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軟件設(shè)計(jì)師每日一練試題(2025/11/6)

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軟件設(shè)計(jì)師每日一練試題內(nèi)容(2025/11/6)

  • 試題1

    通過在出口防火墻上配置(  )功能可以阻止外部未授權(quán)用戶訪問內(nèi)部網(wǎng)絡(luò)。
    A.ACL
    B.SNAT
    C.入侵檢測(cè)
    D.防病毒

    查看答案

    試題參考答案:A

    試題解析與討論:www.xcpkj.com/exam/ExamDay.aspx?t1=4&day=2025/11/6

  • 試題2

    下圖所示的 UML 序列圖中,(1)表示返回消息,Account 應(yīng)該實(shí)現(xiàn)的方法有(2)。

    (1)A.xfer
    B.check
    C.evaluation
    D.minus
    (2)A.xfer0
    B.xfen( )、plus( )和 minus( )
    C.check( )、plus( )和 minus( )
    D.xfar( )、evaluation( )、plus( )和minus( )

    查看答案

    試題參考答案:C、B

    試題解析與討論:www.xcpkj.com/exam/ExamDay.aspx?t1=4&day=2025/11/6

  • 試題3

    程序語言的大多數(shù)語法現(xiàn)象可用上下文無關(guān)文法描述。對(duì)于一個(gè)上下文無關(guān)文法G=(N,T,P,S),其中N是非終結(jié)符號(hào)的集合,T是終結(jié)符號(hào)的集合,P是產(chǎn)生式集合,S是開始符號(hào)。令集合V= N∪T,那么G所描述的語言是 ( ) 的集合。
    A、從S出發(fā)推導(dǎo)出的包含V中所有符號(hào)的串
    B、從S出發(fā)推導(dǎo)出的僅包含T中符號(hào)的串
    C、N中所有符號(hào)組成的串
    D、T中所有符號(hào)組成的串

    查看答案

    試題參考答案:B

    試題解析與討論:www.xcpkj.com/exam/ExamDay.aspx?t1=4&day=2025/11/6

  • 試題4

    常用的虛擬存儲(chǔ)器由( )兩級(jí)存儲(chǔ)器組成。
    A.主存-輔存
    B.主存-網(wǎng)盤
    C.Cache-主存
    D.Cache-硬盤

    查看答案

    試題參考答案:A

    試題解析與討論:www.xcpkj.com/exam/ExamDay.aspx?t1=4&day=2025/11/6

  • 試題5

    已知文法G: S—>A0|B1,A —>  S1|1, B —>  S0|0,其中S是開始符號(hào)。從S出發(fā)可以推導(dǎo)出(  )。
    A.所有由0構(gòu)成的字符串
    B.所有由1構(gòu)成的字符串
    C.某些0和1個(gè)數(shù)相等的字符串
    D.所有0和1個(gè)數(shù)不同的字符串

    查看答案

    試題參考答案:C

    試題解析與討論:www.xcpkj.com/exam/ExamDay.aspx?t1=4&day=2025/11/6

  • 試題6

    為實(shí)現(xiàn)程序指令的順序執(zhí)行,CPU  ( )  中的值將自動(dòng)加 1。
    A、指令寄存器(IR)
    B、程序計(jì)數(shù)器(PC)
    C、地址寄存器(AR)
    D、指令譯碼器(ID)

    查看答案

    試題參考答案:B

    試題解析與討論:www.xcpkj.com/exam/ExamDay.aspx?t1=4&day=2025/11/6

  • 試題7

    Why is(1)fun? What delights may its practitioner espect his reward? First is the
    sheer joy of making things.As the child delights in his mud pie,so the adult enjoys
    building things,especially things of his own design.Secong is the pleasure of making
    things that are useful to other people.Third is the fascinanon of fashioning complex
    puzzle-like objects  of interlocking moving  parts and watching them  work in subtle
    eyeles,playing out the consequences of principies built in from the beginning.Fourth
    is the joy of always learning,which springs from the(2)nature of the task.In one way
    or    another    the    problem    is    ever    new,and    its    solver    learns
    something:sometimes(3),sometimes  theoretical,and  sometimes both.Finally,there  is
    the delight of working in such a tractable medium.The(4),like the poet,works only
    slightly removed from nure thought-stuff.Few media of ereation are so flexible,so easy
    to polish and rework,so readily capable of realizing grand conceptual structures.
    Yet the program(5),unlike the poet’s words,is real in the sense that it moves and
    works,producing  visible  outputs  separate  from  the  comstrct  itself.It  prints
    results,draws pictures,produces sounds,moves arms.Progamming then is fun because it
    gratifies creative longings built deep within us and delights sensibities we hav e in
    common with all men.
    (1)A、programming
    B、composing
    C、working
    D、writing
    (2)A、repeating
    B、basic
    C、non-repeating
    D、advance
    (3)A、semantic
    B、practical
    C、lexical
    D、syntactical
    (4)A、poet
    B、architect
    C、doctor
    D、programmer
    (5)A、construct
    B、code
    C、size
    D、scale

    查看答案

    試題參考答案:A、C、B、D、A

    試題解析與討論:www.xcpkj.com/exam/ExamDay.aspx?t1=4&day=2025/11/6

  • 試題8

    某高校信息系統(tǒng)釆用分布式數(shù)據(jù)庫系統(tǒng),該系統(tǒng)中“本地業(yè)務(wù)能夠在本地正常進(jìn)行而不依賴于其他場(chǎng)地的數(shù)據(jù)庫”和“數(shù)據(jù)在不同場(chǎng)地上的存儲(chǔ)”分別稱為分布式數(shù)據(jù)庫的 ( ) 。
    A.共享性和分布性
    B.自治性和分布性
    C.自治性和可用性
    D.可用性和分布性

    查看答案

    試題參考答案:B

    試題解析與討論:www.xcpkj.com/exam/ExamDay.aspx?t1=4&day=2025/11/6

  • 試題9

    在Windows 系統(tǒng)中,( )不是網(wǎng)絡(luò)服務(wù)組件。
    A、RAS
    B、HTTP
    C、IIS
    D、DNS

    查看答案

    試題參考答案:B

    試題解析與討論:www.xcpkj.com/exam/ExamDay.aspx?t1=4&day=2025/11/6

  • 試題10

    ( ) 軟件成本估算模型是一種靜態(tài)單變量模型,用于對(duì)整個(gè)軟件系統(tǒng)進(jìn)行估算。
    A.Putnam
    B.基本COCOMO
    C.中級(jí)COCOMO
    D.詳細(xì)COCOMO

    查看答案

    試題參考答案:B

    試題解析與討論:www.xcpkj.com/exam/ExamDay.aspx?t1=4&day=2025/11/6

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