操作系统全英文期末考试题带答案

合集下载

《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。

英文版计算机试题库及答案

英文版计算机试题库及答案English Version of Computer Test Bank and AnswersIn today's interconnected world, computer literacy is of utmost importance. Aspiring computer professionals should possess not only theoretical knowledge but also practical skills. To aid in the assessment of such competencies, an English version of the computer test bank, along with comprehensive answers, has been developed.Introduction: The Importance of a Computer Test BankA computer test bank serves as a valuable resource for both teachers and students. It provides a collection of carefully curated questions that cover various topics in computer science and information technology. This enables instructors to assess the depth of understanding and problem-solving abilities of their students while allowing learners to gauge their own proficiency.The Purpose of an English VersionWith the increasing global demand for computer professionals, it is essential to have a standardized assessment tool that can be easily accessed and understood by individuals with different linguistic backgrounds. Thus, the development of an English version of the computer test bank aims to facilitate international exchange and provide a standardized evaluation of computer skills worldwide.Features of the English Version Computer Test Bank1. Inclusive Question Categories: The computer test bank covers a broad range of categories, including programming languages, algorithms, databasemanagement, computer networks, cybersecurity, software engineering, and more. This ensures a comprehensive evaluation of the test-takers' knowledge.2. Multiple Difficulty Levels: The questions in the test bank are categorized into different difficulty levels, such as beginner, intermediate, and advanced. This allows users to select questions that align with their skill level and progressively enhance their understanding.3. Real-world Scenarios: Many questions in the test bank are designed to simulate real-world scenarios, requiring critical thinking and problem-solving skills. This approach helps to bridge the gap between theoretical knowledge and practical application.4. Comprehensive Answer Key: Alongside the test bank, a detailed answer key is provided, offering step-by-step explanations and solutions. This ensures that users not only know the correct answer but also understand the reasoning behind it, fostering deeper comprehension.Benefits of the English Version Computer Test Bank1. Global Standardization: The English version of the computer test bank establishes a global standard for evaluating computer proficiency. Institutions from various countries can adopt the same assessment tool, allowing for accurate comparisons on an international scale.2. Enhancing Teaching: For educators, the test bank serves as an invaluable resource to prepare classroom assessments and track students' progress. It offers a wide range of questions that can be tailored to specific course objectives, facilitating effective teaching methodologies.3. Personal Skill Enhancement: Test bank users can self-administer quizzes to evaluate their understanding, identify knowledge gaps, and focus on areas that require improvement. By regularly practicing with the test bank, individuals can enhance their overall computer skills.4. Career Advancement: The English version of the computer test bank serves as a reliable certification of computer proficiency, augmenting one's resume and potential job prospects. Employers can trust the standardized assessment results, simplifying the hiring process.ConclusionThe availability of the English version of the computer test bank and its accompanying answer key is a significant step forward in standardizing computer proficiency evaluation. Whether used by educators, students, or professionals, it contributes to the enhancement of computer literacy worldwide. By adopting this comprehensive assessment tool, individuals can strengthen their computer skills and open doors to rewarding opportunitiesin the ever-evolving field of technology.。

操作系统期末考试试题及答案PDF

操作系统期末考试试题及答案PDF一、单项选择题（每题2分，共20分）1. 在现代操作系统中，进程和程序的主要区别是（）。

A. 进程是一个动态的概念，程序是一个静态的概念B. 进程是程序的执行过程，程序是进程的代码集合C. 进程是程序的代码和数据的集合，程序是进程的执行过程D. 进程是程序的代码集合，程序是进程的执行过程2. 在操作系统中，进程调度的目的是（）。

A. 决定进程的执行顺序B. 提高CPU的利用率C. 保证进程的公平性D. 以上都是3. 下列关于死锁的描述，不正确的是（）。

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. 死锁产生的四个必要条件是互斥、占有和等待、不可剥夺、循环等待C. 死锁的预防方法是破坏占有和等待条件D. 死锁的避免方法是通过银行家算法来实现4. 在操作系统中，虚拟内存的主要作用是（）。

A. 提高内存的访问速度B. 提高内存的利用率C. 扩大内存的容量D. 以上都是5. 下列关于文件系统的的说法，不正确的是（）。

A. 文件系统是操作系统中负责管理文件的系统B. 文件系统提供了文件的创建、删除、读取和写入等操作C. 文件系统将文件存储在磁盘上，并且可以对文件进行加密D. 文件系统允许多个用户同时访问同一个文件6. 在操作系统中，分页管理方式的主要优点是（）。

A. 减少了内存的碎片B. 简化了内存管理C. 提高了内存的利用率D. 以上都是7. 下列关于进程通信的说法，不正确的是（）。

A. 进程通信是指进程之间交换信息的过程B. 进程通信的方式有共享内存、消息传递、信号量等C. 进程通信可以提高系统的并发性D. 进程通信会导致进程的阻塞8. 在操作系统中，中断处理程序的主要作用是（）。

A. 处理硬件设备发出的中断信号B. 处理用户发出的中断信号C. 处理操作系统发出的中断信号D. 处理进程发出的中断信号9. 下列关于操作系统的用户界面的说法，不正确的是（）。

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。

操作系统全英文期中考试题(带答案)

XX大学2011——2012学年第一学期《操作系统》期中考试试题（A）一、选择（每题1分，共20分）1.Which function does the operating system can not complete directly of the following four options? ( b )A.Managing computer's hard drivepile the programC.Virtual memoryD.Delete files2.Considering the function of the operating system, ( b ) must give timely response for the external request within the specified time.A.multiuser time sharing systemB.real-time operating systemC.batch operating systemwork operating system3. A process can transform from waiting state to ready state relying on ( d )A.programmer commandB.system serviceC.waiting for the next time sliceD.wake-up of the 'cooperation' process4.As we all know,the process can be thought of as a program in execution.We can deal with the the problem about ( b ) easier after importing the concept of process.A.exclusive resourcesB.shared resourcesC.executing in orderD.easy to execute5.CPU-scheduling decisions may take place under the following circumstances except which one？（D ）A.When a process switches from the running state to the waiting stateB.When a process switches from the running state to the ready stateC.When a process switches from the waiting state to the ready stateD.When a process switches from the ready state to the waiting state6.In the four common CPU scheduling algorithm, Which one is the best choice for the time-sharing system in general？（ C ）A.FCFS scheduling algorithmB.Priority scheduling algorithmC.Round-robin scheduling algorithmD.Shortest-job-first scheduling algorithm7.If the initial value of semaphore S is 2 in a wait( ) and signal( ) operation，its current value is -1,that means there are ( B ) processes are waiting。

《操作系统》往年试卷及答案例

华南农业大学期末考试试卷（A 卷）201X 学年第一学期考试科目：操作系统考试类型：（闭卷）考试考试时间： 120 分钟学号姓名年级专业（答案直接写在试卷上，卷面书写必须工整、清晰、规范）一、选择及填空题（本大题共25个空，每空1分，共25分）1. _________操作系统能及时处理由过程控制反馈的数据并响应。

A. 分布式B. 实时C. 分时D. 嵌入式2. 当CPU 处于系统态时，它可以执行的指令是计算机系统的________。

A. 只有访管指令B. 只有特权指令C. 所有指令D. 只有非特权指令3. 在“基址B+限长L ”内存保护方案中，合法的逻辑地址A 应该满足_________条件。

A. 0≤A ＜LB. 0≤A ≤LC. B ≤A ＜LD. B ≤A ≤L4. 分时操作系统的主要目标是提高或改善计算机系统的_________。

A. 实时性B. 资源利用率C. 交互性D. 软件运行速度5. Linux 中的伙伴系统是用于________。

A. 文件目录的查找B. 磁盘空间的管理C. 内存空间的管理D. 文件保护6. 在下列死锁的解决方法中，属于死锁预防策略的是________。

A. 银行家算法B. 资源有序分配C. 剥夺资源D. 资源分配图化简7. 进程创建时，操作系统不需要给新进程执行下面的________工作。

A. 分配唯一的PIDB. 分配内存空间C. 初始化PCBD. 抢占当前进程8. 虚拟存储器的目的是实现________。

A. 存储保护B. 程序迁移C.动态重定位D. 扩充主存容量9. 某分时系统将有50个用户同时上机，为保证2s 的响应时间，时间片最大应为_______。

A. 50msB. 40msC. 100msD. 20ms10. “选一个进程占用CPU ”是_________的功能。

A. 短程调度B. 中程调度C. 长程调度D. 高级调度11. 与系统“抖动”现象无关的原因是__________。

操作系统期末试题(含答案)

操作系统期末考试（A）1、文件系统的主要组成部分是（ D ）A、文件控制块及文件B、I/O文件及块设备文件C、系统文件及用户文件D、文件及管理文件的软件2、实现进程互斥可采用的方法（C）A、中断B、查询C、开锁和关锁D、按键处理3、某页式管理系统中，地址寄存器的低9位表示页内地址，则页面大小为（B）A、1024字节B、512字节C、1024KD、512K4、串联文件适合于（B）存取A、直接B、顺序C、索引D、随机5、进程的同步与互斥是由于程序的（D ）引起的A、顺序执行B、长短不同C、信号量D、并发执行6、信号量的值（D ）A、总是为正B、总是为负C、总是为0D、可以为负整数7、多道程序的实质是（B）A、程序的顺序执行B、程序的并发执行C、多个处理机同时执行D、用户程序和系统程序交叉执行8、虚拟存储器最基本的特征是（A）A、从逻辑上扩充内存容量B、提高内存利用率C、驻留性D、固定性9、飞机定票系统是一个（A ）A、实时系统B、批处理系统C、通用系统D、分时系统10、操作系统中，被调度和分派资源的基本单位，并可独立执行的实体是（C）A、线程B、程序C、进程D、指令二、名词解释（每小题3分，共15分）1.死锁: 多个进程因竞争资源而造成的一种僵局，若无外力作用，这些进程将永远不能再向前推进2.原子操作: 一个操作中的所有动作要么全做，要么全不做，它是一个不可分割的操作。

3.临界区: 在每个进程中访问临界资源的那段代码4.虚拟存储器: 是指仅把作业的一部分装入内存便可运行作业的存储器系统。

也即是具有请求调入功能和置换功能，能从逻辑上进行扩充的一种存储系统。

5.文件系统: 是指含有大量的文件及其属性的说明，对文件进行操纵和管理的软件，以及向用户提供的使用文件的接口等的集合三、判断改错题（判断正误，并改正错误，每小题2分，共20分）1、通道是通过通道程序来对I/O设备进行控制的。

（T）2、请求页式管理系统中，既可以减少外零头，又可以减少内零头。

计算机英语考试题及答案

计算机英语考试题及答案一、选择题（每题2分，共20分）1. Which of the following is not a type of computer hardware?A. CPUB. RAMC. SoftwareD. Hard Disk答案：C2. What does the acronym "USB" stand for?A. Universal Serial BusB. User System BusC. User Storage BusD. Universal Storage Bus答案：A3. What is the primary function of a router in a computer network?A. To store dataB. To process dataC. To connect multiple devicesD. To print documents答案：C4. Which of the following is a programming language?A. HTMLB. CSSC. JavaScriptD. All of the above答案：D5. What does "RAM" stand for in computer terminology?A. Random Access MethodB. Random Access MemoryC. Remote Access MemoryD. Rapid Access Memory答案：B6. What is the term for a collection of data stored on a computer?A. FileB. FolderC. DatabaseD. Memory答案：A7. Which of the following is a type of computer virus?A. WormB. TrojanC. Both A and BD. None of the above答案：C8. What is the purpose of a firewall in a computer system?A. To prevent unauthorized accessB. To speed up internet connectionsC. To store dataD. To print documents答案：A9. What does "GUI" stand for in the context of computer systems?A. Graphical User InterfaceB. General User InterfaceC. Global User InterfaceD. Graphical Universal Interface答案：A10. What is the term for a small computer program that performs a specific task?A. ApplicationB. SoftwareC. UtilityD. Script答案：D二、填空题（每题2分，共20分）1. The basic unit of data in a computer is called a____________.答案：bit2. A computer's operating system is an example of______________.答案：system software3. The process of converting data into a form that can be understood by a computer is called ______________.答案：encoding4. The term used to describe the speed of a computer's processor is ______________.答案：clock speed5. A computer network that spans a large geographical area is known as a ______________.答案：WAN (Wide Area Network)6. The process of recovering lost data is called______________.答案：data recovery7. A computer program that is designed to disrupt or damage a computer system is known as a ______________.答案：malware8. The primary storage medium for a computer's operating system and most frequently used programs is the______________.答案：hard drive9. The term used to describe the process of transferring data from one computer to another is ______________.答案：data transfer10. A computer that is part of a network and shares its resources with other computers is called a ______________.答案：server三、简答题（每题10分，共40分）1. What are the main components of a computer system?答案：The main components of a computer system include the central processing unit (CPU), memory (RAM), storage devices (hard disk, solid-state drive, etc.), input devices (keyboard, mouse, etc.), output devices (monitor, printer, etc.), andthe operating system.2. Explain the difference between hardware and software in a computer system.答案：Hardware refers to the physical components of a computer, such as the CPU, memory, and storage devices. Software, on the other hand, comprises the programs and instructions that run on the hardware, including theoperating system, applications, and utilities.3. What is the role of a firewall in a computer network?答案：A firewall is a network security system that monitors and controls incoming and outgoing network traffic based on predetermined security rules. It acts as a barrier between a trusted internal network and untrusted external networks,such as the Internet, to prevent unauthorized access and protect the internal network from potential threats.4. Describe the process of data encryption and its importance in computer security.答案：Data encryption is the process of converting readable data into an unreadable format, called ciphertext, using an algorithm and a key. This process ensures that only authorized parties with the correct key can access and decrypt the data. Encryption is crucial for protecting sensitive information from unauthorized access, ensuring data privacy and security in computer systems and networks.。

计算机英语试题及答案

计算机英语试题及答案一、选择题（每题2分，共20分）1. Which of the following is not a programming language?A. JavaB. C++C. PythonD. Photoshop2. What does CPU stand for?A. Central Processing UnitB. Central Power UnitC. Central Printing UnitD. Central Programming Unit3. What is the full form of HTML?A. Hyper Text Markup LanguageB. Hyper Text Markup LanguageC. Hyper Text Markup LanguageD. Hyper Text Markup Language4. In computer networking, what does IP stand for?A. Internet ProtocolB. Internet ProgrammingC. Internet ProviderD. Internet Processor5. What is the primary function of a router?A. To print documentsB. To store dataC. To connect multiple networksD. To provide internet access6. Which of the following is a type of database management system?A. MS WordB. MS ExcelC. MS AccessD. MS PowerPoint7. What does the acronym RAM stand for?A. Random Access MemoryB. Rapid Access MemoryC. Remote Access MemoryD. Real Access Memory8. What is the term for a collection of data stored in a computer in a structured format?A. FileB. DocumentC. DatabaseD. Folder9. Which of the following is not a type of computer virus?A. TrojanB. WormC. AntivirusD. Ransomware10. What does the term "bandwidth" refer to in the context ofinternet usage?A. The width of the cableB. The speed of data transferC. The number of usersD. The quality of the connection二、填空题（每题2分，共20分）1. The basic unit of data in computing is called a ________.2. A ________ is a type of software that is designed to prevent, detect, and remove malware.3. The process of converting data into a form that can be easily transmitted or stored is known as ________.4. A ________ is a hardware device that allows a computer to connect to a network.5. The term "cybersecurity" refers to the practice of protecting systems, networks, and programs from ________ and other types of cyber attacks.6. A ________ is a collection of related data and the way it is organized.7. The ________ is a type of computer memory that retains data even when the power is off.8. The process of finding and fixing errors in a program is known as ________.9. A ________ is a type of software that allows users to create and edit documents.10. The ________ is a set of rules that define how data is formatted, transmitted, and received.三、简答题（每题10分，共30分）1. Explain the difference between a server and a client in a network.2. Describe the role of an operating system in a computer system.3. What are the key components of a computer system?四、论述题（每题30分，共30分）1. Discuss the importance of data backup and recovery in a computer system.答案：一、选择题1. D2. A3. A4. A5. C6. C7. A8. C9. C10. B二、填空题1. bit2. Antivirus3. Encoding4. Network Interface Card (NIC)5. unauthorized access6. Database7. Hard disk8. Debugging9. Word processor10. Protocol三、简答题1. In a network, a server is a computer system or device that provides resources or services to other computers, known as clients. The server manages network traffic, centralizes data storage, and provides access to shared resources, while the client is a computer that requests and uses these resourcesor services.2. An operating system is the software that manages computer hardware, software resources, and provides services for computer programs. It acts as an intermediary between theuser and the computer hardware, allowing users to interactwith the computer without needing detailed knowledge of the hardware.3. The key components of a computer system include thecentral processing unit (CPU), memory (RAM), storage devices (hard disk, SSD), input devices (keyboard, mouse), output devices (monitor, printer), and peripheral devices (scanner, webcam).四、论述题1. Data backup and recovery are critical in a computer system because they ensure that data is preserved in case ofhardware failure, data loss, or cyber attacks. Regularbackups allow for the restoration of data to a previous state, minimizing downtime and potential loss of information. This process is essential for maintaining business continuity and protecting against data loss.。

操作系统期末考试题及答案

操作系统期末考试题及答案一、选择题（每题2分，共20分）1. 在操作系统中，进程和线程的主要区别是：A. 进程有独立的内存空间，线程共享内存空间B. 进程和线程没有区别C. 进程和线程共享内存空间D. 线程有独立的内存空间，进程共享内存空间答案：A2. 以下哪个是操作系统的五大基本功能之一？A. 文件管理B. 网络通信C. 用户界面D. 数据加密答案：A3. 在分页存储管理中，页表的作用是：A. 存储进程的代码和数据B. 实现虚拟内存C. 存储页的物理地址D. 映射虚拟地址到物理地址答案：D4. 死锁的必要条件不包括：A. 互斥条件B. 请求和保持条件C. 不剥夺条件D. 循环等待条件答案：B5. 在操作系统中，文件系统的主要作用是：A. 管理进程B. 管理内存C. 管理设备D. 管理文件和目录答案：D（此处省略其他选择题）二、简答题（每题10分，共30分）1. 简述进程和程序的区别。

答案：进程是程序在执行时的状态，包括程序代码、数据、堆栈等，是资源分配的基本单位。

程序是静态的指令集合，是进程执行的指令集合，不包含执行状态信息。

2. 描述死锁的概念及其产生的原因。

答案：死锁是指在多进程系统中，两个或多个进程因争夺资源而造成的一种僵局，每个进程都在等待其他进程释放资源，而其他进程又在等待它释放资源，导致系统无法继续运行。

死锁产生的原因通常包括：互斥条件、请求和保持条件、不剥夺条件和循环等待条件。

3. 解释虚拟内存的概念及其工作原理。

答案：虚拟内存是一种内存管理技术，它允许计算机通过硬盘空间来扩展可用的内存。

操作系统将虚拟内存划分为多个页面，当物理内存不足时，可以将一些不常用的页面从物理内存移动到硬盘上，这个过程称为页面置换。

虚拟内存使得程序可以使用比物理内存更大的地址空间，提高了内存的使用效率。

三、计算题（每题15分，共30分）1. 假设有一个具有4个页面的程序，页面大小为4KB，程序的起始地址为0x1000，页面起始地址为0x0。

操作系统期末考试试题及答案

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. 在操作系统中，进程和程序的区别是什么？A. 进程是程序的执行实例B. 程序是进程的执行实例C. 进程和程序是同一个概念D. 进程是程序的存储介质答案：A2. 死锁的必要条件包括以下哪几个？A. 互斥条件、占有和等待、不可剥夺条件、循环等待条件B. 互斥条件、占有和等待、可剥夺条件、循环等待条件C. 互斥条件、释放和等待、不可剥夺条件、循环等待条件D. 互斥条件、占有和等待、可剥夺条件、非循环等待条件答案：A3. 虚拟内存技术允许计算机执行以下哪项操作？A. 运行比物理内存大的程序B. 存储更多的数据C. 加速程序的执行D. 减少程序的执行时间答案：A4. 在现代操作系统中，分页和分段的区别是什么？A. 分页是连续内存分配，分段是离散内存分配B. 分页是离散内存分配，分段是连续内存分配C. 分页和分段都是连续内存分配D. 分页和分段都是离散内存分配答案：B5. 操作系统中的文件系统的主要功能是什么？A. 存储和管理文件B. 管理进程C. 管理内存D. 管理输入输出设备答案：A...（此处省略其他选择题）二、简答题（每题10分，共30分）1. 简述操作系统的五大基本功能。

答案：操作系统的五大基本功能包括：进程管理、内存管理、文件系统管理、输入/输出设备管理和用户接口管理。

2. 解释什么是时间片轮转调度算法，并说明其优缺点。

答案：时间片轮转调度算法是一种CPU调度算法，它将CPU时间分配给每个进程，每个进程被分配一个固定的时间片。

当一个进程的时间片用完后，CPU将被分配给下一个进程。

优点包括公平性和响应时间的可预测性。

缺点是上下文切换的开销可能会影响性能。

3. 描述文件系统的层次结构，并解释每个层次的功能。

答案：文件系统的层次结构通常包括文件系统管理器、文件目录和文件。

文件系统管理器负责管理整个文件系统，文件目录用于组织文件，而文件则是存储数据的基本单元。

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 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 )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 theresources 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 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 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。

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。

(完整word版)操作系统英文版课后习题答案整理(word文档良心出品)

1.1What are the three main purposes of an operat ing system?⑴ In terface betwee n the hardware and user;(2) man age the resource of hardware and software;(3) abstracti on of resource;Answer;•Tb provide an environment ksr a computer user to execute programs on computer h日rchvare in n convenient and efficient manner.•Tb allocate the separate resources of the computer as needed to st)ke the problem given.The allcxation prtxzess should b? as fair and efficient as possible.•Asa control program it serves two major functions； (1) supervision of the execution 由user programs to prevent errors and improper use of the computer, and (2) management of the operatioji and control of I/O devices.1.2 List the four steps that are necessary to run a program on a completely dedicated machine. Preprocessing > Process ing > Linking > Executi ng.Answer:乩Reserxe machine time*b* Manually load program into memory.u Load starting address and begin exe匚Lition.cL Monitor and control execution of program fr(im console.b. In teractivec. Time shar ingd. Real timee. Networkf. DistributedAnswera” Batch, Jobs w计h similar reeds are batched together and run through the computer as a group by ^r\operator or automatic jc)b s^quenc^r. [^rformanct? i$ incr^a^ed by atteiTipting to keep CPU 3nd I/O devices busv 311 timesi through buffering, off-line operation^ spooling, and multiprogramming. Batch is good for executing 】arge jobs thjit need little interacticm; it can be submitted and piuked up later.b. Interactive. This system is cumpofied of m^nv short transactions where the results of thenext transactiiin may be unpredictablen Respond time needs tn be short (sectwids) since the user submits and waik for the result.u Time sharing. This systems u 船呂匚Pl sch<*duling ^nd multi programming to pmvidu eConoiniCcil interactive use of 蛊system. The CPL switches rapidly iri>m one user tn another Instead of having a job defined by spcxiled card images^ each program readsits next control card from the terminab and output is normally printed immediately to the screen.(_L Real time. Often tisvci in a dedicated application, this system reads information from sensors and must respond within a fixed amount of time to ensure correct performance.work.f.Distributed .This system distributes computation among several physical processors” TheprtKessors do not share menion- or a clock. Instead, each pnxzessor has its own kxzalmemory. They communicate with each other through various communication lines f such asa high-speed bus or telephone line.1.7 Wehave stressed the need for an operating system to make efficient use of the computing hardware. When is it appropriate for the operating system to forsake this principle and to waste ” resources? Why is such a system not really wasteful?Answer Single-user systems should maximize use of the system for the user A GUImight "xvastc * GPL' cycles, but it optimizes the user T s interaction with the system.2.2 How does the distinction between monitor mode and user mode function as a rudimentary form of protecti on (security) system?Answer: By establishing a set of privileged instructions that can be executed only when in tlie m(snitnr mt)de f the tiperating system is assured of controlling the entire system at all times.2.3 What are the differences between a trap and an interrupt? What is the use of eachfun cti on?Answer An interrupt is a ha rd \ v a re-^en era ted change-of-flow within the system. An interrupt handler is summoried to deal with the cause oF the interrupt; control is then re turned to the interrupted context and instruction. A trap is a software-j;enerated interrupt. An interrupt can be uwd to signal the compJrtio n “f an I/O obviate the nevd for polling. A trap can be used ti> call operating svstem routines or to catch arithmetic errors.2.5 Which of the follow ing in structi ons should be privileged?a. Set value of timer.b. Read the clock.c. Clear memory.d. Turn off interrupts.e. Switch from user to monitor mode.3OS Exercise BookClass No. NameAnswer: The following instructions should be privileged:Set value of timer,b.Clear memory.Jc.Turn off interrupts.d.Switch from user to monitor mode*2.8 Protecting the operating system is crucial to ensuring that the computer system operates correctly. Provision of this protection is the reason behind dual-mode operation, memory protection, and the timer. To allow maximum flexibility, however, we would also like to place mini mal con stra ints on the user.The following is a list of operations that are normally protected. What is the minimal setof in structi ons that must be protected?a. Change to user mode.b. Change to mon itor mode.c. Read from mon itor memory.d. Write into mon itor memory.e. Fetch an instruction from monitor memory.f. Tur n on timer in terrupt.g. Turn off timer interrupt.Answer: The minimal 5et of instructions that must be protected are：a.Change to monitor mtxle.b.Read from moni tor memory*c.Write into monitor me mor v.Jd.Turn off timer interrupt3.6 List five services provided by an operat ing system. Explain how each provides convenience to the users. Explain also in which cases it would be impossible for user-level programs to provide these services.Answer;«Program execution. The operating system loads thv contents (or sections) of a file into menidry and begins its execution. A user-level program could not be trusted to properly allocate CPU time.•I/O operations. Disks, tapes, serial lines；and other devices must be communicated with ata very low level. The user need only specify the dev ice and the operation to perform (in it,while the system converts that request into <ie\r ict^ or contr<i11er-spec i fit commands.User-level pnjgrams cannot be trusted to only access devices they should have access to and to only access them when they otherwise unused.«File-system manipulation, fhere are manv details in file creation, deletion/ alkKation, and naming that users should not have to perform. Blocks of disk space are used by files and must be tracked, deleting a file requires remo\ ing the name file information and freeing the <ilkx:ated blocky I^ratections must also be checked to assure proper file access. User prc^grams could neither ensure adherence to protect]on methcxis nor be trusted to allocate only free block吕and deallocate bkxzks on file deletion.J•Communications. Message passing between systems requires messages be turned into packets of information, sent to the network controller, trannmitted across a community tk>ns medium, and reassembled by the destination system.卩acket ordering and data correction must take place. Again, user program吕might not c(sordinate ac cess to the network dev ice, or they might receive packets destined tor other processes^»Error detection. Error detection (occurs at both the hardware and soFtwart? levels. At tlie hardware level, all data transfers must be inspected to ensure that data hax e not beencorrupted in transit All data on media must be checked to be sure they have not changed since they uritten to the media. At the software level, media must bechecked for data ccnsistencj^; for instance, do the number of allocated and unallocated blocks of storage match the total number on the device. There, errors are frequently pnxzess-independent (for instance, the 匚omiption of data on a disk)5 sc there must be a global program (the operating system) that handles all h pes of errors. Also, by having errors pmc essed by the operating system, processes need not contain code to catch and ccjrrect all the ernjrs possible on a system.3.7 What is the purpose of system calls?Answer; Sv>ttn'. dlltnv ustr-levtl lu request str\ ices nt 11 it? uperating svs-tem.3.10 What is the purpose of system programs?J V USWCE Svstcm programs can be thought of as bundle!ti of useful system oils. Thev r provide bcisic functidcaliU users and sci users do not need to wnte their cwn programs to s<>k r e comnicMi problems,4.1 MS-DOS provided no means of con curre nt process ing. Discuss three major complicati onsthat con curre nt process ing adds to an operat ing system.5OS Exercise BookClass No. NameAnswer：*A method of time sharing must be implemented to allow each of several prcxzesses to have access to the system. This method involves the preemption of processes that do notvoluntarily give up the CPU (by using a system ca1]r for instance) and the kernel being reentrant (so more than one prtxzess may be executing kernel code concurrently).・[Vocesses and system resources must have protections and must be protected from each other. Any given process must be limited in the amount of memory it can use and tlie ope Mt ions 让can perform on devices like di^ks.•Care must be taken in the kernel to prevent deadkxzks between prucesses, so processesaren*t waiting for each other's allocated rest>urces,4.6 The correct producer —consumer algorithm in Section 4.4 allows only n-1 buffers to befull at any one time. Modify the algorithm to allow all buffers to be utilized fully.Answer: No answer.5.1 Provide two program ming examples of multithread ing givi ng improve performa nee overa sin gle-threaded soluti on.Answer (1) A Web server that services each request in a separate Lliread. (2) A parallelized application such as matrix multiplication where different parts of the matrix may be worked on in parallel. (3) An intEivictin GUI program such as a debugger where a thread is used to monitor user input, another thread represents the running application, and a third thread monitors performance.5.3 What are two differences between user-level threads and kernel-level threads? Underwhat circumsta nces is one type better tha n the other?Answer: Context switching between user threads is quite sinniliir to switching between kernel threads, although it is dependent on the threads library and how it maps user threads to kernel threads. In general, context switching between user threads involves taking a user thread of its LWP and replacing it with another thread. This act typically involves saving and restoring the stttte of the registers.6.3 Consider the following set of processes, with the length of the CPU-burst time given inmillisec on ds:Process Burst Time PriorityP1103P211P323F414P552The processes are assumed to have arrived in the order P l, F2, F3, F4, P5, all at time 0.a.Draw four Gantt charts illustrati ng the executi on of these processes using FCFS, SJF, a non preemptive priority (a smaller priority n umber implies a higher priority), and RR (qua ntum = 1) scheduli ng.b.What is the turnaround time of each process for each of the scheduling algorithms in part a?c.What is the waiting time of each process for each of the scheduling algorithms in part a?d.Which of the schedules in part a results in the minimal average waiting time (over all processes)?An swer:Answer;a.The four Gantt charts areb.Turnaround timeFCFS RR SJF Priority101919 16112 1 113 741814421919149 6 匚Waiting time (turnaround time minus burst time)FCFS RR SJF Priority卩】0996Pz10100心115216P*133118Ps14941d. Shortest Job First6.4 Suppose that the following processes arrive for execution at the times indicated. Eachprocess will run the listed amou nt of time. In an sweri ng the questi ons, use non preemptive scheduling and base all decisions on the information you have at the time the decisionmust be made.PtXKCSS Ai ri\ al Time Burst Time0.087OS Exercise BookClass No. NameP20.44l.D1a. What is the average turnaround time for these processes with the FCFS scheduling algorithm?b. What is the average turnaround time for these processes with the SJF scheduling algorithm?c. The SJF algorithm is supposed to improve performa nee, but no tice that we chose to run process P1 at time0 because we did not know that two shorter processes would arrive soon. Compute what the average turnaround time will be if the CPU is leftidle for the first 1 un it and the n SJF scheduli ng is used. Remember that processes P1 and P2 are wait ing dur ing this idle time, so their wait ing time may in crease. Thisalgorithm could be known as future-k no wledge scheduli ng.Answera.10.53b.9.53c.6,86Remember that turnaround time LS finishing time minus arrival time, so have to subtract the arrival tinier to compute thu turnaround times. FCFS is 11 if you forget to subtract arrival time.6.10 Explain the differences in the degree to which the following scheduling algorithms discrim in ate in favor of short processes:a.FCFSb.RRc.Multilevel feedback queuesAnswer：a.FCFS—discriminates against short jobs since any short jobs arriving after long jobs willhave a longer waiting time.b.RR一treats all jobs equally (giving them equal bursts of CPU time) so short jobs will beable to leave the system faster since they will finish first.c.Multilevel feedback queues—work similar to the RR algorithm—thev discriminatefa^r i>rably toward slusrt jobs.7.7 Show that, if the wait and sig nal operati ons are not executed atomically,then mutual exclusi on may be violated.Answer No answer.7.8 The Sleepi ng-Barber Problem. A barbershop con sists of a wait ing room with n chairs and the barber room containing the barber chair. If there are no customers to be served,the barber goes to sleep. If a customer en ters the barbershop and all chairs are occupied,then the customer leaves the shop .If the barber is busy but chairs are available, the nthe customer sits in one of the free chairs. If the barber is asleep, the customer wakes up the barber. Write a program to coord in ate the barber and the customers.Answer: Please refer to the support ing Web s its for source code solution,8.2 Is it possible to have a deadlock involving only one single process? Explain your answer.Answer Ncx I'his folknvs directly from the hold-and-wait condition.8.4 Con sider the traffic deadlock depicted in Figure 8.11.a. Show that the four n ecessary con diti ons for deadlock in deed hold in this example.b. State a simple rule that will avoid deadlocks in this system.Answer No answer.8.13 Con sider the follow ing sn apshot of a system:Allocati on Max AvailableA B C D A B C D A B C DP00 0 1 20 0 1 2 1 5 2 0P1 1 0 0 0 1 7 5 0P2 1 3 5 4 2 3 5 6P30 6 3 20 6 5 2P40 0 1 40 6 5 6An swer the follow ing questi ons using th e ban ker s algorithm:a.What is the content of the matrix Need?b.Is the system in a safe state?c.If a request from process P1 arrives for (0,4,2,0), can the request be gran tedimmediately?Answer;A. Deadlcx^k cannot ixrcur because preemption exists,b. Yes. A process may never acquire all the resources 让needs if they are continuouslypreempted by a series of requests such as those of process C.9.5 Given memory partitions of 100K, 500K, 200K, 300K, and 600K (in order), how would each of the First-fit, Best-fit, and Worst-fit algorithms place processes of 212K, 417K, 112K,and 426K (in order)? Which algorithm makes the most efficie nt use of memory?Answer:a. First-fit:b* 212K is put in 500K partitionc. 417K is put in 600K partitiond* 112K is put in 288K partition (new partition 288K = 500K - 212K)e.426K must waitf.Best-fit:g.212K is put in 300K partition9OS Exercise BookClass No. Nameh.417K is put in 500K partitioni.112K is put in 200K partitionj.426K is put in 600K partitionk.Worst-fit:L 212K is put in 600K partitionm. 417K is put in 500K partitionn. 112K is put in 388K partitionc 426K must waitIn this example, Best-fit turns out to be the bE%t*9.8 Con sider a logical address space of eight pages of 1024 words each, mapped onto a physicalmemory of 32 frames.a. How many bits are there in the logical address?b. How many bits are there in the physical address?Answera,l.ogica) address: 13 bitsb.Physical address： 15 bitsJ9.16 Con sider the follow ing segme nt table:Segme nt Base Len gth02196001230014290100313275804195296What are the physical addresses for the follow ing logical addresses?a. 0,430b. 1,10c. 2,500d. 3,400e. 4,112Answer:a.219 + 430 = 649b.2300 + 10 = 2510u ill亡月ed reference, trap ki operating systemd.1327 -b 400 = 1727e.illegal reference, trap to operating system10.2 Assume that you have a page referenee string for a process with m frames (initiallyall empty). The page refere nee stri ng has len gth p with n disti net page n umbers occur init. For any page-replacement algorithms,a. What is a lower bou nd on the n umber of page faults?b. What is an upper bou nd on the n umber of page faults?Answer:a* nb»p1, 2, 3, 4, 2, 1,5, 6, 2, 1,2, 3, 7, 6, 3, 2, 1, 2, 3, 6.How many page faults would occur for the following replacement algorithms, assuming one, two, three, four, five, six, or seven frames? Remember all frames are initially empty, so your first unique pages will all cost one fault each.LRU replaceme ntFIFO replaceme ntOptimal replaceme nt11OS Exercise BookClassNo.NameNumbei of frames1 2 3 4 5 6 711.7 Expla in the purpose of the ope n and close operati ons.Answer ：» The o[fen operation informs the system that the named file is about to bect>me active^ * Tlie cluse operation informs the system that the named file )s nt> lunger in active use by the user who issued the dose operation.11.9 Give an example of an application in which data in a file should be accessed in the followi ng order: a. Seque ntially b. Ran domlyAnswer ：a. Print the 匚 on tent of the file,b. Print the content of record /. This record can be found using hashing or index tech niques.11.12 Con sider a system that supports 5000 users. Suppose that you want to allow 4990 of these users to be able to access one file.a. How would you specify this protection scheme in UNIX?b. Could you suggest ano ther protecti on scheme that can be used more effectively for this purpose tha n the scheme provided by UNIX?Answer:a. There are twc methods for achieving this ： L Create an access control list withtlu? names of all 4990 users.ii. Put these 4990 users in one ^roup and set the group access accordingly. This scheme cannot always be implemented since user groups are restricted by the system. b. The universe access information applies to all users unless their name appears in the access-control list with different access permission. With this scheme you simply put the names of the remaining ten users in the access control list but \v 让h no access pri\ ileges allovcexfLRUFIFOOptimal 20 20 20 18 18 15 15 16 11 10 14 8 8 10 7 7 10 7 7 77Answer12.1 Consider a file currently consisting of 100 blocks. Assume that the file control block (andthe index block, in the case of indexed allocation) is already in memory. Calculate howmany disk I/O operations are required for contiguous, linked, and indexed (single-level)allocati on strategies, if, for one block, the follow ing con diti ons hold. In thecon tiguousallocati on case, assume that there is no room to grow in the beg inning, but there is room to grow in the end. Assume that the block in formatio n to be added is stored in memory.a. The block is added at the beg inning.b. The block is added in the middle.c. The block is added at the end.d. The block is removed from the begi nning.e. The block is removed from the middle.f. The block is removed from the end.AnswerLinked Indexeda. 201 1 1b. 1015211 3 1d. 198 1 0e. 9852 0f. 0 100 013.2 Con sider the follow ing I/O sce narios on a sin gle-user PC.a. A mouse used with a graphical user in terfaceb. A tape drive on a multitasking operating system (assume no device preallocation is available)c. A disk drive containing user filesd. A graphics card with direct bus conn ecti on, accessible through memory-mappedI/OFor each of these I/O scenarios, would you design the operating system to use buffering, spooli ng, cachi ng, or a comb in ati on? Would you use polled I/O, or in terrupt-drive n I/O? Give reas ons for your choices.Answer:a. A mouse used with a graphical user interfaceBuffering may be needed to record mouse movement during times when higher- priority operations are taking place. Spooling and caching are inappropriate. Inter rupt driven I/O is most appropriate^b” A tape drive on a multitasking operating system (assume no device preAlkxzation is available)Buffering may be needed to manage throughput d让ferEria? behveen the tape drive and the sounzt? or destination of the I/O, C臼匚hing can be used to hold copies of that resides on the tape, for faster access. Spooling could be used to stage data to the device whenmultiple users desire to read from or write to it” Interrupt driven [/O is likely to allow the best performance.13OS Exercise BookClass No. Name匸* A disk drive containing user tilesBuffering can be used to hold data while in transit from user space to the disk, and visaversa. Caching can be used to hold disk-resident data for improved perfor mance.Spoc^ling is not necessary because disks are shared-access devices. Interrupt- driven T/O is best for devices such as disks that transfer data at slow rates,d. A graphics card w让h direct bus coi^nection, accessible through mem<irv-mapped I/OBuffering may be needed to control multiple access and for performance (doublebuffering can be used to hold the next screen image while displaying the current tme). Caching and spooling are not necessary r due to the fast and shared-access natures of the device. Polling and interrupts are only useful for input and for【/O completion detec tion f neither of which is needed for a mem()r y-ma pped device.14.2 Suppose that a disk drive has 5000 cylinders, numbered 0 to 4999. The drive is currently serving a request 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, 130Start ing from the curre nt head positi on, what is the total dista nee (in cyli nders) thatthe disk arm moves to satisfy all the pending requests, for each of the follow ingdiskscheduli ngalgorithms?a. FCFSb. SSTFc. SCANd. LOOKe. C-SCANAnswer：乩The FCFS schedule is 143f 86f 1470, 913, 1774, 948, 1509, 1022, 1754), 130. The total seek distance is 7081.b. The SSTF schedule is 143, 130, 86. 913, 948, 1022, 1470, 1509. 1750r 1774. The total seekdistance is 1745.€. The SCAN schedule is 143, 913, 94«f1022f 1470, 1509,1750f 1774, 4999,130, 86. The to怙1 seek distance is 9769.d.The LOOK schedule is 143, 913, 948,1022, 1470,1509,1750,177< 130,86. The total seekdistance is 3319*e.The C-SCAN schedule is 143f 913,948,1022J 470,1509.1750,1774,4999,8& 130. The totalseek distance is 9813.f.(Bonus.) The C-LOOK schedule is 143,913,94& 1022,1470,1509.1750,1774, 86,130. Thetotal seek distance is 3363.1.1 1.62.3 2.53.7 6.3 6。

操作系统期末考试题和标准答案及评分标准

"操作系统"试题〔A卷〕〔考试时间：90分钟〕一、选择题〔共20分，每题1分〕1．〔〕不是根本的操作系统。

A.批处理操作系统B.分时操作系统C.实时操作系统D.网络操作系统2．现代OS具有并发性和共享性，是〔〕的引入导致的。

A.单道程序B. 磁盘C. 对象D.多道程序3．一般来说，为了实现多道程序设计，计算机最需要〔〕。

A.更大的内存B.更多的外设C.更快的 CPUD.更先进的终端4．在下面的I/O控制方式中，需要CPU干预最少的方式是〔〕。

A.程序I/O方式B.中断驱动I/O控制方式C.直接存储器访问DMA控制方式D.I/O通道控制方式5．在进程状态转换时，以下〔〕转换是不可能发生的。

A.就绪态→运行态B.运行态→就绪态C.运行态→阻塞态D.阻塞态→运行态6．一个进程执行过程中不可能对应( )。

A.一个PCBB.一个JCBC.多个PCBD.一个程序7．进程调度又称为低级调度，其主要功能是( )。

A.选择一个作业调入内存B.选择一个主存中的进程调出到外存C.选择一个外存中的进程调入到主存D.将一个就绪的进程投入运行8．如果允许不同用户的文件可以具有一样的文件名，通常采用〔〕来保证按名存取的平安。

A.重名翻译机构B.建立索引表C.建立指针D.多级目录构造9．文件控制块不包括〔〕。

A.文件名B.文件访问权限说明C.文件物理位置信息D.磁盘坏块信息10．为了提高设备分配的灵活性，用户申请设备时应指定〔〕号。

A.设备类相对B.设备类绝对C.相对D.绝对11．假设进程P一旦被唤醒就能够投入运行，系统可能为( )。

A.在分时系统中，进程P的优先级最高。

B.抢占调度方式，就绪队列上的所有进程的优先级皆比P的低。

C.就绪队列为空队列。

D.抢占调度方式，P的优先级高于当前运行的进程。

12．为了照顾紧迫型作业，应采用〔〕。

A.先来效劳调度算法B.短作业优先调度算法C.时间片轮转调度算法D.优先权调度算法13．一作业进入内存后，则所属该作业的进程初始时处于〔〕状态。

英文版计算机试题及答案

英文版计算机试题及答案一、选择题（每题2分，共20分）1. Which of the following is not a function of an operating system?A. Process managementB. Memory managementC. Data storageD. File management2. In a computer network, what does the term "bandwidth" refer to?A. The width of the network cableB. The maximum rate of data transferC. The number of users connectedD. The speed of the network processor3. What is the primary purpose of a firewall?A. To prevent unauthorized access to a networkB. To encrypt dataC. To manage network trafficD. To store user passwords4. Which of the following is a type of software used for creating and editing documents?A. Spreadsheet softwareB. Database softwareC. Word processing softwareD. Graphics software5. What is the term used to describe the process of converting data from one format to another?A. Data migrationB. Data transformationC. Data conversionD. Data translation6. What does the acronym "CPU" stand for in computing?A. Central Processing UnitB. Central Processing UnitC. Computer Processing UnitD. Computing Processing Unit7. What is the function of a router in a network?A. To connect multiple networksB. To store dataC. To provide power to devicesD. To print documents8. What is the process of finding and fixing errors in software called?A. DebuggingB. PatchingC. UpdatingD. Patching9. Which of the following is a type of computer virus that replicates itself by attaching to other programs?A. TrojanB. WormC. RansomwareD. Spyware10. What is the term for the graphical representation of data on a computer screen?A. Data visualizationB. Data representationC. Data graphingD. Data mapping二、填空题（每题2分，共20分）1. The _________ is the primary memory used by a computer to store data and instructions that are currently being processed.2. A _________ is a type of software that allows users to create and edit images.3. The process of converting analog signals to digital signals is known as _________.4. A _________ is a collection of data stored in a structured format.5. The _________ is a hardware component that connects a computer to a network.6. In computer programming, a _________ is a sequence of statements that perform a specific task.7. The _________ is a type of malware that hides its presence and waits for a trigger to activate.8. A _________ is a type of software that is designed to protect a computer from unauthorized access.9. The _________ is the process of organizing and managing data in a database.10. A _________ is a type of software that allows users tocreate and edit spreadsheets.三、简答题（每题10分，共30分）1. Describe the role of a server in a computer network.2. Explain the difference between a compiler and an interpreter in programming.3. Discuss the importance of data backup and recovery in a computing environment.四、编程题（每题15分，共30分）1. Write a simple program in Python that calculates the factorial of a given number.2. Create a function in Java that takes an array of integers and returns the largest number in the array.答案：一、选择题1. C2. B3. A4. C5. C6. A7. A8. A9. B10. A二、填空题1. RAM (Random Access Memory)2. Graphics software3. Analog-to-digital conversion4. Database5. Network interface card (NIC)6. Function or procedure7. Trojan8. Antivirus software9. Database management10. Spreadsheet software三、简答题1. A server in a computer network is a powerful computer or system that manages network resources, including hardware and software, and provides services to other computers on the network, such as file storage, web hosting, and print services.2. A compiler is a program that translates source codewritten in a programming language into machine code that a computer can execute. An interpreter, on the other hand, reads and executes the source code line by line without the need for a separate compilation step.3. Data backup and recovery are crucial in a computing environment to prevent data loss due to hardware failure, software bugs, or malicious attacks. Regular backups ensure that data can be restored to a previous state in case of corruption or deletion.四、编程题1. Python Program for Factorial Calculation:```pythondef factorial(n):if n == 0:return 1 else:。

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。