华工操作系统实验(共33页)

1、一、实验步骤：1. 在linux下编写一个应用程序，命名为an_ch2_1b。这个程序不断地输出如下行：Those output come from child,系统时间另外写一个应用程序，命名为an_ch2_1a。这个程序创建一个子进程，执行an_ch2_1b。这个程序不断地输出如下行：Those output come from child,系统时间观察程序运行的结果，并对你看到的现象进行解释。2. 在linux环境下编写一个控制台应用程序，程序中有一个共享的整型变量shared_var，初始值为0；创建一个线程并使其立即与主线程并发执行。新创建的线程与主线程 均不断地循环，并输出shar

2、ed_var 的值。主线程在循环中不断地对shared_var 进行加1操作，即每次循环shared_var 被加1；而新创建的线程则不断地对shared_var 进行减1 操作，即每次循环shared_var 被减1。观察程序运行的结果，并对你看到的现象进行解释。二、实验数据： an_ch2_1b.cpp文件：#include <iostream>#include <string>#include <stdlib.h>#include <unistd.h>#include <time.h>using namespace std;st

3、ring getTime() /获取系统时间time_t timep;time(&timep);char tmp64;strftime(tmp,sizeof(tmp),"%Y-%m-%d%H:%M:%S",localtime(&timep);return tmp;int main()while (true)string tmn = getTime();cout << "Those output come from child," << tmn << endl;sleep(1); /为了便于截屏使用sle

4、ep()函数延迟输出return 0;an_ch2_1a.cpp文件：#include <iostream>#include <sys/types.h>#include <unistd.h>#include <cstdio>#include <stdlib.h>using namespace std;int main()pid_t pid;pid = fork();if (pid = -1) cout << "fail to create" << endl;else if (pid = 0)

5、 system("./an_ch2_1b");return 0;Consoleapp.c文件:#include <sys/types.h>#include <stdio.h>#include <unistd.h>#include <pthread.h>int shared_var = 0;void * thread(void * arg)while (1)printf("in the thread shared_var:%dn", -shared_var);int main()pthread_t pt;in

6、t ret = pthread_create(&pt, NULL, (void*)thread, NULL);if (ret != 0) printf("fail to create threadn");while (1)printf("in the main shared_var:%dn", +shared_var);pthread_join(pt, NULL);return 0；1. 生产者消费者问题（信号量）参考教材中的生产者消费者算法，创建5个进程，其中两个进程为生产者进程，3个进程为消费者进程。一个生产者进程试图不断地在一个缓冲中写入大

7、写字母，另一个生产者进程试图不断地在缓冲中写入小写字母。3个消费者不断地从缓冲中读取一个字符并输出。为了使得程序的输出易于看到结果，仿照的实例程序，分别在生产者和消费者进程的合适的位置加入一些随机睡眠时间。可选的实验：在上面实验的基础上实现部分消费者有选择地消费某些产品。例如一个消费者只消费小写字符，一个消费者只消费大写字母，而另一个消费者则无选择地消费任何产品。消费者要消费的产品没有时，消费者进程被阻塞。注意缓冲的管理。2. 用线程实现睡觉的理发师问题，（同步互斥方式采用信号量或mutex方式均可）理发师问题的描述：一个理发店接待室有n张椅子，工作室有1张椅子；没有顾客时，理发师睡觉；第一个

8、顾客来到时，必须将理发师唤醒；顾客来时如果还有空座的话，他就坐在一个座位上等待；如果顾客来时没有空座位了，他就离开，不理发了；当理发师处理完所有顾客，而又没有新顾客来时，他又开始睡觉。3. 读者写者问题教材中对读者写者问题算法均有描述，但这个算法在不断地有读者流的情况下，写者会被阻塞。编写一个写者优先解决读者写者问题的程序，其中读者和写者均是多个进程，用信号量作为同步互斥机制。1. 生产者消费者问题（pro_con.c）#include <pthread.h>#include <unistd.h> #include <semaphore.h> #includ

9、e <stdio.h>#include <stdlib.h>#include <time.h>#include <malloc.h>#define N 10 /缓冲区大小为100char *buffer;int capacity = N;sem_t mutex, empty, full;void * produce_1()while (1)sem_wait(&empty);sem_wait(&mutex);int r1 = rand() % 26; int ju = 0; for(int i = 0; i < N; i+)

10、if (bufferi = '0')bufferi = 'A' + r1;printf("生产者1号生产一个产品 : %c 剩余容量为：%dn", bufferi, -capacity); ju = 1; break; if(ju = 0)printf("没有足够容量！n");sem_post(&mutex);sem_post(&full);usleep(r1 * 100000);void * produce_2()while (1)sem_wait(&empty);sem_wait(&mu

11、tex);int r2 = rand() % 26; int ju = 0; for(int i = 0; i < N; i+) if (bufferi = '0')bufferi = 'a' + r2;printf("生产者2号生产一个产品 : %c 剩余容量为：%dn", bufferi, -capacity);ju = 1; break; if(ju = 0)printf("没有足够容量！n");sem_post(&mutex);sem_post(&full);usleep(r2 * 10000

12、0);void * consume_1()while (1)sem_wait(&full);sem_wait(&mutex);int ju = 0; for (int i = 0; i < N; i+)if (bufferi >= 'A'&&bufferi <= 'Z')printf("消费者1号消费一个产品 : %c 剩余容量为：%dn", bufferi, +capacity); bufferi = '0'ju = 1; break;if (ju = 0)printf(&q

13、uot;没有消费者1号所需的产品！n");sem_post(&mutex);sem_post(&empty);int r3 = rand() % 26;usleep(r3 * 100000);void * consume_2()while (1)sem_wait(&full);sem_wait(&mutex);int ju = 0; for (int i = 0; i < N; i+)if (bufferi >= 'a'&&bufferi <= 'z')printf("消费者2

14、号消费一个产品 : %c 剩余容量为：%dn", bufferi, +capacity); bufferi = '0' ju = 1;break;if (ju = 0)printf("没有消费者2号所需的产品！n");sem_post(&mutex);sem_post(&empty);int r4 = rand() % 26;usleep(r4 * 100000);void * consume_3()int ju = 0;while (1)sem_wait(&full);sem_wait(&mutex);int ju

15、 = 0; for (int i = 0; i < N; i+)if (bufferi >= 'A' && bufferi <= 'Z') | (bufferi >= 'a' && bufferi <= 'z')printf("消费者3号消费一个产品 : %c 剩余容量为：%dn", bufferi, +capacity); bufferi = '0'ju = 1; break;if (ju = 0)printf("没有产品

16、可以消费！n");sem_post(&mutex);sem_post(&empty);int r5 = rand() % 26;usleep(r5 * 100000);int main()buffer = (char*)malloc(N * sizeof(char*);for (int i = 0; i < N; i+)bufferi = '0'sem_init(&mutex, 1, 1);sem_init(&empty, 0, N);sem_init(&full, 0, 0); srand(time(0);pthread

17、_t tid5;pthread_attr_t attr;pthread_attr_init(&attr);pthread_create(&tid0,&attr,produce_1,NULL);pthread_create(&tid1,&attr,produce_2,NULL);pthread_create(&tid2,&attr,consume_1,NULL);pthread_create(&tid3,&attr,consume_2,NULL); pthread_create(&tid4,&attr,con

18、sume_3,NULL); for(int i=0;i<5;i+)pthread_join(tidi,NULL);return 0;2. 用线程实现睡觉的理发师问题（barber.c）#include <pthread.h>#include <stdio.h>#include <stdlib.h>#include <time.h>#include <unistd.h>#include <semaphore.h>#define N 5sem_t customer,barber;int chairs,waiting =

19、0,work = 0;pthread_mutex_t mutex;void * Barber ( ) printf("无顾客，理发师睡觉n"); while (1) sem_wait(&customer); pthread_mutex_lock(&mutex); chairs+; pthread_mutex_unlock(&mutex); work = 1; printf ( "理发师正在给一名顾客理发. %d 个顾客正在接待室等待。n", -waiting); sleep(2); printf ( "一名顾客理发完成。

20、n" ); work = 0; sem_post(&barber); if(waiting = 0)printf("无顾客，理发师睡觉n"); void * Customer ( void* arg ) int *p = (int*) arg; int x = *p; pthread_mutex_lock(&mutex); if ( chairs > 0 ) chairs-; sem_post(&customer); if(waiting = 0 && work = 0) printf( "第 %d 个顾客进

21、来 , 唤醒理发师.n" , +x); waiting+; else printf ( "第 %d 个顾客进来 , %d 个顾客正在接待室等待.n" , x+1 , +waiting ); pthread_mutex_unlock(&mutex); sem_wait(&barber); else pthread_mutex_unlock(&mutex); printf ( "第 %d 个顾客进来，没有座位而离开！n" , x+1 ); int main ( ) sem_init (&customer , 0 ,

22、0 ); sem_init (&barber , 0 , 1 ); chairs = N; pthread_t bar; pthread_t cusN*100; int cus_idN*100; pthread_create ( &bar , NULL , Barber , NULL ); int i; srand(time(0); for ( i = 0 ; i < N*100 ; i + ) usleep(100000*(rand() % 30); cus_idi = i; pthread_create ( &cusi , NULL , Customer ,

23、&cus_idi ); pthread_join ( bar , NULL ); for ( i = 0 ; i < N*100 ; i+ ) pthread_join ( cus_idi , NULL ); 3. 读者写者问题（reader_writer.c）# include <stdio.h># include <stdlib.h># include <time.h># include <sys/types.h># include <pthread.h># include <semaphore.h>#

24、include <string.h># include <unistd.h>sem_t RWMutex, mutex1, mutex2, mutex3, wrt;int writeCount, readCount;struct data int id;int lastTime;void* Reader(void* param)int id = (struct data*)param)->id;int lastTime = (struct data*)param)->lastTime;printf("读进程 %d 等待读入n", id);s

25、em_wait(&mutex3);sem_wait(&RWMutex);sem_wait(&mutex2);readCount+;if(readCount = 1) sem_wait(&wrt);sem_post(&mutex2);sem_post(&RWMutex);sem_post(&mutex3);printf("读进程 %d 开始读入，%d 秒后完成n", id, lastTime);sleep(lastTime);printf("读进程 %d 完成读入n", id);sem_wait(&a

26、mp;mutex2);readCount-;if(readCount = 0) sem_post(&wrt);sem_post(&mutex2);pthread_exit(0);void* Writer(void* param) int id = (struct data*)param)->id;int lastTime = (struct data*)param)->lastTime;printf("写进程 %d 等待写入n", id);sem_wait(&mutex1);writeCount+;if(writeCount = 1) s

27、em_wait(&RWMutex);sem_post(&mutex1);sem_wait(&wrt);printf("写进程 %d 开始写入，%d 秒后完成n", id, lastTime);sleep(lastTime);printf("写进程 %d 完成写入n", id);sem_post(&wrt);sem_wait(&mutex1);writeCount-;if(writeCount = 0) sem_post(&RWMutex);sem_post(&mutex1);pthread_exit

28、(0);int main() pthread_t tid;pthread_attr_t attr;pthread_attr_init(&attr); sem_init(&mutex1, 0, 1); sem_init(&mutex2, 0, 1); sem_init(&mutex3, 0, 1); sem_init(&wrt, 0, 1); sem_init(&RWMutex, 0, 1); readCount = writeCount = 0;int id = 0; srand(time(0);while(1) int role = rand(

29、) % 100; int lastTime = rand() % 10; id+;struct data* d = (struct data*)malloc(sizeof(struct data);d->id = id;d->lastTime = lastTime;if(role < 50) /读 printf("创建读进程，PID ： %dn", id);pthread_create(&tid, &attr, Reader, d);else if(role >= 50) /写 printf("创建写进程，PID ： %dn

30、", id);pthread_create(&tid, &attr, Writer, d); sleep(rand() % 8);return 0; 1. 实现一个“difftree”命令，其功能是比较两个目录下的文件结构和文件信息。当在命令行方式下执行“difftree <dir1> <dir2>”命令时，能够比较目录dir1和 目录dir2是否具有相同的结构，对相同的部分，进一步比较相同文件名的文件内容。列出比较的文件系统结构图。本实验是对单个文件比较的扩展，设计中需要考虑目录操作。difftree.c#include <stdio.

31、h>#include <stdlib.h>#include <unistd.h>#include <fcntl.h>#include <errno.h>#include <string.h>#include <dirent.h>int filelevel1 = 0, filelevel2 = 0;char DIRNAME1256, DIRNAME2256;void my_error(const char *strerr) perror(strerr); exit(1);void findfile(char fileN

32、ame1, char direntName1, char direntName2) char command512 = "diff " DIR *p_dir = NULL; struct dirent *p_dirent = NULL; p_dir = opendir(direntName2); if(p_dir = NULL) my_error("opendir error"); while(p_dirent = readdir(p_dir) != NULL) char *backupDirName = NULL; if(p_dirent->d_

33、name0 = '.')continue; int i; if(p_dirent->d_type = DT_DIR) int curDirentNameLen = strlen(direntName2) + 1; backupDirName = (char *)malloc(curDirentNameLen); memset(backupDirName, 0, curDirentNameLen); memcpy(backupDirName, direntName2, curDirentNameLen); strcat(direntName2, "/")

34、; strcat(direntName2, p_dirent->d_name); findfile(fileName1, direntName1, direntName2); memcpy(direntName2, backupDirName, curDirentNameLen); free(backupDirName); backupDirName = NULL; else if(!strcmp(fileName1, p_dirent->d_name) char FileDirName256; strcpy(FileDirName, direntName2); int curDi

35、rentNameLen = strlen(direntName2) + 1; backupDirName = (char *)malloc(curDirentNameLen); memset(backupDirName, 0, curDirentNameLen); memcpy(backupDirName, FileDirName, curDirentNameLen); strcat(FileDirName, "/"); strcat(FileDirName, p_dirent->d_name); strcat(command, direntName1); strca

36、t(command, " "); strcat(command, FileDirName); printf("%s%sn", p_dirent->d_name,"文件相同，比较："); system(command); closedir(p_dir); void comparefile(char direntName1, char direntName2) char command512; DIR *p_dir = NULL; struct dirent *p_dirent = NULL; p_dir = opendir(dir

37、entName1); if(p_dir = NULL) my_error("opendir error"); while(p_dirent = readdir(p_dir) != NULL) char *backupDirName = NULL; if(p_dirent->d_name0 = '.') continue; int i; if(p_dirent->d_type = DT_DIR) int curDirentNameLen = strlen(direntName1) + 1; backupDirName = (char *)mallo

38、c(curDirentNameLen); memset(backupDirName, 0, curDirentNameLen); memcpy(backupDirName, direntName1, curDirentNameLen); strcat(direntName1, "/"); strcat(direntName1, p_dirent->d_name); comparefile(direntName1, direntName2); memcpy(direntName1, backupDirName, curDirentNameLen); free(backu

39、pDirName); backupDirName = NULL; else char FileDirName256; strcpy(FileDirName, direntName1); int curDirentNameLen = strlen(direntName1) + 1; backupDirName = (char *)malloc(curDirentNameLen); memset(backupDirName, 0, curDirentNameLen); memcpy(backupDirName, FileDirName, curDirentNameLen); strcat(File

40、DirName, "/"); strcat(FileDirName, p_dirent->d_name); findfile(p_dirent->d_name, FileDirName, direntName2); closedir(p_dir); void PrintDirentStruct(char direntName, int level) DIR *p_dir = NULL; struct dirent *p_dirent = NULL; p_dir = opendir(direntName); if(p_dir = NULL) my_error(&q

41、uot;opendir error"); while(p_dirent = readdir(p_dir) != NULL) char *backupDirName = NULL; if(p_dirent->d_name0 = '.') continue; int i; for(i = 0; i < level; i+) printf(""); printf(" "); printf(" "); printf("%sn", p_dirent->d_name); /如果目录项

42、仍是一个目录的话，进入目录 if(p_dirent->d_type = DT_DIR) int curDirentNameLen = strlen(direntName) + 1; backupDirName = (char *)malloc(curDirentNameLen); memset(backupDirName, 0, curDirentNameLen); memcpy(backupDirName, direntName, curDirentNameLen); strcat(direntName, "/"); strcat(direntName, p_dir

43、ent->d_name); level += 1; PrintDirentStruct(direntName, level); memcpy(direntName, backupDirName, curDirentNameLen); free(backupDirName); backupDirName = NULL; closedir(p_dir); int main() memset(DIRNAME1, 0, sizeof(DIRNAME1); memset(DIRNAME2, 0, sizeof(DIRNAME2); scanf("%s%s", DIRNAME1,

44、 DIRNAME2); printf("%sn", "第一个目录："); PrintDirentStruct(DIRNAME1, filelevel1); printf("n"); printf("%sn", "第二个目录："); PrintDirentStruct(DIRNAME2, filelevel2); if(filelevel1 = filelevel2)printf("n两个目录文件结构相同n"); comparefile(DIRNAME1, DIRNAME2);

45、 return 0;1. 在linux中实现一个命令执行程序doit，它执行命令行参数中的命令之后统计1) 命令执行占用的CPU时间(包括用户态和系统态时间，以毫秒为单位)，2) 命令执行的时间，3) 进程被抢占的次数，4) 进程主动放弃CPU的次数，5) 进程执行过程中发生缺页的次数2. 在linux中实现一个简单的命令解释程序，功能要求：1) 同时支持内部命令和外部命令，内部命令支持两个（cd、exit）2) 支持后台命令提示：实验中可能用到的系统调用如下： fork() 创建一个新进程 getrusage() 取得进程的资源使用情况 gettimeofday() 取当前的时间 execv

46、e() 装入一个程序并执行 wait() 等待子进程结束 chdir() 改变进程的工作目录 strtok() 字符串解析1. doit.c #include<sys/types.h>#include<sys/time.h>#include<sys/resource.h>#include<string.h>#include<stdlib.h>#include<errno.h>#include<stdio.h>extern int errno;int main(int argc, char*argv)char c

47、ommand200;struct timeval start, end;strcpy(command, argv1);for (int i = 2; i < argc; i+)strcat(command, " ");strcat(command, argvi);gettimeofday(&start, NULL);system(command);gettimeofday(&end, NULL);if (errno != 0)printf("error:%sn", strerror(errno);exit(0);elsestruct

48、 rusage result;memset(&result, 0, sizeof(struct rusage);getrusage(RUSAGE_CHILDREN, &result);double time = (double)end.tv_sec - (double)start.tv_sec) * 1000 + (double)end.tv_usec - (double)start.tv_usec) / 1000;printf("n命令执行占用的CPU时间: 用户态:%f毫秒t系统态:%f毫秒n",(double)result.ru_utime.tv_usec / 1000 + (double)result.ru_utime.tv_sec * 1000, (double)result.ru_stime.tv_usec / 1000 + (double)result.ru_stime.tv_sec * 1000);printf("命令执