1、PV操作（哲學家問題）給每個哲學家編號，規定奇數號的哲學家先拿他的左筷子，然后再去拿他的右筷子；而偶 數號的哲學家則相反。這樣總可以保證至少有一個哲學家可以進餐。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 *);全局變量CRITICAL_SECTION c

2、rout;/這個變量用來保證輸出時不會竟7CRITICAL_SECTION fork5;/定義五個臨界變量，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);初始化臨界變量for(i=0;i5;i+)InitializeCriticalSection(fork + i):/創建五個哲學家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)/如果線程創建失敗返回Tcerr error while create thread i endl;cerr ，zerror code : GetLastError 0 endl;等待所有線程結束retval = WaitForMultipleObjects (5, hthread, true, INFINITE); /等待多個線程 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) /關閉句柄cerr error while close thread iendl;cerr ，zerror code: /zGetLastError0endl;return 0;DWORD WINAPI ph訂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); / /放下右邊的筷子/print(k, give left);LeaveCriticalSection(fork + k); /放下左邊的筷子/./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) ;/獲得左邊的筷子/print(k, take left);EnterCriticalSection(fork + (k + 1)%5);/獲得右邊的筷子/print(k, take right);void print(int who, const char *str)tvpedefint semaphore; /*信號量是一種特殊的整型變量*/semaphore mutex=l; /* 互斥訪問 */semaphore emptv=N;/*記錄緩沖區中空的槽數*/semaphore fiill=0; /*記錄緩沖區中滿的槽數*/semaph

8、ore buffN; /*有N個槽數的緩沖區bufN,并實現循環緩沖隊列*/ 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 滿；*/ 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);/*生產者*/void producer(void) mt item;wlule(l)produce_item(&item);p(&emptv);/*遞減空槽數*/p(&mutex);/*進入臨界區*/enter_item(item); /*將一個新的數據項放入緩沖區*/ V(&mutex);/*離開臨界區*/v(&hill);/*遞增滿槽數/if(fiill=N)/*若緩沖區滿的話，喚醒消費者進程*/consumerQ;/*消費者*/void consumer(void) mt get_

11、item;wlule(l)p(&fbll);/*遞減滿槽數*/p(&mutex);/*進入臨界區/remove_item(&g亡/*從緩沖區中取走一個數據項*/ v(&mutex);/*離開臨界區*/v(&empty);/*遞增空槽數*/consume_item(get_item);/*對數據項進行操作(消費)*/ if(empty=N)/*若緩沖區全空的話，喚生產者進程*/pioducerO;/*調用生產者一消費者進程實現進程間同步*/ 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傲軌拼首豐：: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生產者消費者問題解決方法二：因為實際情況是生產消費速度不會一致，也不會想方法一一樣，消費完后在生產。我們 生產者要保持滿足消費者的需求。調整下面的數值，可以發現，當生產者個數多于消費者個 數時，生產速度快，生產者經常等待消費者：反之，消費者經常等待，可以保證消費者可以 一直消費。實現代碼：?tinclude #include const unsigned short SIZE_OF_BUFFER = 10; /緩沖區長度unsigned short ProductID = 0;unsigned short ConsumelD = 0;/產品號將被消耗的產品號unsi

16、gned short in = 0;產品進緩沖區時的緩沖區下標unsigned short out = 0;產品出緩沖區時的緩沖區下標int g.bufferSIZE.OF.BUFFER;/緩沖區是個循環隊列bool g_continue = true;/控制程序結束HANDLE g_hMutex;/用于線程間的互斥HANDLE g.hFullSemaphore;當緩沖區滿時迫使生產考等待HANDLE g.hEmptySemaphore;/當緩沖區空時迫使消費者等待DWORD WINAPI Producer (LPVOID); /生產者線程DWORD WINAPI Consumer (LPVO

17、ID); /消費者線程int mainO創建各個互斥信號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);調整下而的數值，可以發現，當生產者個數多于消費者個數時，/生產速度快，生產者經常等待消費者：反之，消費者經常等待const unsigned short

18、 PRODUCERS.COUNT = 3;/生產者的個數const unsigned short CONSUMERS.COUNT = 1;/消費者的個數總的線程數const unsigned short THREADS.COUNT = PRODUCERS.COUNT+CONSUNERS.COUNT:HANDLE hThreadsPRODUCERS_COUNT; 各線程的handleDWORD producer ID CONSUMERS.COUNT; /生產者線程的標識符DWORD consumer ID THREADS.COUXT: /消費者線程的標識符創建生產者線程for (int i=O;

19、iPRODUCERS_COUNT;+i)hThreadsiJ =CreateThread(NULL, 0, Producer, NULL, 0, &producerIDi);if (hThreadsi=NULL) return -1;創建消費者線程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、按回車后終止程序運行g_continue = false;return 0;生產一個產品。簡單模擬了一下，僅輸出新產品的ID號void Produce 0std:cerr Producing +ProductID std:cerr Succeed std:endl;/把新生產的產品放入緩沖區void Append0std:cerr ，zAppending a product g_bufferlinZ = ProductID;in = (in+l)%SIZE_OF_BUFFER;std:cerr Succeed std:endl;/輸出緩沖區當前的狀態for (int i=O;iSIZE_OF_

21、BUFFER;+i) std:cout i : g_bufferi; if (i=in) std:cout - 生產； if (i=out) std:cout 消費； std:cout std:endl;從緩沖區中取出一個產品void Take 0std:cerr Taking a product ”; ConsumeID = g_bufferout:out = (out+1)%SIZE_OF_BUFFER;std:cerr Succeed std:endl;/輸出緩沖區當前的狀態for (int i=O;iSIZE_OF_BUFFER;+i) std:cout i : g_bufferi; if (i=in) std:cout - 生產； if (i=out) std:cout 消費； std:cout std:endl:/消耗一個產品voi