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軟件設計師每日一練試題（2025/11/9）

來源：信管網 2025年11月10日【所有評論】

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軟件設計師每日一練試題內容（2025/11/9）

試題1

若每一條指令都可以分解為取指、分析和執(zhí)行三步。已知取指時間t_取指=5△t,分析時間t_分析=2△t，執(zhí)行時間t_執(zhí)行=5△t。如果按順序方式從頭到尾執(zhí)行完500 條指令需（1）△t。如果按照[執(zhí)行]_k、[分析]_k+1 、[取指]_k+2重疊的流水線方式執(zhí)行指令，從頭到尾執(zhí)行完500 條指令需（2）△t。
（1）A、5590
B、5595
C、6000
D、6007
（2）A、2492
B、2500
C、2510
D、2515
查看答案
試題參考答案：C、C
試題解析與討論：www.xomuzic.com/exam/ExamDay.aspx?t1=4&day=2025/11/9
試題2
The project workbook is not so much a separate document as it is a structure imposed on the documents that the project will be producing anyway.
All the documents of the project need to be part of this （?。? This includes objectives ,external specifications , interface specifications , technical standards , internal specifications and administrative memoranda(備忘錄).
Technical prose is almost immortal. If one examines the genealogy ( Ff ) of a customer manual for a piece of hardware or software , one can trace not only the ideas , but also many of the very sentences and paragraphs back to the first （請作答此空） proposing the product or explaining the first design. For the technical writer, the paste-pot is as mighty as the pen.
Since this is so, and since tomorrow's product-quality manuals will grow from today’s memos, it is very important to get the structure of the documentation right. The early design of the project （） ensures that the documentation structure itself is crafted, not haphazard. Moreover, the establishment of a structure molds later writing into segments that fit into that structure.
The second reason for the project workbook is control of the distribution of （）. The problem is not to restrict information, but to ensure that relevant information gets to all the people who need it.
The first step is to number all memoranda, so that ordered lists of titles are available and h worker can see if he has what he wants. The organization of the workbook goes well beyond this to establish a tree-structure of memoranda. The （） allows distribution lists to be maintained by subtree, if that is desirable.
A.objective
B.memoranda
C.standard
D.specification
查看答案
試題參考答案：B
試題解析與討論：www.xomuzic.com/exam/ExamDay.aspx?t1=4&day=2025/11/9
試題3
進程P1、P2、P3、P4、P5和P6的前趨勢圖如下所示，若用PV操作控制進程P1、P2、P3、P4、P5和P6并發(fā) 執(zhí)行的過程，則需要設置7個信號量S1、S2、S3、S 4、S5、S6和S7，且信號量S1～S7的初值都等于零。
P 1—P6的進程執(zhí)行圖中，a和b處應分別填寫（）；c 和d處應分別填寫（）；e和f處應分別填寫（）；g 和h處應分別填寫（）

問題1
A．V（S1）和V（S2）
B．P（S1）和P（S2）
C．P（S1）和V（S2）
D．V（S1）和P（S2）
問題2
A．P（S1）P（S2）和P（S3）和P（S4）
B．V（S1）V（S2）和V（S3）和V（S4）
C．P（S1）P（S2）和V（S3）和V（S4）
D．V（S1）V（S2）和P（S3）和P（S4）
問題3
A．P（S5）P（S6）和P（S4）和P（S5）
B．V（S5）V（S6）和V（S4）和V（S5）
C．P（S5）P（S6）和V（S4）和V（S5）
D．V（S5）V（S6）和P（S4）和P（S5）
問題4
A．P（S7）和P（S6）P（S7）
B．V（S7）和P（S6）P（S7）
C．P（S7）和V（S6）V（S7）
D．V（S7）和V（S6）V（S7）
查看答案
試題參考答案：A、C、D、B
試題解析與討論：www.xomuzic.com/exam/ExamDay.aspx?t1=4&day=2025/11/9
試題4

假設鐵路自動售票系統(tǒng)有n個售票終端,該系統(tǒng)為每個售票終端創(chuàng)建一個進程P_i(i=1,2,…,n)管理車票銷售過程。假設T_j(j=1,2,…,m)單元存放某日某趟車的車票剩余票數(shù),Temp為P_i進程的臨時工作單元,x為某用戶的購票張數(shù)。P_i進程的工作流程如下圖所示,用P操作和Ⅴ操作實現(xiàn)進程間的同步與互斥。初始化時系統(tǒng)應將信號量S賦值為(24)。圖中(a)、(b)和(c)處應分別填入(25)。
A.n-1
B.0
C.1
D.2
A.V(S)、P(S)和P(S)
B.P(S)、P(S)和V(S)
C.V(S)、V(S)和P(S)
D.P(S)、V(S)和V(S)
查看答案
試題參考答案：C、D
試題解析與討論：www.xomuzic.com/exam/ExamDay.aspx?t1=4&day=2025/11/9
試題5
下圖是一個軟件項目的活動圖，其中頂點表示項目里程碑，連接頂點的邊表示包含的活動，則一共有(7）條關鍵路徑，關鍵路徑長度為(8) 。

（7）
A．2
B．4
C．3
D．1
（8）
A．48
B．55
C．30
D．46
查看答案
試題參考答案：B、B
試題解析與討論：www.xomuzic.com/exam/ExamDay.aspx?t1=4&day=2025/11/9
試題6

在程序運行過程中，CPU 需要將指令從內存中取出并加以分析和執(zhí)行。CPU 依據(jù)( )來區(qū)分在內存中以二進制編碼形式存放的指令和數(shù)據(jù)。
A．指令周期的不同階段
B．指令和數(shù)據(jù)的尋址方式
C．指令操作碼的譯碼結果
D．指令和數(shù)據(jù)所在的存儲單元
查看答案
試題參考答案：A
試題解析與討論：www.xomuzic.com/exam/ExamDay.aspx?t1=4&day=2025/11/9
試題7
（）開發(fā)過程模型最不適用開發(fā)初期對軟件需求缺乏準確全面認識的情況。
A．瀑布
B．演化
C．螺旋
D．增量
查看答案
試題參考答案：A
試題解析與討論：www.xomuzic.com/exam/ExamDay.aspx?t1=4&day=2025/11/9
試題8

開發(fā)專家系統(tǒng)時，通過描述事實和規(guī)則由模式匹配得出結論，這種情況下適用的開發(fā)語言是（）。
A、面向對象語言
B、函數(shù)式語言
C、過程式語言
D、邏輯式語言
查看答案
試題參考答案：D
試題解析與討論：www.xomuzic.com/exam/ExamDay.aspx?t1=4&day=2025/11/9
試題9
某類圖如圖所示，下列選項錯誤的是（）。

A.一個A1的對象可能與一個A2的對象關聯(lián)
B.一個A的非直接對象可能與一個A1 的對象關聯(lián)
C.類B1的對象可能通過A2與C1的對象關聯(lián)
D.有可能A的直接對象與B1的對象關聯(lián)
查看答案
試題參考答案：D
試題解析與討論：www.xomuzic.com/exam/ExamDay.aspx?t1=4&day=2025/11/9
試題10
Low-code and no code software development solutions have emerged as viable and convenient alternatives to the traditional development process.
Low-code is a rapid application development (RAD) approach that enables automated code generation through(71) building blocks like drag-and-drop and pull-down menu interfaces.This (72) allows low-code users to focus on the diferentiator rather than the common denominator of programming. Low-code is a balanced middle ground between manual coding and no-code as its users can still add code over auto-generated code.While in low-code there is some hand holding done by developers in the form of scripting or manual coding， no-code has a completely (73) approach, with 100% dependence on visual tools.
A low-code application platform (LCAP)——also called a low-code development platform (LCDP)——contains an integrated development environment (IDE) with (74) features like APIs, code templates, reusable plug-in modules and graphical connectors to automate a significant percentage of the application development process.LCAPs are typically available as cloud-based Platform-as-a-Service (PaaS) solutions.
A low-code platform works on the principle or lowering complexity by using visual to and techniques like process modeling. where users employ visual tools to define workflobusiness rules, user interfaces and the like.Behind the scenes, the complete workflovautomatically converted into code. LCAPs are used predominantly by professional developeautomate the generic aspects of coding to redirect effort on the last mile of (75).
(71)
A．visual
B．component-based
C．object-oriented
D．structural
(72)
A．block
B．automation
C．function
D．method
(73)
A．modern
B．hands-off
C．generic
D．labor-free
(74)
A．reusable
B．built-in
C．existed
D．well-known
(75)
A．delivery
B．automation
C．development
D．success
查看答案
試題參考答案：A、B、B、B、C
試題解析與討論：www.xomuzic.com/exam/ExamDay.aspx?t1=4&day=2025/11/9