試題1

計算機(jī)中CPU的中斷響應(yīng)時間指的是( )的時間
A.從發(fā)出中斷請求到中斷處理結(jié)束
B.從中斷處理開始到中斷處理結(jié)束
C.CPU分析判斷中斷請求
D.從發(fā)出中斷請求到開始進(jìn)入中斷處理程序

查看答案

試題2

王某是某公司的軟件設(shè)計師，每當(dāng)軟件開發(fā)完成后均按公司規(guī)定編寫軟件文檔，并提交公司存檔。那么該軟件文檔的著作權(quán)(  )享有。
A. 應(yīng)由公司
B. 應(yīng)由公司和王某共有
C. 應(yīng)由王某
D. 除署名權(quán)外，著作權(quán)的其它權(quán)利由王某

查看答案

試題3

下面關(guān)于校驗方法的敘述，  （ ）  是正確的。
A、采用奇偶校驗可檢測數(shù)據(jù)傳輸過程中出現(xiàn)一位數(shù)據(jù)錯誤的位置并加以糾正
B、采用海明校驗可檢測數(shù)據(jù)傳輸過程中出現(xiàn)一位數(shù)據(jù)錯誤的位置并加以糾正
C、采用海明校驗，校驗碼的長度和位置可隨機(jī)設(shè)定
D、采用 CRC 校驗，需要將校驗碼分散開并插入數(shù)據(jù)的指定位置中

查看答案

試題4

某程序的程序圖如下圖所示，運(yùn)用 McCabe 度量法對其進(jìn)行度量，其環(huán)路復(fù)雜度是 （ ）  。

A、4
B、5
C、6
D、8

查看答案

試題5

一棵滿二叉樹，其每一層結(jié)點(diǎn)個數(shù)都達(dá)到最大值，對其中的結(jié)點(diǎn)從1開始順序編號，即根結(jié)點(diǎn)編號為1，其左、右孩子結(jié)點(diǎn)編號分別為2和3，再下一層從左到右的編號為4、5、6、7，依此類推，每一層都從左到右依次編號，直到最后的葉子結(jié)點(diǎn)層為止，則用（）可判定編號為m和n的兩個結(jié)點(diǎn)是否在同一層。
A．log₂m=log₂n
B．[log₂m]=[=[log₂n]
C．[log₂m]+1=[log₂n]
D．[log₂m]=[log₂n]+1

查看答案

試題6

如下所示的 UML 狀態(tài)圖中，（ ）時，不一定會離開狀態(tài) B。

A．狀態(tài) B 中的兩個結(jié)束狀態(tài)均達(dá)到
B．在當(dāng)前狀態(tài)為 B2 時，事件 e2 發(fā)生
C．事件 e2 發(fā)生
D．事件 el 發(fā)生

查看答案

試題7

自底向上的集成測試策略的優(yōu)點(diǎn)包括(  )。
A.主要的設(shè)計問題可以在測試早期處理
B.不需要寫驅(qū)動程序
C.不需要寫樁程序
D.不需要進(jìn)行回歸測試

查看答案

試題8

對于大多數(shù)通用程序設(shè)計語言，用（）描述其語法即可。
A．正規(guī)文法
B．上下文無關(guān)文法
C．上下文有關(guān)文法
D．短語結(jié)構(gòu)文法

查看答案

試題9

Software entities are more complex for their size than perhaps any other humanconstruct, because no two parts are alike （at least above the statement level）. If they are, wemake the two similar parts into one, a（ 71 ）, open or closeD． In this respect software systemsdiffer profoundly from computers, buildings, or automobiles, where repeated elements abound．
Digital computers . are themselves more complex than most things people build;they have very large numbers of states This makes conceiving, describing, and testing them harD． Software systems have orders of magnitude more（ 72 ）than computers do
Likewise, a scaling-up of a software entity is not m erely a repetition of the same elementsm larger size; it is necessarily an mcrease in the number of different elements. In most cases, the elements interact with each other in some（ 73 ）fashion: and the complexity of the wholencreases much more than linearly.
The complexity of software is a（an）（ 74 ）property, not an accidental one Hencedescriptions of a software entity that abstract away its complexity often abstract away its essence.Mathematics and the physical sciences made great strides for threecenturies by constructingsimplified models of complex phenomena, deriving properties fiom the models, and verifyingthose properties experimentally. This worked becausethe complexities（ 75 ） in the modelswere not the .essential properties of the phenomena: It does not work when the complexities are the essence.
Many of the classical problems of developing software products derive from this essential complexi and its nonlinear uicreases with size. Not only .technical problems but management problems as well come from the coin plexity.
A． task
B．job
C．subroitune
D．Program
A．states
B．parts
C．conditions
D．Expressions
A．linear
B. nonlinear
C．Parallel
D．Addititive
A. surface
B． Outside
C．exterior
D．Essential
A． fixed
B． Included
C．ignored
D．stabilized

查看答案

試題10

循環(huán)冗余校驗碼(CRC.利用生成多項式進(jìn)行編碼。設(shè)數(shù)據(jù)位為k位，校驗位為r位，則CRC碼的格式為（）。
A．k個數(shù)據(jù)位之后跟r個校驗位
B．r個校驗位之后跟k個數(shù)據(jù)位
C．r個校驗位隨機(jī)加入k個數(shù)據(jù)位中
D．r個校驗位等間隔地加入k個數(shù)據(jù)位中

查看答案