操作系统全英文期末考试题
《linux操作系统》期末试卷A及答案
○题号题号：分值2011－2012 学年度第一学期期末考试试题（ A 卷）( 考试时间: 90 分钟)考试科目：Linux 操作系统总分复查人100C. cat file2.txt file1.txtD. cat file1.txt > file2.txt9．为了达到使文件的所有者有读(r) 和写(w) 的许可，而其他用户只能进行只读访问，在设置文件的许可值时，应当设为：( B ) 。
A. 566B. 644C. 655D. 74410．一个文件的权限是-rw-rw-r-- ，这个文件所有者的权限是（ C ）。
A. read-onlyB. writeC. read-write号座得分○得分评卷人二、多选题（共10 题，每题 2 分，共20 分）得分评卷人线一、单选题（共10 题，每题 2 分，共20 分）1 . Redhat 9 所支持的安装方式有（BCD ）。
A 通过Telnet 进行网络安装B 从本地硬盘驱动器进行安装C 通过NFS进行网络安装D 通过HTTP进行网络安装2 . 下列哪几个符号是Linux 通配符（CD ）。
1．从后台启动进程，应在命令的结尾加上符号（ A ）名 2. 如果执行命令#chmod 746 file.txt ，那么该文件的权限是（ A ）。
姓A. rwxr--rw-B. rw-r--r--C. --xr —r wxD. rwxr--r —封3．若要使用进程名来结束进程，应使用（ A ）命令。
A. killB.psC.pssD.pstree4．以长格式列目录时，若文件test 的权限描述为：drwxrw-r-- ，则文件test 的类型及○文件主的权限是 A 。
A. 目录文件、读写执行B. 目录文件、读写：别 C. 普通文件、读写 D. 普通文件、读班密5．当字符串用单引号（’’）括起来时，SHELL 将 C 。
A. 解释引号内的特殊字符B. 执行引号中的命令C. 不解释引号内的特殊字符D. 结束进程○6．用rm命令删除非空目录需要加上哪个参数？（ A ）A. rB. fC. tD. c7 . 怎样显示当前目录（ A ）。
操作系统期末考试复习题(全)
操作系统期末考试复习题(全)注：本复习题部分参考自清华大学计算机系2019年秋季学期“操作系统”课程终极考试题目以及其他经典操作系统考试。
一、选择题1. 下列哪个不是实现进程间通信（IPC）的方式？A. 信号量B. 共享内存C. 管道D. 文件2. 下列哪个不是让文件描述符指向指定文件的函数？A. open()B. creat()C. close()D. dup()3. 下列哪个不是内存管理中的页面置换算法？A. FIFOB. LRUC. OPTD. COW4. 下列哪个不是Redis的应用场景？A. 缓存B. 计数器C. 分布式锁D. 数据库5. 下列哪个不是Linux中的调度算法？A. Round RobinB. First In First OutC. Shortest Job FirstD. Random6. 下列哪个函数可以在进程中产生子进程？A. exec()B. fork()C. spawn()D. clone()7. 下列哪个文件系统不支持软链接？A. ext2B. ext3C. NTFSD. FAT328. 下列哪个命令可以显示Linux操作系统的系统负载？A. cat /proc/loadavgB. ps -efC. topD. uptime9. 下列哪个不是Linux的文件权限？A. 读B. 写C. 移动D. 执行10. 下列哪个不是进程状态？A. 运行B. 等待C. 冻结D. 暂停二、填空题1. 若线程竞争同一资源，可能会导致______问题。
答案：死锁。
2. 在Linux中，可通过卸载模块的方式动态添加/更新/删除系统中的______。
答案：驱动。
3. Linux系统优先级高的进程通过调度机制可以抢占系统中优先级低的进程的占用资源，这种调度机制被称为______调度。
答案：抢占式。
4. 运行中的进程分为三种状态：就绪状态、运行状态、阻塞（睡眠）状态，又称为三态进程模型，俗称为______模型。
操作系统期末试卷(含答案)1
一、选择题1、在现代操作系统中引入了（），从而使并发和共享成为可能。
A.单道程序B. 磁盘C. 对象D.多道程序2、( )操作系统允许在一台主机上同时连接多台终端，多个用户可以通过各自的终端同时交互地使用计算机。
A.网络B.分布式C.分时D.实时3、从用户的观点看，操作系统是（）。
A. 用户与计算机硬件之间的接口B.控制和管理计算机资源的软件C. 合理组织计算机工作流程的软件D.计算机资源的的管理者4、当CPU处于管态时，它可以执行的指令是（）。
A. 计算机系统中的全部指令B. 仅限于非特权指令C. 仅限于访管指令D. 仅限于特权指令5、用户在程序中试图读取某文件的第100个逻辑块时，使用操作系统提供的（）接口。
A. 系统调用B.图形用户接口C.原语D.键盘命令6、下列几种关于进程的叙述，（）最不符合操作系统对进程的理解？A.进程是在多程序并行环境中的完整的程序。
B.进程可以由程序、数据和进程控制块描述。
C.线程是一种特殊的进程。
D.进程是程序在一个数据集合上运行的过程，它是系统进行资源分配和调度的一个独立单位。
7、当一个进程处于（）状态时，称其为等待（或阻塞）状态。
A. 它正等待中央处理机B. 它正等待合作进程的一个消息C. 它正等待分给它一个时间片D. 它正等待进入内存8、一个进程释放一种资源将有可能导致一个或几个进程（）。
A.由就绪变运行B.由运行变就绪C.由阻塞变运行D.由阻塞变就绪9、下面关于线程的叙述中，正确的是（）。
A.不论是系统支持线程还是用户级线程，其切换都需要内核的支持。
B.线程是资源的分配单位，进程是调度和分配的单位。
C.不管系统中是否有线程，进程都是拥有资源的独立单位。
D.在引入线程的系统中，进程仍是资源分配和调度分派的基本单位。
10、设有3个作业，它们同时到达，运行时间分别为T1、T2和T3,且T1≤T2≤T3，若它们在单处理机系统中按单道运行，采用短作业优先调度算法，则平均周转时间为（）。
操作系统期末考试试卷
操作系统期末考试试卷0250-0252操作系统试卷A0250—0252操作系统试卷A一、简答题（每题5分，共30分）1。
什么是虚拟设备？2。
What’s the differrence between a process and a program？3。
What’s Hyper—Treading technology？4.死锁的必要条件是什么？5.为什么将文件控制块分成主部和次部两部分？6.若系统有同类资源m个,被n个进程共享，问：当m〉n和m〈=n时每个进程最多可以请求多少个这类资源，使系统一定不会发生死锁？为什么?二、填空题(每空1分，共10分)1.操作系统的两个重要特性是：(1) 和（2） .2。
只能在管态下执行的指令称为(3) .处理机状态由目态转换为管态的唯一途径是(4），管态到目态的转换可以通过修改（5) 来实现。
3.进程在其生存期内可以处于如下三种基本状态之一：运行态、就绪态和等待态。
当一个就绪进程（6）时，其状态由就绪变为运行,当一个运行进程被抢占处理机时，其状态由运行变为（7) ,当一个运行进程因某事件受阻时，其状态由运行变为(8) ，当进程所等待的事件已经发生时,该进程状态由（9) 变为就绪。
4。
线程是进程内的一个相对独立的(10）.三、计算题（每题10分，共40分)1.设某计算机系统采用虚拟页式存储管理方法，进程的虚拟地址空间为64KB,页面尺寸为4KB。
假设当前进程的页表如右图所示（页表以二进制形式表示），请将虚拟地址8196和2050转换为物理地址。
2。
设某计算机系统采用虚拟页式存储管理方法，内存中为该进程分配4个物理页架，开始时内存页架为空,假设进程在一段时间内的页面访问序列如下:6,0,1，2,0,3，0,4，2，3，0，3，2，1，2,0,1,7,0，1，请画图表示采用以下页面淘汰算法时的缺页中断次数:(1）最佳页面淘汰算法（OPT)；（2)先进先出页面淘汰算法（FIFO）；（3)使用过最久的先淘汰(LRU）。
unix期末考试题库及答案
unix期末考试题库及答案UNIX期末考试题库及答案一、选择题（每题2分，共20分）1. UNIX操作系统属于哪一类操作系统？A. 实时操作系统B. 分时操作系统C. 批处理操作系统D. 网络操作系统答案：B2. 在UNIX系统中，哪个命令用于查看当前目录下的文件和文件夹？A. lsB. pwdC. cdD. mkdir答案：A3. UNIX系统中，哪个命令用于查看文件内容？A. catB. touchC. echoD. more答案：A4. 在UNIX中，如何查看当前系统的时间？A. timeB. dateC. calD. clock答案：B5. UNIX系统中，哪个命令用于创建新的目录？A. lsB. mkdirC. cdD. rm答案：B6. 在UNIX系统中，哪个命令用于查看当前路径？A. lsB. pwdC. cdD. mkdir答案：B7. UNIX系统中，哪个命令用于删除文件？A. rmB. mvC. cpD. touch答案：A8. 在UNIX中，哪个命令用于复制文件？A. cpB. mvC. rmD. ln答案：A9. UNIX系统中，哪个命令用于移动或重命名文件？A. mvB. cpC. rmD. touch答案：A10. 在UNIX中，哪个命令用于创建链接？A. lnB. cpC. mvD. rm答案：A二、填空题（每题2分，共20分）1. UNIX系统中，文件权限通常由10个字符组成，其中第1个字符表示文件类型，第2-4个字符表示______，第5-7个字符表示______，第8-10个字符表示______。
答案：所有者权限，组权限，其他用户权限2. 在UNIX中，使用______命令可以查看系统帮助信息。
答案：man3. UNIX系统中，______命令用于查看当前登录用户。
答案：who4. UNIX系统中，______命令用于查看进程信息。
答案：ps5. UNIX系统中，______命令用于查看磁盘使用情况。
操作系统期末考试题和答案
操作系统期末考试题和答案一、选择题（每题2分，共20分）1. 在操作系统中，进程和程序的主要区别是（）。
A. 程序是静态的，进程是动态的B. 程序是动态的，进程是静态的C. 程序是操作系统的一部分，进程是用户的一部分D. 程序是用户的一部分，进程是操作系统的一部分答案：A2. 下列关于死锁的描述中，错误的是（）。
A. 死锁是指两个或多个进程在执行过程中，因争夺资源而造成的一种僵局B. 死锁产生的原因是系统资源不足C. 死锁的四个必要条件是互斥、占有和等待、不可剥夺和循环等待D. 死锁可以预防，但无法避免答案：D3. 在分页存储管理中，页表的作用是（）。
A. 将逻辑地址转换为物理地址B. 将物理地址转换为逻辑地址C. 存储进程的执行状态D. 存储进程的资源分配情况答案：A4. 虚拟内存技术的主要目的是（）。
A. 提高CPU的利用率B. 提高内存的利用率C. 提高I/O设备的利用率D. 提高磁盘的利用率答案：B5. 在操作系统中，文件的逻辑结构通常采用（）。
A. 顺序结构B. 链接结构C. 索引结构D. 树形结构答案：A6. 操作系统中，文件的物理结构通常采用（）。
A. 顺序结构B. 链接结构C. 索引结构D. 树形结构答案：B7. 在操作系统中，文件的共享是指（）。
A. 多个进程可以同时访问同一个文件B. 多个进程可以同时修改同一个文件C. 多个进程可以同时创建同一个文件D. 多个进程可以同时删除同一个文件答案：A8. 在操作系统中，文件的保护是指（）。
A. 防止文件被非法访问B. 防止文件被非法修改C. 防止文件被非法删除D. 以上都是答案：D9. 在操作系统中，文件的组织方式通常采用（）。
A. 顺序文件B. 随机文件C. 索引文件D. 以上都是答案：D10. 在操作系统中，文件的存取方式通常采用（）。
A. 顺序存取B. 随机存取C. 直接存取D. 以上都是答案：D二、填空题（每题2分，共20分）1. 操作系统的主要功能包括______、______、文件管理、设备管理和______。
linux操作系统期末考试试题及答案
linux操作系统期末考试试题及答案一、选择题（每题2分，共20分）1. 以下哪个命令可以查看Linux系统的发行版信息？A. cat /etc/issueB. cat /etc/redhat-releaseC. cat /etc/debian_versionD. lsb_release -a答案：A2. 在Linux系统中，以下哪个命令可以用来查看文件权限？A. ls -lB. ls -aC. ls -rD. ls -t答案：A3. 以下哪个命令用于创建一个新的用户？A. useraddB. userdelC. groupaddD. groupdel答案：A4. 在Linux系统中，以下哪个命令可以用来查看系统运行时间？A. uptimeB. topC. psD. free答案：A5. 以下哪个命令可以用来查看系统负载？A. uptimeB. topC. psD. free答案：A6. 以下哪个命令可以用来挂载一个USB设备？A. mountB. umountC. mountpointD. mount | grep答案：A7. 以下哪个命令可以用来查看网络连接信息？A. ifconfigB. ipconfigC. netstatD. ping答案：C8. 以下哪个命令可以用来重启Linux系统？A. rebootB. shutdown -rC. shutdown -hD. init 6答案：A9. 以下哪个命令可以用来杀死一个进程？A. killB. pkillC. killallD. kill -9答案：A10. 在Linux系统中，以下哪个文件包含了系统环境变量？A. /etc/profileB. /etc/bash.bashrcC. ~/.bashrcD. /etc/environment答案：D二、填空题（每题2分，共20分）11. 在Linux系统中，文件权限分为三种类型：读（______）、写（______）和执行（______）。
国开期末考试《操作系统》机考试题及答案(第8套)
国开期末考试《操作系统》机考试题及答案(第8套)一、单选题1. 操作系统的基本功能不包括以下哪项？- (A) 进程管理- (B) 文件管理- (C) 网络管理- (D) 内存管理- Answer: (C)2. 下列哪种文件系统不支持文件共享？- (A) FAT- (B) NTFS- (C) ext4- (D) NFS- Answer: (A)3. 进程同步的经典问题中，以下哪个问题不属于资源分配类问题？- (A) 哲学家就餐问题- (B) 读者写者问题- (C) 指令冲突问题- (D) 吸烟者问题- Answer: (D)4. 以下哪种调度算法不考虑进程优先级？- (A) 先来先服务调度- (B) 短作业优先调度- (C) 时间片轮转调度- (D) 最高响应比优先调度- Answer: (A)5. 下列哪项不属于虚拟内存的功能？- (A) 内存保护- (B) 内存扩充- (C) 内存共享- (D) 内存加密- Answer: (D)二、填空题1. 进程的特征有______(5个字)。
- Answer: 动态性、并发性、独立性、不确定性、制约性2. 磁盘调度算法的目标是______(4个字)。
- Answer: 提高磁盘的利用率3. 在分页存储管理方式下，逻辑地址由______和______两部分组成。
- Answer: 页号、页内偏移4. 进程调度算法中，______是指将处理机从一个进程转移到另一个进程的过程。
- Answer: 上下文切换5. 用户态和核心态的切换是通过______指令实现的。
- Answer: 特权指令三、简答题1. 请简要解释进程和线程的区别。
- Answer: 进程是操作系统进行资源分配和调度的基本单位，具有独立的内存空间和系统资源。
而线程是进程的执行单位，一个进程可以包含多个线程，共享进程的资源。
线程之间的切换比进程之间的切换开销更小，线程间的通信更加方便。
2. 请简要描述页面置换算法中的FIFO算法。
Linux_期末考试试题8套(含答案)
一、选择题 (每小题 2 分，共 50 分)
1.在创建 Linux 分区时，一定要创建（ D ）两个分区
A. FAT/NTFS B. FAT/SWAP C. NTFS/SWAP D.SWAP/根分区
2.在 Red Hat Linux 9 中，系统默认的（A）用户对整个系统拥有完全的控制权。
A. exportfs B. alias C. exports D. export
5. 哪个目录存放用户密码信息( B )
A. /boot
B. /etc
C. /var
D. /dev
6. 默认情况下管理员创建了一个用户，就会在( B )目录下创建一个用户主目录。
A. /usr B. /home
C. /root D. /etc
7. . 当使用 mount 进行设备或者文件系统挂载的时候，需要用到的设备名称位于( D )目录。
A. /home
B. /bin
C. /etc
D. /dev
8. 如果要列出一个目录下的所有文件需要使用命令行( C )。
A. ls –l B. ls
C. ls –a(所有)
D. ls –d
9. 哪个命令可以将普通用户转换成超级用户(D )
三、操作题(每小题 5 分，共 40 分)
40.简述在虚拟机中安装 Red Hat Linux 9.0 的过程 答;1。下载操作系统的镜像 ISO 文件 2．下载虚拟机并安装 3．通过 ISO 文件安装操作系统 4．执行相关配置即可
武汉大学计算机学院
2009-2010 学年第一学期期末考试
《Linux 原理与应用 》期末考试试卷（ ）
19.在 Red Hat Linux 9 中，一般用（ D ）命令来查看网络接口的状态
linu 期末考试参考卷
linux操作系统一、单项选择题1.关于Linux内核版本的说法，以下错误的是（ C ）。
A．表示为主版本号.次版本号.2.自由软件的含义是（ B ）。
A．用户不需要付费B．软件可以自由修改和发布C．只有软件作者才能向用户收费D．软件发行商不能向用户收费3.系统引导的过程一般包括如下几步：a．MBR中的引导装载程序启动；b．用户登录；c．Linux内核运行；d．BIOS自检。
正确的顺序是（ B ）。
A．d,b,c,a B．d,a,c,b C．b,d,c,a D．a,d,c,b4.字符界面下使用shutdown命令重启计算机时所用的参数是（ D ）。
A．-h B．-t C．-k D．-r5.下面哪个不是桌面环境软件？（ B ）A．KDE B．twm C． Gnome D．CDE6.cd命令可以改变用户的当前目录，当用户键入命令“cd”并按Enter键后，（ C ）。
A．当前目录改为根目录B．当前目录不变，屏幕显示当前目录C．当前目录改为用户主目录D．当前目录改为上一级目录7.在UNIX/Linux系统添加新用户的命令是（ D ）A. groupaddB. usermodC. userdelD. useradd8.Android系统是基于Linux操作系统开发的手机系统，因此在进行刷机、删除系统自带软件时需要获得管理员账户（ C ）的权限。
A.administratorB.adminC.rootD.liveuser9.修改用户自身的密码可使用（ A ）A. passwdB. passwd -d mytestC. passwd mytestD. passwd -lC、20和21D、22和2310.已知当前系统umask的值为022，请问这时新建目录文件的权限为（ A ）。
A. 755B. 644C.775D.66419.已知用户sarwar在虚拟终端1和虚拟终端2及图形用户界面下登录，请问sarwar 在虚拟终端1中执行ls >tty2后，命令输出到哪里？（ D ）A.虚拟终端1B.虚拟终端2C.图形终端D.无输出20.在vi中强制退出而不保存编辑内容的命令是？（ D ）A. :qB. :wC. :wqD. :q!二、填空题（每空1分，共15分）1.默认情况下，超级用户和普通用户的登录提示符分别是：‘#’和‘$’Linux内核引导时，从文件 /etc/fstab中读取要加载的文件系统。
计算机操作系统_期末复习考试题库
操作系统负责为方便用户管理计算机系统的（）。
在单一处理机上，将执行时间有重叠的几个程序称为（）。
操作系统是一组（）。
以下（）项功能不是操作系统具备的主要功能。
单机操作系统的共享资源是指（）财务软件是一种（）在分时系统中，时间片一定，（），响应时间越长。
UNIX命令的一般格式是（）并发是并行的不同表述，其原理相同。
[参考答案] 错误多用户操作系统一定是具有多道功能的操作系统。
[参考答案] 正确并发和_________是操作系统的两个最基本的特征，两者之间互为存在条件。
[参考答案]共享在多道程序设计的计算机系统中，CPU（）。
现代操作系统的两个基本特征是（）和资源共享。
从总体上说，采用多道程序设计技术可以（）单位时间的算题量，但对每一个算题，从算题开始到全部完成所需的时间比单道执行所需的时间可能要（）。
世界上第一个操作系统是（）。
（）没有多道程序设计的特点引入多道程序的目的是（）。
各进程向前推进的速度是不可预知的，体现出“走走停停”的特征，称为进程的（）。
并发是并行的不同表述，其原理相同。
[参考答案] 错误具有多道功能的操作系统一定是多用户操作系统。
[参考答案] 错误并发和_________是操作系统的两个最基本的特征，两者之间互为存在条件。
[参考答案]共享在批处理方式下，操作员把一批作业组织成（）向系统成批输入。
系统调用是由操作系统提供的内部调用，它（）。
从系统的角度出发，希望批处理控制方式下进入输入井的作业（）尽可能小。
作业在系统中存在与否的唯一标志是（）。
作业调度程序从处于（）状态的队列中选取适当的作业调入主存运行。
作业调度是（）。
处于后备状态的作业存放在（）中多用户操作系统离开了多终端硬件支持，则无法使用[参考答案] 错误设有3个作业J1，J2，J3，其运行时间分别是1，2，3小时。
假设这些作业同时到达，并在一台处理机上按单道运行，采用短作业优先调度算法，则平均周转时间由小到大的执行序列是J1，J2，J3。
Linnux期末考试题目及答案
Linnux期末考试题目及答案Linux期末考试题目及答案一、选择题（每题2分，共20分）1. Linux操作系统属于以下哪一种类型？A. 单用户单任务操作系统B. 多用户多任务操作系统C. 单用户多任务操作系统D. 多用户单任务操作系统答案：B2. 在Linux系统中，以下哪个命令用于查看当前目录下的文件和文件夹？A. lsB. pwdC. cdD. mkdir答案：A3. 如果需要查看Linux系统的运行时间和平均负载，可以使用以下哪个命令？A. uptimeB. topC. psD. who答案：A4. 在Linux中，以下哪个文件是系统的启动配置文件？A. /etc/fstabB. /etc/inittabC. /etc/passwdD. /etc/sysconfig答案：B5. 下面哪个选项是正确的Linux文件权限设置？A. -rwxr-xr--B. drwxr-xr-xC. -rwxrwxrwxD. drwxrwxrwx答案：B6. 在Linux中，哪个命令可以查看当前系统的内核版本？A. uname -aB. cat /etc/os-releaseC. lsb_release -aD. cat /proc/version答案：A7. 在Linux系统中，以下哪个命令用于压缩文件？A. gzipB. tarC. zipD. bzip2答案：B8. 在Linux中，以下哪个命令可以查看网络连接状态？A. ifconfigB. netstatC. routeD. nslookup答案：B9. 在Linux中，以下哪个命令用于查找文件？A. findB. grepC. whichD. whereis答案：A10. 在Linux系统中，以下哪个命令用于显示当前路径？A. cdB. pwdC. lsD. mkdir答案：B二、填空题（每空2分，共20分）1. Linux系统中的文件系统是以_________为树形结构组织的。
操作系统期末考试及参考答案
考试试卷：A√卷、B卷（请在选定项上打√）考试形式：闭、开√卷（请在选定项上打√），允许带 入场考试日期： 年 月 日,考试时间： 120 分钟 任任课教师：__诚信考试，沉着应考，杜绝违纪。
考生姓名：学号：所属院系：_题序1-701234总分得分评卷人注意：本考卷分P a r t1、P a r t2两部分，第一部分为选择题共70分；第二部分为问答题共30分。
选择题答案请填入以下表格，只答在题目上的不给分。
问答题答案请写在答题纸上。
题号12345678910 1－1011－2021－3031－4041－5051－6061－70Part One: Multiple Choice Questions (one mark each.)Choose the best answer for the following questions. There is only one best answer for each question.1.Operating systems provide certain levels of interfaces. However, is in general not provided by OS.A. Application programming interface (API)B. Command line interpreterC. Graphic user interface (GUI)D. System call2. A distributed system could be _________.A. A client-server systemB. A peer-to-peer systemC. A clustering systemD. All the above3.When operating system says Resource, it could beA.Memory spaceB.Global variableswork bandwidthD.All the abovepared to the OSes with microkernel, the monolithic counterpart sometimes shows advantage in .A. ScalabilityB. ModularityC. PerformanceD. Readability5.Which of the functionalities listed below must be supported by the operating system for handheld devices.A. Batch programmingB. Virtual memoryC. Time sharingD. Networking6.Which of the following types of operating systems has the best job throughput ?A. Time sharingB. InteractiveC. BatchD. Real time7. A CPU scheduler focuses on __________ scheduling.A. mixture-termB. short-termC. medium-termD. long-term8.The context-switch causes overhead by OS. The action affects many objects, but is not included.A. registerB. global variableC. stackD. memory9.Which of the following process state transitions is impossible to happen?A. from ready state to running stateB. from ready state to waiting stateC. from running state to ready stateD. from waiting state to ready state10. A process will change its state from “waiting” to “ready” when _____.A.it has been selected for execution by schedulerB.the event it had been waiting for has occurredC.its time slice is finishedD.it waits for some event11.The main difference between a process and a program is that _________.A. a process has its life cycle while a program has not.B. a program has its life cycle while a process has not.C. a program can own resources while a process cannot.D. a process can own resources while a program cannot.ing the program shown as following:#include <sys/types.h>#include <stdio.h>#include <unistd.h>int value = 10;int main() {pid_t pid;pid = fork();value += 10;if(pid == 0) { /* child process */value += 5; }else if (pid > 0) { /* parent process */wait(NULL);printf("PARENT: value = %d", value); /* LINE A */exit(0);}}Which string will be output at Line A?A. PARENT: value =20B. PARENT: value =10C. PARENT: value =15D. PARENT: value =2513. A semaphore array in Linux is often used as ______.A. a kind of direct communicationB. a kind of low-level communicationC. a kind of symmetrical communicationD. a kind of inter-process communication14.Which of the following statement is true ?A. Sometimes multithreading does not provide better performance than a single-threaded solutionB. Sometimes multithreading does the same performance as a single-threaded solutioC. Sometimes multithreading provides better performance than a single-threaded solutionD. All the above are true15.Threads in a process share the _________.A. Stack memoryB. Heap memoryC. Register valuesD. Global variables16.In general, multithreading shows some features benefiting user applications. Even though an operatingsystem does not support multithreading, those features could be brought with by use of __________.A. One to One ModelB. Kernel level threadC. User level threadD. None17.Which of the following scheduling algorithms could result in starvation ?A. First come first servedB. Round robinC. Shortest job firstD. Highest response_ratio next18.Consider a variant of the RR scheduling algorithm in which the entries in the ready queue are pointers tothe PCBs. If there are two pointers to the same PCB:A. It would not be the RR algorithm and be illegal.B. The time slice would have to be adjusted in order to rebalance the CPU load.C. The pointed process always gains twice the CPU time.D. The time interrupt should be smart enough which makes the OS kernel more complicated.19.Suppose the system is dominated by processes with short burst-time, ____ is the most appropriate choice.A. Multilevel queuesB. Multilevel feedback queuesC. First come first servedD. Round robin20.Sometimes two scheduling criteria are conflict with each other, and not satisfied both. Which of thefollowing pairs of scheduling criteria are ALWAYS non-conflicting?A. CPU utilization and response timeB. Average turnaround time and average waiting timeC. Average turnaround time and maximum waiting timeD. I/O device utilization and CPU utilization21.Talking about the scheduling for CPU burst cycle vs I/O burst cycle, which statement is true.A. A scheduler does not care the process either in CPU burst cycle or I/O burst cycleB. A process is either CPU burst or I/O burstC. A process with CPU burst cycle is preferredD. A process with I/O burst cycle is preferredFor the next 3 questions, considering the following set of processes, with the length of the CPU-burst time given in milliseconds:Process Arrival time Burst timeP103P225P341P454P58122.For the FCFS scheduling algorithm, the average waiting time is_____.A. 14/5B. 25/5C. 44/5D. 33/523.For the SJF scheduling algorithm, the average waiting time is _____.A. 11/5B. 22/5C. 41/5D. 30/524.For the Round Robin (quantum is 2) scheduling algorithm, the average waiting time is ___.A. 44/5B. 34/5C. 29/5D. 15/525.The critical section in an OS is ________.A. a process schedulerB. a data sectionC. a synchronization mechanismD. a segment of code26.Which of the following statements is incorrect regarding Busy Waiting?A. Busy waiting makes worse CPU throughput.B. Busy waiting could be avoided by proper CPU scheduling.C. Busy waiting does not just come with Critical Section Problem.D. If a solution to the Critical Section Problem causes busy waiting, the solution is incorrect.27.Which of the following statements is correct?A. Critical section is a piece of code in a process for mutual exclusion.B. Critical section is a piece of code in a process for process synchronization.C. Critical section is a piece of code in a process for inter-process communication.D. Critical section is a piece of code in a process for accessing critical (shared) resources.28.For two-process Critical Section(CS) problem solution, the Progress condition does not mean thatA. Only processes wish to enter the section are the candidates.B. If and only if there are some processes wish to enter the critical section, the Progress condition applies.C.The decision to enter the critical section should be made within limited time, even though there exists aprocess running in its critical section.D. A process is allowed to enter its critical section many times while the others keep waiting.29.Which one of the following statements is correct about spinlock?A. Spinlock is appropriate for single-processor systemsB. Spinlock is often used in multiprocessor systemsC.Spinlock could be used in single-processor systemsD. Spinlock is not often used in multiprocessor systems30.Which one of the following statements is correct about synchronization primitives ?A. The primitive could be implemented by disabling interrupts, even in single-processor systems.B. The primitive could not implemented by disabling interrupts, neither in single-processor systems nor inmultiprocessor systems.C. The primitive could only be implemented by disabling interrupts for multiprocessor systems.D. If used in user-level programs, the primitive could be implemented by disabling interrupts.31.Hope the server limits its number to be concurrently connected no more than N clients. One solution willbeA. A semaphore for resource sharing purpose, with the initial value NB. A semaphore for resource sharing purpose, with the initial value 1C. A semaphore for synchronization purpose, with the initial value ND. A semaphore for synchronization purpose, with the initial value 132.Which one of the following is not the necessary condition for a deadlock to occur?A. StarvationB. Mutual exclusionC. Hold and waitD. NO Preemption33.Which of the following methods can prevent the deadlock from the very beginning?A. Resource allocation in an increasing order of enumerationB. Banker’s algorithmC. Deadlock detectionD. Deadlock avoidance34.Consider the following snapshot of a system:Allocation Max AvailableABCD ABCD ABCD0 0 1 2 0 0 1 2 1 5 2 0P1 0 0 0 1 7 5 0P1P1 3 5 423 5 620 6 3 2 0 6 5 2P3P0 0 1 4 0 6 5 64Which one is the safe sequence for the system?A. < P0, P3, P4, P2, P1>B. < P1, P2, P4, P3, P0>C. < P2, P0, P4, P1, P3>D. < P4, P3, P1, P2, P0>35.The address binding could be by the way ofA. The variables in source codes converted to the binaryB. The variables in source codes compiled into object modulesC. Several object modules are linked together into a single programD. All the above36.An unsafe state implies ________.A. the existence of deadlockB. that deadlock will eventually occurC. that some unfortunate sequence of events might lead to a deadlockD. The scenario that the Dining Philosophers Problem described37.In a real computer system, neither the resources available nor the demands of processes for resources areconsistent over long periods (months). Resources break or are replaced, new processes come and go, new resources are bought and added to the system. If deadlock is controlled by the banker’s algorithm, which of the following changes can be made safely (without introducing the possibility of deadlock) ?A. Increase the number of processes.B. Increase Max for one process (the process needs more resources than allowed, it may want more)C. Increase Available (new resources added).D. Decrease Available (resource permanently removed from system)38.Which of the following memory management method helps to share a code segment across processes?A. Contiguous memory allocationB. Pure segmentationC. Pure pagingD. None of above39.Which of the following memory management method has no impact in terms of internal fragmentation?A. Two-level pagingB. SegmentationC. PagingD. Linux paging strategy40.To apply the demand paging memory management, the CPU with powerful MMU is a must. However,is NOT a necessity.A. InterruptB. Present bit defined in the segment table entryC. TLBD. Page table41.Assume that you are monitoring the rate at which the pointer in the clock algorithm (which indicates thecandidate page for replacement) moves. What can you say about the system if you notice the pointer is moving fast ?A. the program is accessing a large number of pages simultaneouslyB. the operation finding candidate pages for replacement is efficientC. the virtual memory system is extremely efficientD. that indicates many of the resident pages are not being accessed42.Suppose that a machine provides instructions that can access memory locations using the one-levelindirect addressing scheme. How many page faults incurred when all of the pages of a program are currently non-resident and the first instruction of the program is an indirect memory load operation ?A. 3B. 2C. 1D. 043. A certain computer provides its users with a virtual-memory space of 232bytes. The computer has 218bytes of physical memory. The virtual memory is implemented by paging, and the page size is 4096 bytes.A user process generates the virtual address 11123456, actually its page number is ___A. 69923B. 2715C. 1110D. 1112345644.Consider a demand-paging system with the following time-measured utilizations:CPU utilization 20%Paging disk 97.7%Other I/O devices 5%Which of the following will improve CPU utilization ?A. Increase the degree of multiprogrammingB. Decrease the degree of multiprogrammingC. Install a faster CPUD. Install a bigger paging disk45.Which of the following indicates that the system performs wellA. A process suffers deadlockB. A process suffers starvationC. A process suffers bad turnaround timeD. A process suffers thrashing46.Virtual memory management with paging does not requireA. the page replacementB. to process the page fault interruptC. to load some code or data into the contiguous memory spaceD. none of the above47.In order for a virtual memory management performing well, it is preferred thatA. processes do not have too much I/O operationsB. the program size should not be bigger than the whole memory spaceC. there are some large size contiguous memory spaceD. the locality of processes is well featuredFor the next 2 questions, consider a paging system with the page table stored in memory.48.If a memory reference takes 200 nanoseconds, how long does a paged memory reference take ?A. 200 nanoseconds + the time to process the page entries which is a bit over 200 nanosecondsB. depends, sometimes 400 nanoseconds, and sometimes 200 nanosecondsC.200 nanosecondsD. 400 nanoseconds49.If we add TLBs, and 75 percent of all page-table references are found in the TLBs, what is the effectivememory reference time ? (Assume that finding a page-table entry in the TLBs takes zero time, if the entry is there.)A. 250 nanosecondsB. 350 nanosecondsC. 400 nanosecondsD. 200 nanoseconds50.If an OS is not facilitated with virtual memory management, thenA. A process has to be loaded fully in memory before execution and kept staying in memoryB. A process does not have to be loaded fully in memory before execution, neither does stay in memoryC. A process does not have to be loaded fully in memory before execution, however has to stay in memoryduring executionD. A process has to be loaded fully in memory before execution, however does not have to stay in memoryduring executionFor the next 3 questions, assume that five memory partitions are of size 100KB, 500KB, 200KB, 300KB, and 600KB (in order). The processes of 212KB, 417KB, 112KB and 426KB(in order) are applied to be put in place.51.If the first-fit algorithm is equipped, the process of 112KB will be put in the partition withA. 200KBB. 300KBC. 500KBD. 600KB52.If the best-fit algorithm is equipped, the process of 112KB will be put in the partition withA. 500KBB. 200KBC. 300KBD. 600KB53.If the worst-fit algorithm is equipped, the process of 112KB will be put in the partition withA. 500KBB. 200KBC. 300KBD. 600KB54.Which of the following designs is not to share a file located in a remote machine.A. Maintaining several replicasB. IP address followed by a pathC. URLD. Virtual address55. A file’s attributes vary from one operating system to another, but typically consist ofA. File nameB. SectorsC. TypeD. The name of the creating program56.The per-process open-file table isA. unique and maintained by OS for all usersB. one of OS data structure for better performance of file system managementC. claimed by each process and for its own purposeD. an accounting data structure to tell how many files opened by the process57.As to the way accessing data of a file,A. The sequential manner is better than the random one.B. The random access is better than the sequential one.C. Both the sequential and the random access are the right way.D. Either the sequential or the random access is replaced by DBMS.58.Which of the following design is practical by operating systems.A. Automatically open a file while referenced for the first time, and close the file when the job terminates.B. The user has to open and close the file explicitly.C. The user has to open the file explicitly, but the file is closed automatically when the job terminates.D. All the above59.Regarding the file access permission, which of the following statements is NOT correct.A. The “Cloud” providing SaaS is an example facilitated the file system to write once but read many timesB. Some file systems are read only, not allowing any write operation.C. Some file systems are dedicated to write only once but read many timesD. The web site providing search service is an example facilitated the file system to write once but read manytimesB flash drive is popular nowadays. Usually it is not formatted withA. btrfsB. ISO 9660C. FA TD. EXT261.None of the disk-scheduling disciplines could avoid starvation, except ____.A. FCFSB. SSTFC. C-SCAND. C-LOOK62.Overheads are always associated with an interrupt service, resulting in worse performance. However, theydo not include the cost of ________A. saving process stateB. executing the instruction just next the interrupt pointC. restoring process stateD. flushing the instruction pipeline63.Look at the fact that requests are not usually uniformly distributed. For example, a cylinder containing thefile system FAT can be expected to be accessed more frequently than a cylinder that only contains files.And the fact that file systems typically find data blocks via an indirection table, such as a FAT in DOS.Which of the following ways would take advantage of this indirection to improve disk performance ?A. Keep the metadata in the nearest corner of cylindersB. Cache the metadata in primary memoryC. Back up the metadataD. Redesign the file system by discarding the indirection64.Which scheme of disk array provides no data redundancy ?A. RAID 3B. RAID 0C. RAID 1D. RAID 265.Suppose that a disk drive has 5000 cylinders, numbered 0 to 4999. The drive is currently serving arequest at cylinder 143, and the previous request was at cylinder 125. The queue of pending requests, in FIFO order, is86, 1470, 913, 1774, 948, 1509, 1022, 1750, 130Starting from the current head position, what is the sequence of cylinder number that the disk arm moves to satisfy all the pending requests, for the SSTF algorithms?A. 143, 913, 948, 1022, 1470, 1509, 1750, 1774, 4999, 130, 86B. 143, 913, 948, 1022, 1470, 1509, 1750, 1774, 130, 86C. 143, 130, 86, 913, 948, 1022, 1470, 1509, 1750, 1774D. 143, 86, 1470, 913, 1774, 948, 1509, 1022, 1750, 13066.Some file systems allow disk storage to be allocated at different levels of granularity. For example, a filesystem could allocate 4KB of disk space as a single block, or as four 1024-byte blocks. Do you think the example isA. Absolutely nonsenseB. Probably true, but it is only for academic researchC. Practical, because there exists a popular file system using the schemeD. Absolutely true, all file systems use the scheme67.Which of the following statements is wrong from the operating system view?A. Memory sometimes is used as a diskB. Memory sometimes is used as read onlyC. Memory sometimes is used as a USB flash driveD. Memory sometimes is used as a disk cache68. A logical address is ________A. the address in an object fileB. the address in an executable fileC. the address in a CPU instruction together with operatorD. All the above69. A file system uses a scheme with support for extents. A file is a collection of extents, with each extentcorresponding to a contiguous set of blocks. This file system is calledA. Contiguous allocationB. Linked allocationC.Indexed allocationD. None of above70.Which reason does not make sense: The operating system generally treats removable disks as shared filesystems but assigns a tape drive to only one application at a time, becauseA. Disks have fast random-access time, so they give good performance for interleaved access streams. Bycontrast, tapes have high positioning timeB. The owner of the Tape cartridge may wish to store the cartridge off-site (far away from the computer) tokeep a copy of the data safe from a fire at the location of the computerC. Historically tape cartridges are often used to send large volumes of data from a producer to the consumer.Such a tape cartridge is reserved for that particular data transferD. None of abovePart Two: (30 marks)1. ( 12 marks) Consider the following set of processes, with the length of the CPU-burst time given in milliseconds:Process Burst Time PriorityP1 10 3P2 1 1P3 2 3P4 1 4P5 5 2The processes are assumed to have arrived in the order P1, P2, P3, P4, P5, all at time 0. Suppose it is used FCFS, SJF, a non-preemptive priority (a smaller priority number implies a higher priority), and RR (quantum = 1) scheduling.a. What is the turnaround time of each process for each of the scheduling algorithms ?b. Which of the schedules results in the minimal average waiting time (over all processes)?2. ( 6 marks) Lamport's bakery algorithm is intended to improve the safety in the usage of shared resources among multi ple threads by means of mutual exclusion, briefed as following.1 lock(integer i) {2 Choosing[i] = true;3 Number[i] = 1 + max(Number[1], ..., Number[NUM_THREADS]);4 Choosing[i] = false;5 f f o r (j = 1; j <= NUM_THREADS; j++) {6 // Wait until thread j receives its number:7 w w h i l e (Choosing[j]) { /* nothing */ }8 // Wait until all threads with smaller numbers or with the same9 // number, but with higher priority, finish their work:10 w w h i l e ((Number[j] != 0) && ((Number[j], j) < (Number[i], i))) {11 /* nothing */12 }13 }14 }15 //critical section16 unlock(integer i) {17 Number[i] = 0;18 }1920 Thread(integer i) {21 w w h i l e (true) {22 lock(i);23 // The critical section goes here...24 unlock(i);25 // n o n-c r i t i c a l s e c t i o n...26 }27 }(1) Suppose Thread(i) and Thread(j) are running concurrently, will they occasionally get the same Number[] ?i.e. Number[i] == Number[j]. If so, please give a scenario.(2) If the Choosing array is not applied, i.e. Line 2, Line 4 and Line 7 are deleted, will the algorithm conflict with Critical Section Requirements of Mutual Exclusion, Progress and Bounded Waiting. If so, please give a scenario.3. ( 4 marks) Assume we have a demand-paged memory. The page table is held in registers. It takes 8 milliseconds to service a page fault if an empty page is available or the replaced page is not modified, and 20 milliseconds if the replaced page is modified. Memory access time is 100 nanoseconds. Assume that the page to be replaced is modified 70 percent of the time. What is the maximum acceptable page-fault rate for an effective access time of no more than 200 nanoseconds?4. ( 8 marks) The EXT2 file system defines an index array with 15 pointers to locate all data blocks of a file.(1) The first 12 items of the index array accommodate the locations of the first 12 data block.(2) The 13th item points to an index block called the indirect block, which contains index entries, each being a pointer to a data block.(3) The 14th item points to an index block containing entries, where each entry is a pointer to yet another indirect block as described in (2).(4) The 15th item points to an index block containing entries where each entry is a pointer to another index block as described in (3).Suppose the EXT2 data block is of size 4096 bytes, and an index entry is of size 4 bytes. Please answer how would be the maximal size of a file ?操作系统试卷参考答案和评分标准Part 1. Answer Sheet:（每小题1分，共计70分）12345678910A D D C C CB B B B11121314151617181920A A D D D C C C C B21222324252627282930A A A D D D D CB B31323334353637383940A A A A D C CB B B41424344454647484950A AB BC CD D A A51525354555657585960A B C D A B C D A B61626364656667686970A B B B C C C D D DPart 2.1．Answer: 12 分a. Turnaround time（共10分，错1个扣0.5分）FCFS RR SJF PriorityP1 10 19 19 16P211 2 1 1P3 13 7 4 18P4 14 4 2 19P5 19 14 96b. Shortest Job First（2分）2．Answer: 6 分（2小题各 3 分）(1) Number[i] = 1 + max(Number[1], ..., Number[NUM_THREADS]);This line of statement is not an atomic operation.If there is a breakout by the scheduler before and after the assignment operation, “=”, it may result in the snapshot, in which Number[i] == Number[j](2) Demo/* we have no choosing mechanism.Pi Pj*/->reg = max(.....) + 1;->reg = max(...) + 1;->set number[j] = reg;->for(index = 0; index < n; index++)->run when index == i;->while((number[index]!=0&&(number[index,index]< number[j,j]));/******* NOTE! the process i hasn't set number[i],so thecondition is false *********/-> enter critical section->set number[i] = reg;->while((number[index]!=0 && (number[index],index) < number[i],i)));/*-------------------------------------------------------------*//** here we can conclude that Pi can enter critical section as well as Pj **/3．Answer: 4 分0.2 _sec = (1 − P) ×0.1 _sec + (0.3P) ×8 millisec + (0.7P) ×20 millisec0.1 = −0.1P + 2400 P + 14000 P0.1 =16,400 PP =0.000006（答案正确的，给4分。
《操作系统》期末考试试题(含答案)
考试科目：《操作系统原理》（A卷）学年学期：2014-2015学年第三学期姓名：学院/系：计算机系学号：考试方式：闭卷年级专业：考试时长：120分钟班别：------------以下为试题区域，共五道大题，总分100分,考生请在答题纸上作答------------一、单项选择题（共10小题，每小题1分，共10分）在每小题列出的四个备选项中只有一个是最符合题目要求的，请将其代码填写在题后的括号内。
错选、多选或未选均无分。
1、操作系统通过（）来管理计算机系统的软硬件资源。
A．中断向量B．CPU指令集C．设备驱动程序D．信号量2、在六状态进程模型中，处于就绪态的进程，只能转入的状态为（）。
A．运行B．阻塞C．挂起D．退出3、在下列并发模型中，不包含同步的问题为（）。
A．生产者-消费者问题B．读者-写者问题C．理发店问题D．哲学家就餐问题4、在进程短程调度的下列算法中，最公平的算法为（）。
A．最短进程优先SPN B．最短剩余时间SRT C．最高响应比优先HRRND．虚拟时间片轮转VRR5、在内核级线程模型中，同一进程所产生的多个线程不共享进程的（）。
A．代码B．上下文寄存器保存区C．打开的文件D．数据6、在动态分区的放置算法中，性能最差的算法通常是( )。
A．最佳适配B．首次适配C．下次适配D．末次适配7、在虚拟内存管理的页面替换算法中，较实用的算法为（）。
A．先进先出FIFO B．最优OPT C．最近最少使用LRU D．时钟Clock8、64位的操作系统一般采用（）级页表。
A．一B．二C．三D．四9、较实用的磁盘调度算法为（）。
A．先进先出FIFO B．后进先出LIFO C．最短服务时间优先SSTF D．电梯扫描Scan 10、现代主流操作系统所采用的文件存储方式多为（）。
A．连续分配B．链接分配C．索引分配D．哈希分配二、多项选择题（共5小题，每小题2分，共10分）在每小题列出的五个备选项中至少有两个是符合题目要求的，请将其代码填写在题后的括号内。
操作系统全英文期末考试题(带答案)
一．选择题（20分，每题1分）1. Generally speaking, which one is not the major concern for a operating system in the following four options?( D )A.Manage the computerB.Manage the system resourcesC.Design and apply the interface between user's program and computer hardware systemD.High-level programming language complier2.The main disadvantage of batch system is ( C )A.CPU utilization is lowB.Can not concurrentck of interactionD.Low degree of automation3.A process transforms from waiting state to ready state is caused by the ( B )A.Interrupt eventB.Process schedulingC.Create a process for a programD.Waiting for some events4.The concurrent process is refers to ( C )A.The process can be run in parallelB.The process can be run in orderC.The process can be run in the same timeD.The process can not be interrupted5.In multi-process system, in order to ensure the integrity of public variables, the processes should be mutually exclusive access to critical areas. The so-called critical area is ( D )A.A bufferB.A date areaC.Synchronization mechanismD.A program6.The orderly use of resources allocation strategy can destroy the condition ( D ) to avoid deadlock.A.Mutual exclusiveB.Hold and waitC.No preemptionD.Circular waiter's applications use the system resources to complete its operation by thesupport and services of ( C )A.clicking the mouseB.Keyboard commandC.System callD.Graphical user interface8.There are four jobs arrived at the same time and the execution time of each job is 2h. Now they run on one processor at single channel,then the average turnaround time is ( B )A.1hB.5hC.2.5hD.8h9.Among the job scheduling algorithms, ( B ) is related to the job's estimated running time.A.FCFS scheduling algorithmB.Short-job-first scheduling algorithmC.High response ratio algorithmD.Balanced scheduling10.In memory management, the purpose of using the overlay and swapping is ( C )A.Sharing main memoryB.Expanding main memory physicallyC.Saving main memory spaceD.Improving CPU utilization11.In the page-replacement algorithm,which one can cause the Belady phenomenon? ( A )A.FIFOB.LRUC.CLOCKINGD.OPT12.The following description of the system in safe state,which one is correct?( B )A.It must cause deadlock if the system is in insecure stateB.It may cause deadlock if the system is in insecure stateC.It may cause deadlock if the system is in secure stateD.All are wrong13.Generally, when we talk about"Memory Protection", the basic meaning is ( C )A.Prevent hardware memory from damagingB.Prevent program from losing in memoryC.Prevent the cross-border call between programsD.Prevent the program from being peeped14.The actual capacity of virtual memory is equal to ( B )A.The capacity of external memory(disk)B.The sum of the capacity of external memory and main memoryC.The space that the CPU logical address givesD.The smaller one between the option B and C15.Physical file's organization is determined by ( D )A.ApplicationsB.Main memory capacityC.External memory capacityD.Operating system16.A computer system is configured with two plotters and three printers,in order to properly drive these devices,system should provide ( C ) device driver program.A.5B.3C.2D.117.When there are fewer number of channels in system ,it may cause "bottlenecks".To solve this problem,which of the follow options is not the effective way?( A )A.improving the speed of CPUing the virtual device technologyC.Adding some hardware buffer on the devicesD.Increasing the path between devices and channels18.When I/O devices and main memory are exchanging data, it can be achieved without CPU's frequently intervention,this way of exchanging data is called ( C )A.PollingB.InterruptsC.Direct memory accessD.None of them19.The following description of device management, which one is not correct?( B )A.All external devices are managed by the system in uniformB.Channel is a software of controlling input and outputC.The I/O interrupt events from the I/O channel are managed by device managementD.One of the responsibility of the operating system is to use the hardware effectively20.The operating system used ( A ), it realized a mechanism that we can use more space to save more time.A.SPOOLINGB.Virtual storageC.ChannelD.Overlay二．填空题（20分，每空1分）1.Software may trigger an interrupt by executing a special operation called a system call .(P7)2.If there is only one general-purpose CPU,then the system is a single-processor system.(p12)3. A process can be thought of as a program in execution. (p79)4.As a process executes,it changes state.Each process may be in one of the following states:new,running,waiting,ready or terminated .(p83)5.Long-term(job) scheduling is the selection of processes that will beallowed to contend for the CPU.And Short-term(CPU) scheduling is the selection of one process from the ready queue. (p116)6.The process executing in the operating system may be either independent processes or cooperating processes. Cooperating processes require an interprocess communication mechanism to communicate with each other.Principally,communication is achieved through two schemes: share memory and message passing. (p116)7.In modern operating systems, resource allocation unit is process, processor scheduling unit is thread .(p127)8.Most modern operating systems provide kernel support for threads;among these are Windows,as well as Solaris and Linux .(p146)9.CPU scheduling is the basis of multiprogrammed operating systems.(p153)10.The FCFS algorithm is nonpreemptive;the RR algorithm is preemptive.11.Sometimes,a waiting process is never again able to change state,because the resources it has requested are held by other waiting processes.This situation is called deadlock . (p245)12.The main purpose of a computer system is to execute programs.These programs,together with the data they access,must be in main memory(at least partially) during execution.(P274)13. The various memory-management algorithms differ in may aspects.In comparing different memory-management strategies,we use the follow considerations:Hardware support,Performance,Fragmentation,Relocation, Swapping,Sharing and protection . (p310)14.A process is thrashing if it is spending more time paging than executing.15.Virtual memory is a technique that enables us to map a large logical addressspace onto a smaller physical memory.(p365)16.When we solve the major problems of page replacement and frame allocation,the proper design of a paging system requires that we consider page size,I/O,locking,process creation,program structure,and other issues.(p366) 17.The operating system abstracts from the physical properties of its storage devices to define a logical storage unit,the file . (p373)18.Since files are the main information-storage mechanism in most computer system,file protection is needed.(p408)19.The seek time is the time for the disk arm to move the heads to the cylinder containing the desired sector.(P457)20.The hardware mechanism that enables a device to notify the CPU is called an interrupt .(p499)三．简答题（30分，每题6分）1.What is the operating system?What role does the operating system play in a computer?开放题，解释操作系统概念，操作系统可以实现哪些基本功能？关键词：a.管理系统资源，控制程序运行，改善人机界面，为其他应用软件提供支持。
《linux操作系统》期末试卷A及答案
《linux操作系统》期末试卷A及答案《Linux操作系统》期末试卷A及答案一、选择题（共20题，每题2分，共40分）1.Linux是一种开源操作系统，它的内核是由谁开发的？A.Richard StallmanB.Linus TorvaldsC.Bill GatesD.Steve Jobs答案：B2.如何在Linux命令行中查看当前目录下的文件列表？A.dirB.lsC.cdD.pwd答案：B3.在Linux中，如何将文件从一个目录复制到另一个目录？A.cpB.mvC.rmD.mkdir答案：A20.Linux中的bash是什么？A.文件管理器B.编辑器C.命令行解释器D.图形界面工具答案：C二、填空题（共10题，每题2分，共20分）1.在Linux中，用于创建目录的命令是__________。
答案：mkdir2.在Linux中，用于复制文件的命令是__________。
答案：cp10.在Linux中，用于退出当前用户会话的命令是__________。
答案：exit三、简答题（共5题，每题10分，共50分）1.请简要描述Linux操作系统的特点。
答案：Linux是一种开源操作系统，具有稳定性、安全性、多用户支持、多任务支持等特点。
它采用模块化设计，支持多种硬件平台，并且具备强大的网络功能。
同时，Linux还拥有丰富的应用软件和工具，广泛应用于服务器、嵌入式设备等领域。
2.Linux中的文件权限是如何设置和管理的？答案：Linux中的文件权限通过chmod命令来设置和管理。
该命令可以用数字或符号来表示文件的权限，并可以设置用户、组或其他人的读、写、执行权限。
同时，使用chown和chgrp命令可以更改文件的所有者和所属组。
五、附件本文档无附件。
六、法律名词及注释1.开源：开源是指软件的源代码可以被公开查看、使用和修改的授权方式。
开源软件具备更高的透明度和可定制性，且常常拥有更好的安全性和稳定性。
2.内核：内核是操作系统的核心部分，负责管理计算机硬件和软件资源，提供基本的系统服务和功能。
linux操作系统期末考试题及答案
linux操作系统期末考试题及答案一、选择题（每题2分，共20分）1. Linux操作系统的内核作者是谁？A. Linus TorvaldsB. Dennis RitchieC. Ken ThompsonD. Bill Gates答案：A2. 在Linux系统中，哪个命令用于查看当前目录下的文件和文件夹？A. lsB. pwdC. cdD. mkdir答案：A3. Linux系统中，文件权限的表示方法中，'r'代表什么？A. 读B. 写C. 执行D. 所有权限答案：A4. 在Linux中，以下哪个命令用于下载文件？A. wgetB. getC. fetchD. pull答案：A5. 在Linux系统中，哪个命令用于查看当前系统的运行时间和平均负载？A. uptimeB. loadC. timeD. duration答案：A6. Linux系统中，如何查看当前登录用户？A. whoB. userC. loginD. users答案：A7. 在Linux中，哪个命令用于查看当前路径？A. pathB. pwdC. cdD. ls答案：B8. 在Linux系统中，以下哪个命令用于创建一个新文件夹？A. mkdirB. makedirC. newdirD. create答案：A9. Linux系统中，哪个命令用于查看当前系统的磁盘使用情况？A. dfB. diskC. diskusageD. space答案：A10. 在Linux中，哪个命令用于查看当前路径下的隐藏文件？A. ls -aB. ls -hC. ls -lD. ls -d答案：A二、填空题（每题2分，共20分）1. 在Linux系统中，文件的权限通常由三组数字表示，分别代表______、______和______的权限。
答案：所有者、组、其他2. 在Linux中，使用______命令可以查看当前系统的版本信息。
答案：uname -a3. Linux系统中，______命令用于解压tar.gz文件。
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一．选择题（20分，每题1分）1. Generally speaking, which one is not the major concern for a operating system in the following four options( D )the computerthe system resourcesand apply the interface between user's program and computer hardware systemprogramming language complier2.The main disadvantage of batch system is ( C )A.CPU utilization is lowB.Can not concurrentck of interactionD.Low degree of automation3.A process transforms from waiting state to ready state is caused by the ( B )A.Interrupt eventB.Process schedulingC.Create a process for a programD.Waiting for some events4.The concurrent process is refers to ( C )A.The process can be run in parallelB.The process can be run in orderC.The process can be run in the same timeD.The process can not be interrupted5.In multi-process system, in order to ensure the integrity of public variables, the processes should be mutually exclusive access to critical areas. The so-called critical area is ( D )A.A bufferB.A date areaC.Synchronization mechanismD.A program6.The orderly use of resources allocation strategy can destroy the condition ( D ) to avoid deadlock.A.Mutual exclusiveB.Hold and waitC.No preemptionD.Circular waiter's applications use the system resources to complete itsoperation by the support and services of ( C )A.clicking the mouseB.Keyboard commandC.System callD.Graphical user interface8.There are four jobs arrived at the same time and the execution time of each job is 2h. Now they run on one processor at single channel,then the average turnaround time is ( B )9. the job scheduling algorithms, ( B ) is related to the job's estimated running time.A.FCFS scheduling algorithmB.Short-job-first scheduling algorithmC.High response ratio algorithmD.Balanced scheduling10.In memory management, the purpose of using the overlay and swapping is ( C )A.Sharing main memoryB.Expanding main memory physicallyC.Saving main memory spaceD.Improving CPU utilization11.In the page-replacement algorithm,which one can cause the Belady phenomenon ( A )A.FIFOB.LRUC.CLOCKINGD.OPT12.The following description of the system in safe state,which one is correct( B )A.It must cause deadlock if the system is in insecure stateB.It may cause deadlock if the system is in insecure stateC.It may cause deadlock if the system is in secure stateD.All are wrong13.Generally, when we talk about"Memory Protection", the basic meaning is ( C )A.Prevent hardware memory from damagingB.Prevent program from losing in memoryC.Prevent the cross-border call between programsD.Prevent the program from being peeped14.The actual capacity of virtual memory is equal to ( B )A.The capacity of external memory(disk)B.The sum of the capacity of external memory and main memoryC.The space that the CPU logical address givesD.The smaller one between the option B and C15.Physical file's organization is determined by ( D )A.ApplicationsB.Main memory capacityC.External memory capacityD.Operating system16.A computer system is configured with two plotters and three printers,in order to properly drive these devices,system should provide ( C ) device driver program.17.When there are fewer number of channels in system ,it may cause "bottlenecks".To solve this problem,which of the follow options is not the effective way( A )A.improving the speed of CPUing the virtual device technologyC.Adding some hardware buffer on the devicesD.Increasing the path between devices and channels18.When I/O devices and main memory are exchanging data, it can be achieved without CPU's frequently intervention,this way of exchanging data is called ( C )A.PollingB.InterruptsC.Direct memory accessD.None of them19.The following description of device management, which one is not correct( B )A.All external devices are managed by the system in uniformB.Channel is a software of controlling input and outputC.The I/O interrupt events from the I/O channel are managed by device managementD.One of the responsibility of the operating system is to use the hardware effectively20.The operating system used ( A ), it realized a mechanism that we can use more space to save more time.A.SPOOLINGB.Virtual storageC.ChannelD.Overlay二．填空题（20分，每空1分）1.Software may trigger an interrupt by executing a special operation called a system call .(P7)2.If there is only one general-purpose CPU,then the system is a single-processor system.(p12)3. A process can be thought of as a program in execution. (p79)4.As a process executes,it changes process may be in one of the following states:new,running,waiting,ready or terminated .(p83)5.Long-term(job) scheduling is the selection of processes that will beallowed to contend for the Short-term(CPU) scheduling is the selection of one process from the ready queue. (p116)6.The process executing in the operating system may be either independent processes or cooperating processes. Cooperating processes require an interprocess communication mechanism to communicate with each ,communication is achieved through two schemes: share memory and message passing. (p116)modern operating systems, resource allocation unit is process, processor scheduling unit is thread .(p127)modern operating systems provide kernel support for threads;among these are Windows,as well as Solaris and Linux .(p146)scheduling is the basis of multiprogrammed operating systems.(p153)FCFS algorithm is nonpreemptive;the RR algorithm is preemptive. ,a waiting process is never again able to change state,because the resources it has requested are held by other waiting situation is called deadlock . (p245)main purpose of a computer system is to execute programs,together with the data they access,must be in main memory(at least partially) during execution.(P274)13. The various memory-management algorithms differ in may comparing different memory-management strategies,we use the follow considerations:Hardwaresupport,Performance,Fragmentation,Relocation,Swapping,Sharing and protection . (p310)14.A process is thrashing if it is spending more time paging than executing.15.Virtual memory is a technique that enables us to map a largelogical address space onto a smaller physical memory.(p365)16.When we solve the major problems of page replacement and frame allocation,the proper design of a paging system requires that we consider page size,I/O,locking,process creation,program structure,and other issues.(p366)17.The operating system abstracts from the physical properties of its storage devices to define a logical storage unit,the file . (p373)18.Since files are the main information-storage mechanism in most computer system,file protection is needed.(p408)19.The seek time is the time for the disk arm to move the heads to the cylinder containing the desired sector.(P457)20.The hardware mechanism that enables a device to notify the CPU is called an interrupt .(p499)三．简答题（30分，每题6分）1.What is the operating systemWhat role does the operating system play in a computer开放题，解释操作系统概念，操作系统可以实现哪些基本功能关键词：a.管理系统资源，控制程序运行，改善人机界面，为其他应用软件提供支持。