JAVA基础面试题-3(答案版).doc

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1、JAVA语言基础笔试题-3Question1Given:1.publicclassTeamextendsjava.util.LinkedList{2.publicvoidaddPlayer(Playerp){3.add(p);4.}5.publicvoidcompete(Teamopponent){/*morecodehere*/}6.}7.classPlayer{/*morecodehere*/}Whichtwoaretrue?(Choosetwo.)A.Thiscodewillcompile.B.Thiscodedemonstratesproperdesignof

2、anis-arelationship.C.Thiscodedemonstratesproperdesignofahas-arelationship.D.AJavaprogrammerusingtheTeamclasscouldremovePlayerobjectsfromaTeamobject.答案：AD考点：List接口实现，对象间泛化和关联关系说明：isa关系一般用继承来表示，但此题这个继承从逻辑上不是太恰当.Hasa关系一般来说表示某对象体内有其它对象的存在，该对象体现为属性形态，在UML中叫做关联关系。本题Team中虽然可以保存Player，但并非显式以属性形

3、式存在。由于LinkedList自带remove方法，可以通过Team.remove(p)来删除player对象。Question2Whichfouraretrue?(Choosefour.)A.Has-arelationshipsshouldneverbeencapsulated.B.Has-arelationshipsshouldbeimplementedusinginheritance.C.Has-arelationshipscanbeimplementedusinginstancevariables.D.Is-arelationshipscanbeimplem

4、entedusingtheextendskeyword.E.Is-arelationshipscanbeimplementedusingtheimplementskeyword.F.TherelationshipbetweenMovieandActressisanexampleofanis-arelationship.G.Anarrayoracollectioncanbeusedtoimplementaone-to-manyhas-arelationship.答案：CDEG考点：集合类型，对象间泛化和关联关系的理解Hasa关系一般表示为一个类拥有一个属性，属性被封装是

5、常见的事情。Isa关系一般用继承来表示，子类体内拥有父类部分。接口的实现，也适用于isa关系来理解，因为接口从本质来说，也属于类的形态。Question3Whichtwoaretrueabouthas-aandis-arelationships?(Choosetwo.)A.Inheritance(继承)representsanis-arelationship.B.Inheritancerepresentsahas-arelationship.C.Interfacesmustbeusedwhencreatingahas-arelationship.D.Instancev

6、ariablescanbeusedwhencreatingahas-arelationship.答案：AD考点：对象间泛化和关联关系的理解Question4Given:10.interfaceJumper{publicvoidjump();}......20.classAnimal{}......30.classDogextendsAnimal{31.Tailtail;32.}......40.classBeagleextendsDogimplementsJumper{41.publicvoidjump(){}42.}.......50.classCatimpleme

7、ntsJumper{51.publicvoidjump(){}52.}Whichthreearetrue?(Choosethree.)A.Catis-aAnimalB.Catis-aJumperC.Dogis-aAnimalD.Dogis-aJumperE.Cathas-aAnimalF.Beaglehas-aTailG.Beaglehas-aJumper答案：BCF考点：对象间泛化和关联关系的理解Question5Given:1.importjava.util.*;2.publicclassExample{3.publicstaticvoidmai