高級(jí)軟件工程近三年試題及翻譯word版-

1、The MidtermExamination of Software EngineeringThe FirstTerm in 2009-2010 Academic YearFor the Undergraduates at Yunnan University(120 minutesInstructor: Li TongStudent Number:_ Name:_ School:_ 1. Multiple Choices (10 Points,Each 0.5Point:(1Which of the items listed below is not one of the software e

2、ngineering layers?a. Processb. Manufacturingc. Methodsd. Tools(2 What are the three generic phases of software engineering?a. definition, development, supportb. what, how, wherec. programming, debugging, maintenanced. analysis, design, testing(3 The prototyping model of software development isa. A r

3、easonable approach when requirements are well defined.b. A usefulapproach when a customer cannot define requirements clearly.c. The best approach to use for projects with large development teams.d. A risky model that rarely produces a meaningful product.(4 The formal methods model of software develo

4、pment makes use of mathematical methods toa. Define the specification for computer-based systemsb. Develop defect free computer-based systemsc. Verify the correctness of computer-based systemsd. All of the above(5 Use-cases are scenarios that describea. how software is to be used in a given situatio

5、n.b. how CASE tools will be used to construct the system.c. the build plan for a software product.d. the test cases for a software product.(6 What types of models are created during software requirements analysis?a. functional and behavioralb. algorithmic and data structure2 c. architectural and str

6、ucturald. usability and reliability(7 The data flow diagrama. depicts relationships between data objectsb. depicts functions that transform the data flowc. specified major logical decisions as they occurd. indicates system reactions to external events(8 Which of the following is not an area of conce

7、rn in the design model?a. architectureb. datac. interfacesd. project scope(9 To achieve high modularity of software components, you needa. high coupling and high cohesionb. high coupling and low cohesionc. low coupling and high cohesiond. low coupling and low cohesion(10 Which of the following is no

8、t part of software architecture?a. algorithm detailsb. databasesc. data designd. program structure(11 Which of the following is not a fundamental structured programming construct?a. recursionb. conditionc. repetitiond. sequence(12 Which of these are objectives for software testing?a. determine the p

9、roductivity of programmersb. eliminate the need for future program maintenancec. eliminate every error prior to released. uncover software errors(13 Encapsulation of attributes and operations within an objecta. allows for easy reuse of this information.b. increases the cost of program maintenance.c.

10、 is a poor programming practice.d. none of the above(14 A generalized description of a collection of similar objects is aa. classb. instancec. subclassd. super class3 (15 Which of the following should be considered as candidate objects ina problem space?a. eventsb. peoplec. structuresd. all of the a

11、bove(16 The first step in any OOA process model are toa. build an object-relationship modelb. define collaborations between objectsc. elicit customer requirementsd. select a representation language(17 Which of the following items does not appear on a CRC card?a. class collaboratorsb. class namec. cl

12、ass reliabilityd. class responsibilities(18 The UML approach to object-oriented design has two major activities.a. architectural design and object designb. interface design and message designc. message design and system designd. system design and object design(19 The correctness of the OOA and OOD m

13、odels is judged based onthea. models conformance to real world problem domainb. review of the connections between classesc. review of the modeling conventions usedd. both a and c(20 What are the two parallel engineering activities found in the CBSE process model?a. component-based development and li

14、brary developmentb. domain engineering and component-based developmentc. domain engineering and process developmentd. none of the above2. True or False(15 Points,Each 1 Point:(1 The architectural model provides the software engineer with a view of the systemas a whole.a. Trueb. False(2 Modern softwa

15、re engineers believe that the only design notation useful for procedural representation is pseudocode.4 a. Trueb. False(3 The cyclomatic complexity of a program can be computed directly from a PDL representation of an algorithm without drawing a program flow graph.a. Trueb. False(4 Boundary value an

16、alysis can only be used during white-box testing.a. Trueb. False(5 Drivers and stubs are not needed for unit testing because the modulesare tested independently of one another.a. Trueb. False(6 Debugging is not testing, but always occurs as a consequence of testing.a. Trueb. False(7 Operations are o

17、bject procedures that are invoked when an object receives a message.a. Trueb. False(8 Inheritance provides a mechanism by which changes to lower level classes can be propagated to all super classes quickly.a. Trueb. False(9 Object-oriented design achieves low module coupling and provides better info

18、rmation hiding than traditional approaches.a. Trueb. False(10 Object-oriented designs do not need to be implemented using object-oriented programming techniques.a. Trueb. False(11 Class testing for OO software is equivalent to unit testing in conventional software testing.a. Trueb. False(12 Test cas

19、e design for OO software is driven by the algorithmic detailof the individual operations.a. Trueb. False5 (13 Software is a product and can be manufactured using the same technologies used for other engineering artifacts.a. Trueb. False(14 Adding more people to a project that is already behind sched

20、ule is a good way to catch up.a. Trueb. False(15 When using structured design methodologies the process of stepwise refinement is unnecessary.a. Trueb. False3. Answer the following questions briefly (45 Points, Each 15 Points:(1What is a formal technical review and why are users conducted?(2Explain

21、why encapsulation, inheritance, and polymorphism are three important characteristics of object-oriented systems.(3What are the names and contents of the five levels of the SEI Capability Maturity Model?4. (30PointsAnalyse and design a simple information system of air ticket booking using one of the

22、structured methodology and the object-oriented methodology. The system allows the clients to book and buy air tickets face to face with the sellers by cash. Write out the following specifications briefly:(1 Requirements specification,(2 Design specification.該軟件MidtermExamination工程在2009-2010學(xué)年中FirstT

23、erm在云南大學(xué)的大學(xué)生(120分鐘講師:黎通學(xué)生編號(hào):_姓名:_學(xué)校:_ 1。多重選擇(10分,每0.5Point:(1下面列出的項(xiàng)目是不是一個(gè)軟件工程層?答:進(jìn)程灣制造業(yè)角方法四工具(2什么是軟件工程的三個(gè)階段的通用?答:定義,發(fā)展,支持灣什么,如何,在何處角編程,調(diào)試,維修四分析,設(shè)計(jì),測(cè)試(3原型的軟件開(kāi)發(fā)模式答:一個(gè)合理的做法時(shí),要求明確界定。灣阿usefulapproach當(dāng)客戶(hù)無(wú)法定義的要求清楚。角最好的方法用于大型發(fā)展項(xiàng)目隊(duì)。四有風(fēng)險(xiǎn)的模式,很少產(chǎn)生一個(gè)有意義的產(chǎn)品。(4形式方法的軟件開(kāi)發(fā)模式,使使用以數(shù)學(xué)方法答:定義為基于計(jì)算機(jī)的系統(tǒng)規(guī)范灣開(kāi)發(fā)無(wú)缺陷的計(jì)算機(jī)為基礎(chǔ)的系統(tǒng)角驗(yàn)證

24、計(jì)算機(jī)系統(tǒng)的正確性四以上所有(5用例描述的情景答:軟件是如何被使用在特定情況下。灣如何CASE工具將被用來(lái)構(gòu)建系統(tǒng)。角構(gòu)建一個(gè)軟件產(chǎn)品的計(jì)劃。四一個(gè)軟件產(chǎn)品的測(cè)試案例。(6什么類(lèi)型的模型建立在軟件要求分析?答:功能和行為灣算法與數(shù)據(jù)結(jié)構(gòu)2角建筑和結(jié)構(gòu)四可用性和可靠性(7數(shù)據(jù)流圖答:描述數(shù)據(jù)對(duì)象之間的關(guān)系灣描繪函數(shù)變換的數(shù)據(jù)流角指定的主要邏輯的決定,因?yàn)樗麄兂霈F(xiàn)四顯示系統(tǒng)外部事件的反應(yīng)(8以下哪一項(xiàng)不是在設(shè)計(jì)模型關(guān)注的地方?答:建筑灣數(shù)據(jù)角接口四項(xiàng)目范圍(9為了達(dá)到高的軟件組件模塊化,你需要答:高耦合,高凝聚力灣高耦合,低凝聚力角低耦合,高凝聚力四低耦合,低凝聚力(10以下哪一項(xiàng)不是軟件架構(gòu)的一

25、部分?答:算法的細(xì)節(jié)灣數(shù)據(jù)庫(kù)角數(shù)據(jù)設(shè)計(jì)四程序結(jié)構(gòu)(11以下哪一項(xiàng)不是一個(gè)基本結(jié)構(gòu)規(guī)劃建設(shè)?答:遞歸灣條件角重復(fù)四序列(12哪些是對(duì)軟件測(cè)試的目標(biāo)是什么?答:確定程序員生產(chǎn)率灣消除了對(duì)未來(lái)的方案需要維修角消除一切錯(cuò)誤之前釋放四發(fā)現(xiàn)軟件錯(cuò)誤(13的屬性和操作封裝在一個(gè)對(duì)象答:允許對(duì)這些信息很容易重用。灣增加了程序維護(hù)的成本。角是一個(gè)貧窮的編程實(shí)踐。四以上皆非(14一個(gè)類(lèi)似對(duì)象的集合,是一個(gè)廣義的描述答:類(lèi)灣例如角子類(lèi)四超類(lèi)3(15以下哪一項(xiàng)應(yīng)考慮作為候選對(duì)象一個(gè)問(wèn)題空間?答:事件灣人角結(jié)構(gòu)四所有上述(16在任何面向?qū)ο蠓治鲞^(guò)程模型的第一步是答:建立一個(gè)對(duì)象關(guān)系模型灣定義對(duì)象之間的協(xié)作角征求客戶(hù)的要

26、求四選擇一個(gè)表示語(yǔ)言(17下列哪個(gè)項(xiàng)目沒(méi)有出現(xiàn)在一個(gè)CRC卡嗎?答:類(lèi)合作者灣類(lèi)名角級(jí)的可靠性四類(lèi)的職責(zé)(18UML的方法,面向?qū)ο蟮脑O(shè)計(jì)有兩個(gè)主要活動(dòng)。答:建筑設(shè)計(jì)和對(duì)象設(shè)計(jì)灣界面設(shè)計(jì)和信息設(shè)計(jì)角信息設(shè)計(jì)和系統(tǒng)設(shè)計(jì)四系統(tǒng)設(shè)計(jì)和對(duì)象設(shè)計(jì)(19的面向?qū)ο蠓治龊兔嫦驅(qū)ο竽Ｐ偷幕A(chǔ)上正確判斷的答:模型的一致性,以現(xiàn)實(shí)世界問(wèn)題域?yàn)硨彶轭?lèi)之間的聯(lián)系角審查建模使用的約定四同時(shí)A和C(20什么是兩個(gè)并行工程活動(dòng)中發(fā)現(xiàn)的CBSE過(guò)程模型?答:基于組件的開(kāi)發(fā)與圖書(shū)館的發(fā)展灣域工程和基于組件的開(kāi)發(fā)角領(lǐng)域工程與工藝開(kāi)發(fā)四以上皆非2。真或假(15分,每1點(diǎn):(1建筑模型提供了一個(gè)查看軟件工程師在systemas一個(gè)整

27、體。答:真灣假(2現(xiàn)代的軟件工程師認(rèn)為,唯一的設(shè)計(jì)符號(hào)由于程序上的代表性有用的是偽代碼。答:4真灣假(3圈復(fù)雜度的程序,可直接計(jì)算從代表性的算法客運(yùn)專(zhuān)線(xiàn)無(wú)圖紙方案流圖。答:真灣假(4邊界值分析法只能用于在白箱測(cè)試。答:真灣假(5驅(qū)動(dòng)程序和存根不是必須的,因?yàn)閱卧獪y(cè)試的模塊測(cè)試相互獨(dú)立。答:真灣假(6調(diào)試沒(méi)有測(cè)試,但一直作為結(jié)果發(fā)生測(cè)試。答:真灣假(7操作對(duì)象時(shí)被調(diào)用程序的對(duì)象接收消息。答:真灣假(8繼承提供了一種機(jī)制,以較低水平的變化類(lèi)可以傳播到所有超類(lèi)迅速。答:真灣假(9面向?qū)ο笤O(shè)計(jì)實(shí)現(xiàn)低耦合模塊,并提供比傳統(tǒng)方法更好的信息隱藏。答:真灣假(10面向?qū)ο蟮脑O(shè)計(jì)不需要使用實(shí)施面向?qū)ο缶幊碳夹g(shù)。

28、答:真灣假(11班,面向?qū)ο筌浖y(cè)試是單元測(cè)試相當(dāng)于在傳統(tǒng)的軟件測(cè)試。答:真灣假(12面向?qū)ο筌浖y(cè)試案例設(shè)計(jì)的動(dòng)力是詳細(xì)的算法個(gè)人的行動(dòng)。答:真灣假5(13軟件是一種產(chǎn)品,可以使用相同的制造其他工程構(gòu)件中使用的技術(shù)。答:真灣假(14添加更多的人到一個(gè)項(xiàng)目,該項(xiàng)目已落后時(shí)間表是一個(gè)好辦法,迎頭趕上。答:真灣假(15當(dāng)使用結(jié)構(gòu)化設(shè)計(jì)方法的逐步過(guò)程細(xì)化是不必要的。答:真灣假3。簡(jiǎn)要回答下列問(wèn)題(45分,每15分:(1什么是一個(gè)正式的技術(shù)審查,為什么用戶(hù)進(jìn)行?(2解釋為什么封裝,繼承和多態(tài)性3面向?qū)ο笙到y(tǒng)的重要特征。(3什么是姓名和SEI的五個(gè)層次的內(nèi)容能力成熟度模型?4。(30Points分析和設(shè)

29、計(jì)一個(gè)簡(jiǎn)單的信息系統(tǒng)的機(jī)票預(yù)訂使用結(jié)構(gòu)化方法和面向?qū)ο蟮囊环椒?。該系統(tǒng)允許客戶(hù)預(yù)訂和購(gòu)買(mǎi)機(jī)票面對(duì)面與現(xiàn)金賣(mài)家。寫(xiě)出下列規(guī)格簡(jiǎn)述:(一要求的規(guī)格,(2設(shè)計(jì)規(guī)范。The MidtermExamination of Software EngineeringThe FirstTerm in 2008-2009 Academic YearFor the Undergraduatesat Yunnan University(120 minutesTeacher: Li TongStudent No.:_Name:_School:_I. Multiple Choices (15Points,Each1Poi

30、nt:(1 The linear sequential model of software development is also known as thea. classical life cycle modelb. fountain modelc.spiral modeld. chaos model(2 Use-cases are scenarios that describea. how software is to be used in a givensituation.b. how CASE tools will be used to construct the system.c.

31、the build plan for a software product.d. the test cases for a software product.(3 In the context ofsoftware design, two types ofarchitecture partitioning area. bottom-up and top-downb. horizontal and verticalc. subordinate and superordinate2 d. none of the above(4 Which of the following is not an ob

32、jective for building an analysis model?a. define set of software requirementsb. describe customer requirementsc. develop an abbreviated solution for the problemd. establish basis for software design(5 The data flow diagrama. depicts relationships between data objectsb. depicts functions that transfo

33、rm the data flowc. specified major logical decisions as they occurd. indicates system reactions to external events(6 The data dictionary contains descriptions of each softwarea. configuration itemb. data objectc. diagramd. notation(7 Coupling is a qualitative indication of the degree to which a modu

34、lea. can be written more compactly.b. focuses on just one thing.c. is able to complete its function in a timely manner.d. is connected to other modules and the outside world.(8 Which of the following interface design principles reduce the user's memory load?3 a. define intuitive shortcutsb. disc

35、lose information in a progressive fashionc. establish meaningful defaultsd. all of the above(9 Which of the following is not a fundamental structured programming construct?a. recursionb. conditionc. repetitiond. sequence(10 Which of these are objectives for software testing?a. determine the producti

36、vity of programmersb. eliminate the need for future program maintenancec. eliminate every error prior to released. uncover software errors(11 The best reason for using independent software test teams is thata. software developers do not need to do any testingb. strangers will test the software merci

37、lesslyc. testers do not get involved with the project until testing beginsd. the conflicts of interest between developers and testers is reduced(12 The first step in any OOA process model are toa. build an object-relationship modelb. define collaborations between objectsc. elicit customer requiremen

38、tsd. select a representation language4 (13 Software procedure focuses on thea. control hierarchy in a more abstract sense.b. processing details of each module individually.c. processing details of each the set of modules collectively.d. relationship between control and procedure.(14 A generalized de

39、scription of a collection of similar objects is aa. classb. instancec. subclassd. super class(15 The UML approach to object-oriented design has two major activities.a. architectural design and object designb. interface design and message designc. message design and system designd. system design and

40、object designII.True or False (15 Points,Each 1 Point:(1Software is a product and can be manufactured using the sametechnologies used for other engineeringartifacts.(2Adding more people to a project that is already behind schedule is a good way to catch up.(3The process specification used to describ

41、e all flow processes that appear in the final DFD must be written using a program design language.5 (4Software engineers do not need to consider hardware when designinga computer-based system.(5One of the most difficult parts of software requirements analysis is ensuring the developer understands th

42、e customer's needs.(6 The use of context free questions by themselves provides an effective means of eliciting requirements information from the customer.(7For software prototyping to be effective technique, tools are required to develop prototypes rapidly to keep the schedule on track.(8When us

43、ing structured design methodologies the process of stepwise refinement is unnecessary.(9Information hiding makes program maintenance easier by hiding data and procedure from unaffected parts of the program.(10 If past interactive models have created certain user expectations it is not generally good

44、 to make changes to the model.(11 Since a program design language is not a real programming language the designer is free to write the procedural design without worrying about syntax errors.(12 Equivalence testing divides the input domain into classes of data from which test cases can be derived to

45、reduce the total number of test cases that must be developed.(13 Units and stubs are not needed for unit testing because the modules are tested independently of one another.(14 To review a complete CRC model the reviewers only need to walk through one or two representative use-cases.(15 Object-orien

46、ted design achieves low module coupling and provides better information hiding than other approaches.6 III. Answer theFollowing QuestionsBriefly (40Points,Each10Points:(1Explainwhy encapsulation, inheritance and polymorphism are three important characteristics of object-oriented systems.(2What is a

47、good modular from the points of view of functional independence?(3How is a transaction center different from a transform center in a data flow diagram?(4Describe four software architectural styles.IV. Software Engineering Practice (30PointsAnalyse and design a simple management information system of

48、 your school using one of the structured methodology and the OO methodology. The system allows a user to manage school information, including:personnel, teaching, research etc. in his/her school. Write out the following specifications briefly:(1 Requirements specification,(2 Design specification.該軟件

49、MidtermExamination工程該FirstTerm在2008-2009學(xué)年云南大學(xué)對(duì)于Undergraduatesat(120分鐘老師:黎通學(xué)生學(xué)號(hào):_姓名:_ :_ 一,(15Points,Each1Point多選擇:(1線(xiàn)性順序的軟件開(kāi)發(fā)模式也被稱(chēng)為作為答:經(jīng)典的生命周期模型灣噴泉模型c.spiral模型四混沌模型(2用例描述的情景答:軟件是如何被用在givensituation。灣如何CASE工具將被用來(lái)構(gòu)建系統(tǒng)。角構(gòu)建一個(gè)軟件產(chǎn)品的計(jì)劃。四一個(gè)軟件產(chǎn)品的測(cè)試案例。(3在設(shè)計(jì)方面ofsoftware,兩種ofarchitecture劃分等,答:自下而上和自上而下灣橫向和縱向角下

50、屬和上級(jí)2四以上皆非(4以下哪一項(xiàng)不是建立一個(gè)分析的目的模型?答:一套軟件需求定義灣客戶(hù)需求描述角制定一個(gè)解決問(wèn)題的縮寫(xiě)四建立軟件設(shè)計(jì)的基礎(chǔ)(5數(shù)據(jù)流圖答:描述數(shù)據(jù)對(duì)象之間的關(guān)系灣描繪函數(shù)變換的數(shù)據(jù)流角指定的主要邏輯的決定,因?yàn)樗麄兂霈F(xiàn)四顯示系統(tǒng)外部事件的反應(yīng)(6數(shù)據(jù)字典中包含的每個(gè)軟件的描述答:配置項(xiàng)灣數(shù)據(jù)對(duì)象角圖表四符號(hào)(7耦合是一個(gè)質(zhì)的程度,可顯示模塊答:可以寫(xiě)更多的緊湊。灣專(zhuān)注于一件事。角能夠及時(shí)完成其功能。四連接到其他模塊和外面的世界。(8以下哪界面設(shè)計(jì)原則減少用戶(hù)的記憶負(fù)荷?3答直觀(guān)快捷鍵定義灣以漸進(jìn)的方式披露信息角建立有意義的默認(rèn)值四所有上述(9以下哪一項(xiàng)不是一個(gè)根本性的結(jié)構(gòu)化編

51、程興建?答:遞歸灣條件角重復(fù)四序列(10,這是軟件測(cè)試的目標(biāo)?答:確定程序員生產(chǎn)率灣消除了對(duì)未來(lái)的方案需要維修角消除一切錯(cuò)誤之前釋放四發(fā)現(xiàn)軟件錯(cuò)誤(11使用獨(dú)立的軟件測(cè)試團(tuán)隊(duì)的最好的理由是,答:軟件開(kāi)發(fā)人員不需要做任何測(cè)試灣陌生人將毫不留情地測(cè)試軟件角測(cè)試人員不得到與該項(xiàng)目涉及到測(cè)試開(kāi)始四開(kāi)發(fā)人員和測(cè)試人員之間的利益矛盾會(huì)減少(12在任何面向?qū)ο蠓治鲞^(guò)程模型的第一步是答:建立一個(gè)對(duì)象關(guān)系模型灣定義對(duì)象之間的協(xié)作角征求客戶(hù)的要求四選擇一個(gè)表示語(yǔ)言4(13軟件程序的重點(diǎn)是答:控制在一個(gè)更抽象的層次感。灣每個(gè)模塊單獨(dú)處理的細(xì)節(jié)。角每個(gè)模塊的處理細(xì)節(jié)設(shè)置集體。四控制和程序的關(guān)系。(14一個(gè)類(lèi)似對(duì)象的集

52、合,是一個(gè)廣義的描述答:類(lèi)灣例如角子類(lèi)四超類(lèi)(15UML的方法,面向?qū)ο蟮脑O(shè)計(jì)有兩個(gè)主要活動(dòng)。答:建筑設(shè)計(jì)和對(duì)象設(shè)計(jì)灣界面設(shè)計(jì)和信息設(shè)計(jì)角信息設(shè)計(jì)和系統(tǒng)設(shè)計(jì)四系統(tǒng)設(shè)計(jì)和對(duì)象設(shè)計(jì)II.True或False(15分,每1點(diǎn):(1軟件是一種產(chǎn)品,可以使用相同的制造其他engineeringartifacts使用的技術(shù)。(2添加更多的人到一個(gè)項(xiàng)目,該項(xiàng)目已落后時(shí)間表是一個(gè)好辦法,迎頭趕上。(3過(guò)程規(guī)范用于描述所有流量過(guò)程出現(xiàn)在最后的DFD的使用必須寫(xiě)一個(gè)程序設(shè)計(jì)語(yǔ)言。5(4軟件工程師們無(wú)需考慮硬件設(shè)計(jì)時(shí)基于計(jì)算機(jī)的系統(tǒng)。(5軟件需求分析中最困難的部分之一,是確保發(fā)展商了解客戶(hù)的需求。(6范圍內(nèi)自由使用

53、自己的問(wèn)題提供了一個(gè)有效的征求手段從客戶(hù)需求的資料。(7對(duì)于軟件原型是有效的技術(shù),工具需要快速原型開(kāi)發(fā)繼續(xù)跟蹤進(jìn)度。(8當(dāng)使用結(jié)構(gòu)化設(shè)計(jì)方法的逐步過(guò)程細(xì)化是不必要的。(9信息隱藏使得程序的維護(hù)更容易隱藏?cái)?shù)據(jù)和程序從方案的影響,部分。(10如果過(guò)去的互動(dòng)模式造成了一些用戶(hù)的期望是不普遍良好,使該模型的變化。(11由于程序設(shè)計(jì)語(yǔ)言是不是一個(gè)真正的編程語(yǔ)言的設(shè)計(jì)師可以自由地編寫(xiě)程序設(shè)計(jì)不擔(dān)心語(yǔ)法錯(cuò)誤。(12等價(jià)測(cè)試分為數(shù)據(jù)類(lèi)的輸入域從中可以得出的測(cè)試案例,以減少測(cè)試的總?cè)藬?shù)必須發(fā)展情況。(13單位和存根不是必須的,因?yàn)閱卧獪y(cè)試的模塊測(cè)試相互獨(dú)立。(14審查模式的一個(gè)完整的啟聯(lián)資源中心只需步行評(píng)論通過(guò)

54、一個(gè)或兩個(gè)有代表性的用例。(15面向?qū)ο笤O(shè)計(jì)實(shí)現(xiàn)低耦合模塊,并提供優(yōu)于其他方法的信息隱藏。6三。回答以下幾部件QuestionsBriefly(40Points,每10Points:(1Explainwhy封裝,繼承和多態(tài)性是3面向?qū)ο笙到y(tǒng)的重要特征。(2這是不錯(cuò)的模塊化從功能的觀(guān)點(diǎn)是獨(dú)立?(3如何從一個(gè)交易中心在數(shù)據(jù)中心的不同轉(zhuǎn)換流程圖?(4描述4個(gè)軟件的建筑風(fēng)格。四。軟件工程實(shí)踐(30Points分析并設(shè)計(jì)了一個(gè)簡(jiǎn)單的管理信息系統(tǒng)的學(xué)校使用的結(jié)構(gòu)化方法和面向?qū)ο蟮姆椒ㄖ?。該系統(tǒng)允許用戶(hù)來(lái)管理學(xué)校的信息,包括:人事,教學(xué),科研等,他/她的學(xué)校。寫(xiě)出下面簡(jiǎn)要規(guī)格:(一要求的規(guī)格,(2設(shè)計(jì)規(guī)范

55、。The MidtermExamination of Software EngineeringThe FirstTerm in 2007-2008 Academic YearFor the Undergraduates Yunnan University(120 minutesTeacher: Li TongStudent Number:_ Name:_ School:_ 1. Multiple Choices (10 Points,Each0.5Point:(1 Which of the following criteria appears in object-oriented design models, but not conventional design models?a. representation of module hierarchyb. specification of data definitionsc. specification of message connectionsd. specification of procedural logic(2 The UML (unified modeling language appr