*enoueitem m from buf 2 Losume tlie item : tlie screen tt*enoue_iten m from buf 3 cosune the item :the screen tt*enoueitem m from buf4 cosune the item :the screen tt*enoue_iten m from buf 5 cosune the item :the screen tt*enoue_iten m from buf 6 cosune the item :the screen kemoue_item m f i*o

13、m hif T71 cosune the item :the screen tt*enoueitem m from buf 8 cosune the item :the screen tt*enoueitem m from buf 9 cosune the item :the screen 卜emoue_item m from huf0 item to to to to tocosune the enter_item enter_item enter_item enter_item enter_item傲轨拼首丰：:the screen buf 1 buf 12 buf L3 bufL4 bu

14、fL4I printnso thebufL2changedtoemptyki printinso thebuf3changedtoemptyki printnso thebuf4changedtoemptyki printnso thebufL5changedtoemptyki printnso thebuf6changedtoemptyki printnso thebufT71changedtoemptyki printnso thebuf8Jchangedtoemptyki printnso thebuf9changedtoemptyki printnso thebuf0changedto

15、emptyki printn生产者消费者问题解决方法二：因为实际情况是生产消费速度不会一致，也不会想方法一一样，消费完后在生产。我们 生产者要保持满足消费者的需求。调整下面的数值，可以发现，当生产者个数多于消费者个 数时，生产速度快，生产者经常等待消费者：反之，消费者经常等待，可以保证消费者可以 一直消费。实现代码：?tinclude #include const unsigned short SIZE_OF_BUFFER = 10; /缓冲区长度unsigned short ProductID = 0;unsigned short ConsumelD = 0;/产品号将被消耗的产品号unsi

16、gned short in = 0;产品进缓冲区时的缓冲区下标unsigned short out = 0;产品出缓冲区时的缓冲区下标int g.bufferSIZE.OF.BUFFER;/缓冲区是个循环队列bool g_continue = true;/控制程序结束HANDLE g_hMutex;/用于线程间的互斥HANDLE g.hFullSemaphore;当缓冲区满时迫使生产考等待HANDLE g.hEmptySemaphore;/当缓冲区空时迫使消费者等待DWORD WINAPI Producer (LPVOID); /生产者线程DWORD WINAPI Consumer (LPVO

17、ID); /消费者线程int mainO创建各个互斥信号g.hMutex = CreateMutex(XULL, FALSE, NULL);g.hFullSemaphore = CreateSemaphore(NULL, SIZE.OF.BUFFER-l, SIZE_OF_BUFFER-1,NULL); g_hEmp15rSemaphore = CreateSemaphore (NULL, 0, SIZE_OF_BUFFER-1, NULL);调整下而的数值，可以发现，当生产者个数多于消费者个数时，/生产速度快，生产者经常等待消费者：反之，消费者经常等待const unsigned short

18、 PRODUCERS.COUNT = 3;/生产者的个数const unsigned short CONSUMERS.COUNT = 1;/消费者的个数总的线程数const unsigned short THREADS.COUNT = PRODUCERS.COUNT+CONSUNERS.COUNT:HANDLE hThreadsPRODUCERS_COUNT; 各线程的handleDWORD producer ID CONSUMERS.COUNT; /生产者线程的标识符DWORD consumer ID THREADS.COUXT: /消费者线程的标识符创建生产者线程for (int i=O;

19、iPRODUCERS_COUNT;+i)hThreadsiJ =CreateThread(NULL, 0, Producer, NULL, 0, &producerIDi);if (hThreadsi=NULL) return -1;创建消费者线程for ( int i=0;iC0NSl3ERS_C0UNT:+i)hThreadsPRODUCERS_COUNT+i=CreateThread(NULL, 0, Consumer, NULL, 0, &consumerIDi);if (hThreadsi=NULL) return -1;while(g_continue) if(getchar()

20、按回车后终止程序运行g_continue = false;return 0;生产一个产品。简单模拟了一下，仅输出新产品的ID号void Produce 0std:cerr Producing +ProductID std:cerr Succeed std:endl;/把新生产的产品放入缓冲区void Append0std:cerr ，zAppending a product g_bufferlinZ = ProductID;in = (in+l)%SIZE_OF_BUFFER;std:cerr Succeed std:endl;/输出缓冲区当前的状态for (int i=O;iSIZE_OF_

21、BUFFER;+i) std:cout i : g_bufferi; if (i=in) std:cout - 生产； if (i=out) std:cout 消费； std:cout std:endl;从缓冲区中取出一个产品void Take 0std:cerr Taking a product ”; ConsumeID = g_bufferout:out = (out+1)%SIZE_OF_BUFFER;std:cerr Succeed std:endl;/输出缓冲区当前的状态for (int i=O;iSIZE_OF_BUFFER;+i) std:cout i : g_bufferi; if (i=in) std:cout - 生产； if (i=out) std:cout 消费； std:cout std:endl:/消耗一个产品voi