操作系統實驗二：進程、線程之間的同步

二、進程、線程之間的同步

1。生產者消費者問題（信號量+mutex）

參考教材中的生產者消費者算法，創建5個進程，其中兩個進程爲生產者進程，3個進程爲消費者進程。一個生產者進程試圖不斷地在一個緩衝中寫入大寫字母，另一個生產者進程試圖不斷地在緩衝中寫入小寫字母。3個消費者不斷地從緩衝中讀取一個字符並輸出。爲了使得程序的輸出易於看到結果，仿照閱讀材料中的實例程序，分別在生產者和消費者進程的合適的位置加入一些隨機睡眠時間。

可選的實驗：在上面實驗的基礎上實現部分消費者有選擇地消費某些產品。例如一個消費者只消費小寫字符，一個消費者只消費大寫字母，而另一個消費者則無選擇地消費任何產品。消費者要消費的產品沒有時，消費者進程被阻塞。注意緩衝的管理。

int i; hofempty=CreateSemaphore(NULL,nSize,nSize,"empty"); //創建信號量empty if(hofempty == false ) { printf("MainPro create empty failed.../n"); exit(1); } hoffull=CreateSemaphore(NULL,0,nSize,"full"); //創建信號量full if(hoffull == false ) { printf("MainPro create full failed.../n"); exit(1); } hofmutex=CreateMutex(NULL,false,"mutex"); //創建互斥變量mutex if(hofmutex == false ) { printf("MainPro create mutex failed.../n"); exit(1); } // 創建共享緩衝區 (HANDLE)0xFFFFFFFF HANDLE hbuffer=CreateFileMapping(INVALID_HANDLE_VALUE,NULL,PAGE_READWRITE,0,nSize*sizeof(char), "Buffer"); if(hbuffer == NULL ) { printf("Create Shared Buffer failed !/n"); exit(1); } HANDLE hiter = CreateFileMapping((HANDLE)0xFFFFFFFF,NULL,PAGE_READWRITE,0,sizeof(UINT),"Iter"); if(hiter == NULL ) { printf("Create Shared Iterater failed !/n"); exit(1); } printf("Init iter !/n"); int* iter=(int *)MapViewOfFile(hiter,FILE_MAP_ALL_ACCESS,0,0,sizeof(UINT)); *iter = 0; UnmapViewOfFile(hiter); PROCESS_INFORMATION pi[5]; STARTUPINFO si0={sizeof(si0)}; CreateProcess(NULL,"producer.exe 0", NULL,NULL,FALSE,0,NULL,NULL,&si0,&pi[0]); //producer_0 STARTUPINFO si1={sizeof(si1)}; CreateProcess(NULL,"producer.exe 1", NULL,NULL,FALSE,0,NULL,NULL,&si1,&pi[1]); //producer_1 STARTUPINFO si2={sizeof(si2)}; CreateProcess(NULL,"consumer.exe 0", NULL,NULL,FALSE,0,NULL,NULL,&si2,&pi[2]); //consumer_0 STARTUPINFO si3={sizeof(si3)}; CreateProcess(NULL,"consumer.exe 1", NULL,NULL,FALSE,0,NULL,NULL,&si3,&pi[3]); //consumer_1 STARTUPINFO si4={sizeof(si4)}; CreateProcess(NULL,"consumer.exe 2", NULL,NULL,FALSE,0,NULL,NULL,&si4,&pi[4]); //consumer_2 for(i=0;i<5;i++) hofp[i]=pi[i].hProcess; CloseHandle(hofempty); CloseHandle(hoffull); CloseHandle(hofmutex);

2。用信號量和mutex方式實現睡覺的理髮師問題

//創建等待服務的顧客信號量 h_cust=CreateSemaphore(NULL, 0, CHAIRS, "customer"); if(h_cust == NULL ) { printf("Create Semaphore failed .../n"); exit(1); } //創建等待顧客的理髮師信號量 h_barb=CreateSemaphore(NULL, 1, 1, "barber"); if(h_barb == NULL ) { printf("Create Semaphore failed .../n"); exit(1); } //控制互斥訪問 h_mutex=CreateMutex (NULL,FALSE,"mutex"); if(h_mutex == NULL ) { printf("Create Mutex failed .../n"); exit(1); } //理髮師線程 HANDLE h_pbarb=CreateThread(NULL,0,b,NULL,0,NULL); if(h_pbarb== NULL) { printf("Create Thread failed .../n"); exit(1); } HANDLE h_pcust=NULL; while(1) { h_pcust=CreateThread(NULL,0,c,NULL,0,NULL); Sleep(rand()% 2000); if(h_pcust== NULL) { printf("Create Thread failed .../n"); exit(1); } } WaitForSingleObject(h_pbarb,INFINITE);

3。讀者寫者問題

教材和相關的閱讀材料中對讀者寫者問題算法均有描述，但這個算法在不斷地有讀者流的情況下，寫者會被阻塞。編寫一個寫者優先解決讀者寫者問題的程序，其中讀者和寫者均是多個進程，用信號量作爲同步互斥機制。

//創建各互斥信號量------------------------------------------------------ HANDLE h_mutexWriterCount,h_mutexReaderCount, h_libAccess,h_writerAccess; h_mutexWriterCount = CreateSemaphore(NULL,1,1, "mutexWriterCount"); h_mutexReaderCount = CreateSemaphore(NULL,1,1, "mutexReaderCount"); h_libAccess = CreateSemaphore(NULL,1,1, "libAccess");; h_writerAccess = CreateSemaphore(NULL,1,1, "writerAccess"); //創建進程共享的字符lib,並初始化未'#'------------------------------------ HANDLE h_lib; h_lib = CreateFileMapping((HANDLE)0xFFFFFFFF,NULL,PAGE_READWRITE,0,sizeof(char),"lib"); if(h_lib == NULL) { printf("Create shared lib failed ...!/n"); exit(1); } char * lib=(char*)MapViewOfFile(h_lib, FILE_MAP_ALL_ACCESS, 0, 0, sizeof(char)); (*lib)= '#'; UnmapViewOfFile(h_lib); //創建進程共享的整形讀者計數器readerCount,並初始化未0------------------------------------ HANDLE h_readerCount; h_readerCount = CreateFileMapping((HANDLE)0xFFFFFFFF,NULL,PAGE_READWRITE,0,sizeof(UINT),"readerCount"); if(h_readerCount== NULL) { printf("Create readerCount failed ...!/n"); exit(1); } UINT * readerCount =(UINT *)MapViewOfFile(h_readerCount, FILE_MAP_ALL_ACCESS, 0, 0, sizeof(UINT)); (*readerCount)= 0 ; UnmapViewOfFile(h_readerCount); //創建進程共享的整形寫者計數器readerCount,並初始化未0------------------------------------ HANDLE h_writerCount; h_writerCount = CreateFileMapping((HANDLE)0xFFFFFFFF,NULL,PAGE_READWRITE,0,sizeof(UINT),"writerCount"); if(h_readerCount== NULL) { printf("Create writerCount failed ...!/n"); exit(1); } UINT * writerCount =(UINT *)MapViewOfFile(h_readerCount, FILE_MAP_ALL_ACCESS, 0, 0, sizeof(UINT)); (*writerCount)= 0 ; UnmapViewOfFile(h_writerCount); //創建3個讀者進程和2個寫者進程------------------------------------------- bool succ=false; HANDLE h_p[5]; PROCESS_INFORMATION pi[5]; STARTUPINFO si_r1={sizeof(si_r1)}; succ= CreateProcess("reader.exe",//讀者1 "1", NULL, NULL, FALSE, 0, NULL, NULL, &si_r1, &pi[0]); if(!succ){printf("Create Process 1 failed/n"); exit(1);} STARTUPINFO si_r2={sizeof(si_r2)}; succ= CreateProcess("reader.exe",//讀者2 "2", NULL, NULL, FALSE, 0, NULL, NULL, &si_r2, &pi[1]); if(!succ){printf("Create Process 2 failed/n"); exit(1);} STARTUPINFO si_r3={sizeof(si_r3)}; succ= CreateProcess("reader.exe",//讀者3 "3", NULL, NULL, FALSE, 0, NULL, NULL, &si_r3, &pi[2]); if(!succ){printf("Create Process 3 failed/n"); exit(1);} STARTUPINFO si_w1={sizeof(si_w1)}; succ= CreateProcess("writer.exe",//寫者1 "1", NULL, NULL, FALSE, 0, NULL, NULL, &si_w1, &pi[3]); if(!succ){printf("Create Process 4 failed/n"); exit(1);} STARTUPINFO si_w2={sizeof(si_w2)}; succ= CreateProcess("writer.exe",//寫者2 "2", NULL, NULL, FALSE, 0, NULL, NULL, &si_w2, &pi[4]); if(!succ){printf("Create Process 5 failed/n"); exit(1);} for(int i=0;i<5;i++) h_p[i]=pi[i].hProcess; getchar(); WaitForMultipleObjects(5,h_p,TRUE,INFINITE);