04-09软件设计师专业英语真题译文及答案
2004年上半年软件设计师级答案
2004年上半年软件设计师级答案
上午答案
下午答案
试题一
[问题1]
A:传感器: B:报警器
[问题2)
监控传感器、显示信息和状态
[问题3]
1、“读传感器”添加输入数据流“传感器状态”
2、“显示格式”添加输出数据流“TV信号”
3、“拨号”添加输出数据流“电话拨号”
[问题4]
逻辑数据流图说明应该具有哪些加工,而不关心这些加工是如何实现的:物理数据流图则要说明加工是如何实现的。
试题二
[问题1]
(1)l (2)col (3)row (4)2 (5)col (6)row (7)k
[问题2]
判断条件1:b
判断条件2:e
判断条件3:f
试题三
[问题1]
A:浏览客户信息 B:修改个人信息 C:登录 D:删除客户信息
[问题2]
(1)1 (2)0..* (3)0..1 (4)0..1
[问题3]
(1)4种关系的含义:
依赖表示类之间的使用关系。
概括表示一般类和特殊类之间的关系。
关联和聚集都表示实例之间的结构关系。
(2)关联和聚集的区别:
关联指明一个类的对象与另一类的对象间的联系:两个类之间的关联表示了两个同等地位类之间的结构关系,这两个类在概念上是同级别的。
聚集是一种特殊的关联,它表示整体/部分关系。
试题四
(1)JMP LAST
(2)SUB GRO,WDT,GR1
(3)ST GR2,BTASC,GR1
(4)ADD GR0,C48
(5)ST GR0,BTASC,GR1
试题五
(1)p&&p->data !=e 或 p&&(*p).data !=e
(2)p->Lchild 或 (*p).Lchild
(3)s->Rchild 或 (*s).Rchild
09年软件设计师上午试题答案
2009年上半年软考软件设计师上午试卷
● 海明校验码是在n 个数据位之外增设k 个校验位,从而形成一个k+n 位的新的码字,使新的码字的码距比较均匀地拉大。n与k的关系是(1)。
(1)A. 2K-1 ≥n+k B. 2n-1≤ n+k C. n =k D. n -1 ≤ k
● 假设某硬盘由5个盘片构成(共有8个记录面),盘面有效记录区域的外直径为30cm,内直径为10cm,记录位密度为250位/mm,磁道密度为16道/mm,每磁道分16个扇区,每扇区512字节,则该硬盘的格式化容量约为(2)MB。
●(3)是指按内容访问的存储器。
(3)A. 虚拟存储器 B. 相联存储器
C. 高速缓存(Cache)
D. 随机访问存储器
● 处理机主要由处理器、存储器和总线组成,总线包括(4)。
(4)A. 数据总线、地址总线、控制总线 B. 并行总线、串行总线、逻辑总线
C. 单工总线、双工总线、外部总线
D. 逻辑总线、物理总线、内部总线
● 计算机中常采用原码、反码、补码和移码表示数据,其中,±0 编码相同的是(5)。
(5)A. 原码和补码 B. 反码和补码
C. 补码和移码
D. 原码和移码
● 下面关于漏洞扫描系统的叙述,错误的是(7)。
(7)A. 漏洞扫描系统是一种自动检测目标主机安全弱点的程序
B. 黑客利用漏洞扫描系统可以发现目标主机的安全漏洞
C. 漏洞扫描系统可以用于发现网络入侵者
D. 漏洞扫描系统的实现依赖于系统漏洞库的完善
● 网络安全包含了网络信息的可用性、保密性、完整性和网络通信对象的真实性。其中,数字签名是对(8)的保护。
09下半年系统架构设计师英文真题
09下半年系统架构设计师英文真题、译文及答案An architectural style defines as a family of such systems in terms of a(71)of structural organization.More specifically, an architectural style defines a vocabulary of (72)and connector types, and a set of (73)on how they can be combined. For many styles there may also exist one or more (74)that specify how to determine a system's overall properties from the properties of its parts.Many of architectural styles have been developed over the years.
The best-known examples of(75)architectures are programs written in the Unix shell.
(71)A. pattern
B. data flow
C. business process
D. position level
(72)A. metadata
B. components
C. models
计算机专业英语的第四版课后习题翻译
中译英参考答案
Unit 1
A computer system consists of hardware system and software system. The hardware of the computer is usually divided into three major parts or three primary subsystems: the CPU, the memory subsystem, and the I/O subsystem.
The CPU performs many operations and controls computer. The memory subsystem is used to store program being executed by the CPU, along with the program’s data. The I/O subsystem allows the CPU to interact with input and output devices such as the keyboard and monitor of a personal computer. The components of the computer are connected to the buses.
The part of the computer that performs the bulk of data processing operations is called the central processing unit and is referred to as the CPU. In microcomputer, it is often called the microprocessor. The CPU is made up of three major parts: control unit, ALU, and register set.
计算机专业英语翻译及课后答案
Unit Nine: The Internet
Unit Nine/Section A
I. Fill in the blanks with the information given in the text:
1. research
2. ICANN或the Internet Corporation for Assigned Names and Numbers
3. router; gateway
4. temporary/dial-up; permanent/dedicated
5. ISP或Internet service provider
6. network; host
7. decimal
8. mnemonic
II. Translate the following terms or phrases from English into Chinese and vice versa:
1. cell phone 蜂窝电话,移动电话,手机
2. IP address 网际协议地址,IP地址
3. autonomous system 自主系统
4. dial-up connection 拨号连接
5. network identifier 网络标识符
6. binary notation 二进制记数法
7. mnemonic name 助记名,缩写名
8. Internet-wide directory system 因特网范围的目录系统
9. name server 名称服务器
10. Internet infrastructure 因特网基础结构
软件测试专业英语翻译题.doc
原文:
11 Steps for Software Testing Process
Step 1: Asses Development Plan and Status
This first step is a prerequisite to building the VV&T Plan used to evaluate the implemented software solution. During this step, testers challenge the completeness and correctness of the development plan. Based on the extensiveness and completeness of the Project Plan the testers can estimate the amount of resources they will need to test the implemented software solutio n.
Step 2: Develop the Test Plan
Forming the plan for testing will follow the same pattern as any software planning process. The structure of all plans should be the same, but the con tent will vary based on the degree of risk the testers perceive as associated with the software being developed.
软件设计师考试大纲+英语试题
计算机软件水平考试-软件设计师考试大纲(一)
计算机与软件工程知识
1.计算机科学基础
1.1 数制及其转换
·二进制、十进制和十六进制等常用制数制及其相互转换
1.2 数据的表示
·数的表示(原码、反码、补码、移码表示,整数和实数的机内表示,精度和溢出)
·非数值表示(字符和汉字表示、声音表示、图像表示)
·校验方法和校验码(奇偶校验码、海明校验码、循环冗余校验码)
1.3 算术运算和逻辑运算
·计算机中的二进制数运算方法
·逻辑代数的基本运算和逻辑表达式的化简
计算机软件水平考试-软件设计师考试大纲(二)
1.4 数学基础知识
·命题逻辑、谓词逻辑、形式逻辑的基础知识
·常用数值计算(误差、矩阵和行列式、近似求解方程、插值、数值积分)
·排列组合、概率论应用、应用统计(数据的统计分析)
·运算基本方法(预测与决策、线性规划、网络图、模拟)
1.5 常用数据结构
·数组(静态数组、动态数组)、线性表、链表(单向链表、双向链表、循环链表)、队列、栈、树(二叉树、查找树、平衡树、线索树、线索树、堆)、图等的定义、存储和操作
· Hash(存储地址计算,冲突处理)
1.6 常用算法
·排序算法、查找算法、数值计算方法、字符串处理方法、数据压缩算法、递归算法、图的相关算法
·算法与数据结构的关系、算法效率、算法设计、算法描述(流程图、伪代码、决策表)、算法的复杂性
计算机软件水平考试-软件设计师考试大纲(三)
2.计算机系统知识
2.1 硬件知识
2.1.1 计算机系统的组成、体系结构分类及特性
· CPU和存储器的组成、性能和基本工作原理
·常用I/O设备、通信设备的性能,以及基本工作原理
04-09程序员专业英语真题译文及答案
04 上
One use of networks is to let several computers share (66)such as file system,printers,and tape drives.
(66)A.CPU B.memory
C.resources
D.data
参考译文
计算机网络的作用之一就是让多台计算机共享文件系统、打印机和磁带机等资源。
参考答案
(66)C
A firewall is a (67)system designed to(用来)(68)an organization’s network against threats.
(67)A. operating B. programming C. security D. service
(68)A. prevent B. protect保护 C. develop D. exploit
参考译文
防火墙是一个安全系统,用来保护一个组织的网络不受到威胁。
参考答案
(67)C (68)B
The (69)has several major(主要)components(组成), including the system kernel, a memory management system, the file system manager, device drivers, and the system libraries.
(69)A. application B. information system
C. network
D. operating system
软件设计师-计算机专业英语(一)
软件设计师-计算机专业英语(一)
(总分:80.00,做题时间:90分钟)
一、
(总题数:16,分数:80.00)
Originally introduced by Netscape Communications, (1) are a general mechanism which HTFP Server side applications, such as CGI (2) , can use to both store and retrieve information on the HTTP (3) side of the connection. Basically, Cookies can be used to compensate for the (4) nature of HTTP. The addition of a simple, persistent, client-side state significantly extends the capabilities of WWW-based (5) .
(分数:5.00)
A.Browsers
B.Cookies √
C.Connections
D.Scripts
解析:
A.graphics
B.processes
C.scripts √
D.texts
解析:
A.Client √
B.Editor
C.Creator
D.Server
解析:
A.fixed
B.flexible
C.stable
D.stateless √
解析:
A.programs
B.applications √
软件架构师 英语题
软件架构师英语题
As a software architect, you are expected to have a strong command of English, as it is the global language of technology and business. Here are some English language topics that a software architect should be familiar with:
1. Technical Vocabulary: As a software architect, you should be familiar with technical vocabulary related to software development, such as design patterns, algorithms, data structures, and software development methodologies.
2. Communication Skills: Software architects often need to communicate with a diverse range of stakeholders, including developers, project managers, and business leaders. This requires proficiency in written and spoken English to effectively convey complex technical concepts and requirements.
2009下半年软件设计师试题及答案(全部试题)
2009下半年软件设计师试题
●以下关于CPU的叙述中,错误的是(1)。
(1)A.CPU产生每条指令的操作信号并将操作信号送往相应的部件进行控制
B.程序计数器PC除了存放指令地址,也可以临时存储算术/逻辑运算结果
C.CPU中的控制器决定计算机运行过程的自动化
D.指令译码器是CPU控制器中的部件
●以下关于CISC(Complex Instruction Set Computer,复杂指令集计算机)和RISC(Reduced Instruction Set Computer,精简指令集计算机)的叙述中,错误的是(2)。
(2)A.在CISC中,其复杂指令都采用硬布线逻辑来执行
B.采用CISC技术的CPU,其芯片设计复杂度更高
C.在RISC中,更适合采用硬布线逻辑执行指令
D.采用RISC技术,指令系统中的指令种类和寻址方式更少
●浮点数的一般表示形式为N=2E×F,其中E为阶码,F为尾数。以下关于浮点表示的叙述中,错误的是(3)。两个浮点数进行相加运算,应首先(4)。
(3)A.阶码的长度决定浮点表示的范围,尾数的长度决定浮点表示的精度
B.工业标准IEEE754浮点数格式中阶码采用移码、尾数采用原码表示
C.规格化指的是阶码采用移码、尾数采用补码
D.规格化表示要求将尾数的绝对值限定在区间[0.5,1)
(4)A.将较大的数进行规格化处理B.将较小的数进行规格化处理
C.将这两个数的尾数相加D.统一这两个数的阶码
●以下关于校验码的叙述中,正确的是(5)。
(5)A.海明码利用多组数位的奇偶性来检错和纠错B.海明码的码距必须大于等于1 C.循环冗余校验码具有很强的检错和纠错能力D.循环冗余校验码的码距必定为1
软件工程师考题-中英文
软件工程师考题-中英文
软件工程师考题
1JA V A 基础 (1)
2JSP&SERVLET (7)
3JA V ASCRIPT (8)
4数据库(ORACLE) (9)
一Java 基础
(一)Given the following:
1. interface Base {
2. boolean m1 ();
3. byte m2(short s);
4. }
Which code fragments will compile? (Choose all that apply.)
A interface Base2 implements Base { }
B abstract class Class2 extends Base {
public boolean ml() { return true; }
}
C abstract class Class2 implements Base { }
D abstract class Class2 implements Base {
public boolean m1() { return (true); }
}
E class Class2 implements Base {
boolean m1( ) { return false; }
byte m2(short s) { return 42; }
}
Answer:
(二)Given the following:
public abstract interface Frobnicate { public void twiddle(String s) ; }
软件设计师-计算机专业英语(四)
软件设计师-计算机专业英语(四)
(总分:30.00,做题时间:90分钟)
一、综合知识试题
(总题数:6,分数:30.00)
It should go without saying that the focus of UML is modeling. However, what that means, exactly, can be an open-ended question. (1) is a means to capture ideas, relationships,decisions, and requirements in a well-defined notation that can be applied to many different domains. Modeling not only means different things to different people, but also it can use different pieces of UML depending on what you are trying to convey. In general, a UML model is made up of one or more
(2) . A diagram graphically represents things, and the relationships between these things. These
(3) can be representations of realworld objects,pure software constructs, or a description of the behavior of some other objects. It is common for an individual thing to show up on multiple diagrams; each diagram represents a particular interest, or view, of the thing being modeled. UML 2.0 divides diagrams into two categories: structural diagrams and behavioral diagrams. (4) are used to capture the physical organization of the things in your system, i.e., how one object relates to another. (5) focus on the behavior of elements in a system. For example, you can use behavioral diagrams to capture requirements, operations, and internal state changes for elements.
软考英语
PART I COMPUTER HARDW ARE
CHAPTER 1 PRINCIPLES OF COMPUTER
ORGANIZATION
1.1 COMPUTER HARDW ARE
We build computer to solve problems.Early computer solved mathematical and engineering problems,and later computers emphasized information processing for business applications.T- oday,computers also control machines as diverse as automobile engines,robots,and microwave ovens. A computer system solves a problem from any of these domains by accepting input,processing it,and producing output. Fig. 1-1 illustrates the function of a computer system.
Computer systems consist of hardware and software. Hardware is the physical part of the system. Once designed,hardware is difficult and expensive to change. Software is the set of programs that instruct the hardware and is easier to modify than hardware. Computers are valuable because they are general-purpose machines that can solve many different kinds of problems,as opposed to special-purpose machines that can each solve only one kind of problem. Different problems can be solved with the same hardware by supplying the system with a different set of instructions.that is,with different software.
软件工程专业英语考试翻译
Translation
Unit-1
III. (1) 由于计算机与信息系统的结合以及复杂的图形用户界面而产生的新技术对软件工程师提出了新的要求。
(2) 不同类型系统和使用这些系统的不同类型组织的广泛的多样性,意味着我们需要软件开发方法的多样性。…………………. 第8页翻译
III. (1) 当芯片设计者为微处理器开发各种各样的指令集的同时,他们增加了越来越复杂的指令,每个指令的执行都需要若干时钟周期。………………………..第13页翻译
Unit-2
III. (1) 对于详细说明软件需求中的困难,根本原因源于其涉及客户、最终用户、以及软件开发人员的有关三方.。第22页翻译
III. (1) 正如我们所知,指示一台计算机如何执行一项任务的指令被称为计算机程序。(2) 这类软件包含大量数据——不像数据库软件,数据库软件销售时不带任何数据。…………………………….第27页翻译
Unit-3
III. (1) 为了即使在风险存在的情况下能够达到项目目标,需要适当的风险管理。
(2) 通过适当的监控,这些情况能够被识别,从而可以相应地改变计划。
………………………………….第38页翻译
III. (1) 不过,主要区别在于,多用户操作系统排定在一个集中式计算机上的处理请求,而网络操作系统仅仅是将数据和程序路由到每个用户的本地计算机,实际处理是发生在每个用户的本地计算机上。……………………….第42页翻译
Unit-4
III. (1) 详细进度安排只有在实际的人员分配已经完成之后才进行,因为任务分配需要团队成员能力的相关信息。
软件工程 选择题 (中英文)
软件工程选择题 (中英文)
软件工程选择题 (中英文)
=========================
1.What is the purpose of software engineering?
软件工程的目的是什么?
A.To develop high-quality software efficiently and effectively.
高效、有效地开发高质量的软件。
B.To solve complex problems using computer programs.
使用计算机程序来解决复杂的问题。
C.To design user-friendly interfaces for software applications.
设计用户友好的软件界面。
D.To ensure the security and privacy of software systems.
确保软件系统的安全和隐私。
2.What is the software development life cycle (SDLC)?
软件开发生命周期(SDLC)是什么?
A.It is a process for designing and building software systems.
这是一种设计和构建软件系统的过程。
B.It is a set of software engineering practices and methodologies.
这是一套软件工程实践和方法论。
C.It is a framework for managing and controlling software projects.
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04上
(66)data effectively is crucial for success in today’s competitive environment. Managers must know how to use a variety of tools.Integrated data takes information from different sources and puts it together in a meaningful and useful way. One of the difficulties of this is the (67)in hardware and software (68)integration uses a base document that contains copies of other objects. (69)integration uses a base document that contains the current or most recent version of the source document it contains.
(70)provides an overview of the program written in “plain” English , without the computer syntax.
(66)A. Generalizing B. Sharing C. General-using D.Globalizing
(67)A. similarity B. interoperability C. diversity D.interaction
(68)A. Simulated B. Duplicated C.Dynamic D.Static
(69)A. Linked B. pointed C.Dynamic D.Static
(70)A.High-level language B.Decision tree
C.Pseudocode
D.Flowchart
参考译文
在当今的竞争环境下,要想取得成功,有效地共享数据是十分重要的。管理人员必须知道如何使用一系列不同的工具。数据集成使来自不同数据源的数据以一种有效和有用的方式组合在一起。这种集成面临的一个困难就是软硬件的多样性问题。静态集成使用一个包含其他主题的基础文档,动态集成使用它所包含的源文档的当前或最近的版本。
伪代码通过使用简单易懂的,不带有计算机语法的英语,为程序提供了一个概貌。
参考答案
(66)B (67)C (68)D (69)C (70)C
04下
Networks can be interconnected by different devices. In the physical layer, networks can be connected by (66)or hubs,which just move the bits from one network to an identical network. One layer up we find bridges and switches,which operate at data link layer.They can accept (67), examine the MAC address,and forward the frames to a different network while doing minor protocol translation in the process. In the network layer,we have routers that can connect two networks. If two networks have (68)network layer,the router may be able to translate between the packet formats. In the transport layer we find transport gateway, which can interface between two transport connections. Finally,in the application layer,application gateways translate message (69).As an example,gateways between Internet e-mail and X.400 e-mail must (70)the e-mail message and change various header fields.
(66)A.reapers B.relays C.connectors D.modems
(67)A.frames B.packets C.packages D.cells
(68)A.special B.dependent C.similar D.dissimilar
(69)A.syntax B.semantics nguage D.format
(70)A.analyze B.parse C.delete D.create
参考译文
网络可以用不同的设备互连。在物理层,用中继器或集线器互连,这些设备只是在相同的网络之间传送比特串。在上面的数据链路层,可以使用网桥或交换机,这些设备接收数据帧,检查MAC地址,并可以实现少量的协议转换,把数据帧转发到不同的网络中。在网络层,我们使用路由器连接两个网络。如果两上网络的网络层不同,路由器能够对分组格式进
行转换。在传输层,我们使用传输风头,它可以在两个传输连接之间建立接口。最后,在应用层,应用网关实现消息语法之间的翻译。例如,在Internet邮件和X.400邮件之间的网关可以对邮件报文进行语法分析,对报文的各个报头字段做出改变。
参考答案
(66)A (67)A (68)D (69)A (70)B
05上
DOM is a platform-and language- (66)API that allows programs and scripts to dynamically access and update the content, structure and style of WWW documents(currently, definitions for HTML and XML documents are part of the specification). The document can be further processed and the results of that processing can be incorporated back into the presented (67). DOM is a (68)-based API to documents,which requires the whole document to be represented in (69)while processing it.A simpler alternative to DOM is the event-based SAX,which can be used to process very large (70)documents that do not fit info the memory available for processing.
(66)A.specific B.neutral C.contained D.related
(67)A.text B.image C.page D.graphic
(68)A.table B.tree C.control D.event
(69)A.document B.processor C.disc D.memory
(70)A.XML B.HTML C.script D.web
参考译文
DOM是一种与平台和语言无关的应用程序接口(API),它可以动态地访问程序和脚本,更新其内容、结构和WWW文档的风格(目前,HTML和XML文档是通过说明部分定义的)。文档可以进一步被处理,处理的结果可以加入到当前的页面。DOM是一种基于树的API文档,它要求在处理过程中整个文档都表示在存储器中。另外一种简单的API是基于事件的SAX,它可以用于处理很大的XML文档,由于大,所以不适合全部放在存储器中处理。参考答案
(66)B (67)C (68)B (69)D (70)A
05下
MIDI enables people to use (66)computers and electronic musical instruments. There are actually three components to MIDI, the communications " (67)", the Hardware Interface and a distribution (68)called "Standard MIDI Files". In the context of the WWW, the most interesting component is the (69)Format. In principle, MIDI files contain sequences of MIDI Protocol messages. However, when MIDI Protocol (70)are stored in MIDI files, the events are also time-stamped for playback in the proper sequence. Music delivered by MIDI files is the most common use of MIDI today.
(66)A. personal B. electronic C. multimedia D. network
(67)A. device B. protocol C. network D. controller
(68)A. format B. text C. wave D. center