2022年江西软件水平考试考试模拟卷（2）.docx

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1、2022年江西软件水平考试考试模拟卷（2）本卷共分为1大题50小题，作答时间为180分钟，总分100分，60分及格。一、单项选择题（共50题，每题2分。每题的备选项中，只有一个最符合题意） 1.为了限制路由信息传播的范围，OSPF协议把网络划分成4种区域(area)，其中_的作用是连接各个区域的传输网络，_不接收本地自治系统之外的路由信息。A不完全存根区域B标准区域C主干区域D存根区域2.给定一个C类网络192.168.1.0/24，要在其中划分出3个60台主机的网段和2个30台主机的网段，则采用的子网掩码应该分别为_。A255.255.255.128和255.255.255.224B255.

2、255.255.128和255.255.255.240C255.255.255.192和255.255.255.224D255.255.255.192和255.255.255.2403.网络200.105.140.0/20中可分配的主机地址数是_。A1022B2046C4094D81924.RIPv2对RIPv1协议有三方面的改进。下面的选项中，RIPv2的特点不包括_。在RIPv2中，可以采用水平分割法来消除路由循环，这种方法是指_。A不能向自己的邻居发送路由信息B不要把一条路由信息发送给该信思的来源C路由信息只能发送给左右两边的路由器D路由信息必须用组播而不是广播方式发送5.下面给出的网络

3、地址中，属于私网地址的是_。A119.12.73.214B192.32.146.23C172.34.221.18 D10.215.34.124 6.ARP表用于缓存设备的IP地址与MAC地址的对应关系，采用ARP表的好处是_。A便于测试网络连接数B减少网络维护工作量C限制网络广播数量D解决网络地址冲突7.IEEE 802.11i所采用的加密算法为_。ADESB3DESCIDEADAES8.为了限制路由信息传播的范围，OSPF协议把网络划分成4种区域(area)，其中_的作用是连接各个区域的传输网络，_不接收本地自治系统之外的路由信息。A不完全存根区域B标准区域C主干区域D存根区域9.IPv6地

4、址33AB:0000:0000:CD30:0000:0000:0000:0000/60可以表示成各种简写形式，以下写法中，正确的是_。A33AB:0:0:CD30:/60B33AB:0:0:CD3/60C33ABCD30/60D33ABCD3/6010.A transport layer protocol usually has several responsibilities. One is to create a process-to-processcommunication UDP uses（1）numbers to accomplish this. Another responsibi

5、lity is to provide controlmechanisms at the transport level, UDP does this task at a very minimal level. There is no flow control mechanismand there is no（2）for received packet. UDP, however, does provide error control to some extent. If UDPdetects an error in the received packet, it will silently d

6、rop it.The transport layer also provides a connection mechanism for the processes. The （3）must be able to sendstreams of data to the transport layer. It is the responsibility of the transport layer at（4） station to make theconnection with the receiver chop the stream into transportable units, number

7、 them, and send them one by one. Itis the responsibility of the transport layer at the receiving end to wait until all the different units belonging to thesame process have arrived, check and pass those that are （5）free, and deliver them to the receiving process asa stream.（1）是（）A.hopB.portC.routeD.

8、packet11.自动专用IP地址(Automatic Private IP Address，APIPA)是IANA(Internet AssignedNumbers Authority)保留的一个地址块，它的地址范围是_。当_时，使用APIPA。AA类模块地址10.254.0.010.254.255.255BA类模块地址100.254.0.0100.254.255.255CB类模块地址168.254.0.0168.254.255.255DB类模块地址169.254.0.0169.254.255.25512.IPSec的加密和认证过程中所使用的密钥由_机制来生成和分发。AESPBIKECTGS

9、DAH13.在进行域名解析过程中，由_获取的解析结果耗时最短。A主域名服务器B辅域名服务器C缓存域名服务器D转发域名服务器14.DNS服务器在名称解析过程中正确的查询顺序为_。A本地缓存记录区域记录转发域名服务器根域名服务器B区域记录本地缓存记录转发域名服务器根域名服务器C本地缓存记录区域记录根域名服务器转发域名服务器D区域记录本地缓存记录根域名服务器转发域名服务器15.DNS服务器进行域名解析时，若采用递归方法，发送的域名请求为（）。A1条B2条C3条D多条16.ISP分配给某公司的地址块为199.34.76.64/28，则该公司得到的地址数是_。A8B16C32D6417.A transp

10、ort layer protocol usually has several responsibilities. One is to create a process-to-processcommunication UDP uses（1）numbers to accomplish this. Another responsibility is to provide controlmechanisms at the transport level, UDP does this task at a very minimal level. There is no flow control mecha

11、nismand there is no（2）for received packet. UDP, however, does provide error control to some extent. If UDPdetects an error in the received packet, it will silently drop it.The transport layer also provides a connection mechanism for the processes. The （3）must be able to sendstreams of data to the tr

12、ansport layer. It is the responsibility of the transport layer at （4） station to make theconnection with the receiver chop the stream into transportable units, number them, and send them one by one. Itis the responsibility of the transport layer at the receiving end to wait until all the different u

13、nits belonging to thesame process have arrived, check and pass those that are （5）free, and deliver them to the receiving process asa stream.（2）是（）A.connectionB.windowC.ackmowledgementD.destination18.自动专用IP地址(Automatic Private IP Address，APIPA)是IANA(Internet AssignedNumbers Authority)保留的一个地址块，它的地址范围是

14、_。当_时，使用APIPA。A通信对方要求使用APIPA地址B由于网络故障而找不到DHCP服务器C客户机配置中开启了APIPA功能DDHCP服务器分配的租约到期19.Kerberos由认证服务器(AS)和票证授予服务器(TGS)两部分组成，当用户A通过Kerberos向服务器V请求服务时，认证过程如图6-22所示，图中处为（1），处为（2）。（2）是（）AKAV(t+1)BKS(t+1)CKS(t)DKAV(t)20.A transport layer protocol usually has several responsibilities. One is to create a proce

15、ss-to-processcommunication UDP uses（1）numbers to accomplish this. Another responsibility is to provide controlmechanisms at the transport level, UDP does this task at a very minimal level. There is no flow control mechanismand there is no（2）for received packet. UDP, however, does provide error contr

16、ol to some extent. If UDPdetects an error in the received packet, it will silently drop it.The transport layer also provides a connection mechanism for the processes. The （3）must be able to sendstreams of data to the transport layer. It is the responsibility of the transport layer at （4） station to

17、make theconnection with the receiver chop the stream into transportable units, number them, and send them one by one. Itis the responsibility of the transport layer at the receiving end to wait until all the different units belonging to thesame process have arrived, check and pass those that are （5）

18、free, and deliver them to the receiving process asa stream.（3）是（）A.jobsB.processesC.programsD.users21.A transport layer protocol usually has several responsibilities. One is to create a process-to-processcommunication UDP uses（1）numbers to accomplish this. Another responsibility is to provide contro

19、lmechanisms at the transport level, UDP does this task at a very minimal level. There is no flow control mechanismand there is no（2）for received packet. UDP, however, does provide error control to some extent. If UDPdetects an error in the received packet, it will silently drop it.The transport laye

20、r also provides a connection mechanism for the processes. The （3）must be able to sendstreams of data to the transport layer. It is the responsibility of the transport layer at （4） station to make theconnection with the receiver chop the stream into transportable units, number them, and send them one

21、 by one. Itis the responsibility of the transport layer at the receiving end to wait until all the different units belonging to thesame process have arrived, check and pass those that are （5）free, and deliver them to the receiving process asa stream.（4）是（）A.sendingB.routingC.switchingD.receiving22.A

22、 transport layer protocol usually has several responsibilities. One is to create a process-to-processcommunication UDP uses（1）numbers to accomplish this. Another responsibility is to provide controlmechanisms at the transport level, UDP does this task at a very minimal level. There is no flow contro

23、l mechanismand there is no（2）for received packet. UDP, however, does provide error control to some extent. If UDPdetects an error in the received packet, it will silently drop it.The transport layer also provides a connection mechanism for the processes. The （3）must be able to sendstreams of data to

24、 the transport layer. It is the responsibility of the transport layer at （4） station to make theconnection with the receiver chop the stream into transportable units, number them, and send them one by one. Itis the responsibility of the transport layer at the receiving end to wait until all the diff

25、erent units belonging to thesame process have arrived, check and pass those that are （5）free, and deliver them to the receiving process asa stream.（5）是（）A.callB.stateC.costD.error23.下面关于交换机的说法中，正确的是_。A以太网交换机可以连接运行不同网络层协议的网络B从工作原理上讲，以太网交换机是一种多端口网桥C集线器是一种特殊的交换机D通过交换机连接的一组工作站形成一个冲突域24.POP3采用_模式，当客户机需要服

26、务时，客户端软件(Outlook Express或FoxMail)与POP3服务器建立_连接。ABrowser/ServerBClient/ServerCPeertoPeerDPeertoServer25.POP3采用_模式，当客户机需要服务时，客户端软件(Outlook Express或FoxMail)与POP3服务器建立_连接。ATCPBUDPCPHPDIP26.地址编号从80000H到BFFFFH且按字节编址的内存容量为（）KB，若用16K4bit的存储器芯片构成该内存，共需（）片。A128B256C512D102427.地址编号从80000H到BFFFFH且按字节编址的内存容量为（）K

27、B，若用16K4bit的存储器芯片构成该内存，共需（）片。A8B16C32D6428.若计算机存储数据采用的是双符号位(00表示正号、11表示负号)，两个符号相同的数相加时，如果运算结果的两个符号位经_运算得1，则可断定这两个数相加的结果产生了溢出。A逻辑与B逻辑或C逻辑同或D逻辑异或29.以下关于数的定点表示或浮点表示的叙述中，不正确的是_。A定点表示法表示的数(称为定点数)常分为定点整数和定点小数两种B定点表示法中，小数点需要占用一个存储位C浮点表示法用阶码和尾数来表示数，称为浮点数D在总位数相同的情况下，浮点表示法可以表示更大的数30.若某整数的16位补码为FFFFH(H表示十六进制)，

28、则该数的十进制值为_。A0B-1C216-1D-216-131.若计算机采用8位整数补码表示数据，则（）运算将产生溢出。A-127+1B-127-1C127+1D127-132.指令寄存器的位数取决于_。A存储器的容量B指令字长C数据总线的宽度D地址总线的宽度33.在CPU中用于跟踪指令地址的寄存器是_。A地址寄存器(MAR)B数据寄存器(MDR)C程序计数器(PC)D指令寄存器(IR)34.若某条无条件转移汇编指令采用直接寻址，则该指令的功能是将指令中的地址码送入_。APC(程序计数器)BAR(地址寄存器)CAC(累加器)DALU(算逻运算单元)35.编写汇编语言程序时，下列寄存器中，程序员

29、可访问的是_。A程序计数器(PC)B指令寄存器(IR)C存储器数据寄存器(MDR)D存储器地址寄存器(MAR)36.在CPU中，_不仅要保证指令的正确执行，还要能够处理异常事件。A运算器B控制器C寄存器租D内部总线37.X、Y为逻辑变量，与逻辑表达式X+Y等价的是（）。AX+B+C+YDX+Y38.在程序执行过程中，Cache与主存的地址映像由_。A硬件自动完成B程序员调度C操作系统管理D程序员与操作系统协同完成39.常用的虚拟存储器由_两级存储器组成。A主存-辅存B主存-网盘CCache-主存DCache-硬盘40.若某计算机字长为32位，内存容量为2GB，按字编址，则可寻址范围为_。A10

30、24MB1GBC512MBD2GB41.若某计算机系统的I/O接口与主存采用统一编址，则输入输出操作是通过_指令来完成的。A控制B中断C输入输出D访存42.在程序的执行过程中，Cache与主存的地址映像由_。A专门的硬件自动完成B程序员进行调度C操作系统进行管理D程序员和操作系统共同协调完成43.位于CPU与主存之间的高速缓冲存储器Cache用于存放部分主存数据的拷贝，主存地址与Cache地址之间的转换工作由_完成。A硬件B软件C用户D程序员44.相联存储器按_访问。A地址B先入后出的方式C内容D先入先出的方式45.为了便于实现多级中断，使用_来保护断点和现场最有效。AROMB中断向量表C通用

31、寄存器D堆栈46.计算机中主存储器主要由存储体、控制线路、地址寄存器、数据寄存器和_组成。A地址译码电路B地址和数据总线C微操作形成部件D指令译码器47.中断向量可提供_。AI/O设备的端口地址B所传送数据的起始地址C中断服务程序的入口地址D主程序的断点地址48.DMA工作方式下，在（）之间建立直接的数据通信。ACPU与外设BCPU与主存C主存与外设D外设与外设49.在输入输出控制方法中，采用_可以使得设备与主存间的数据块传送无需CPU干预。A程序控制输入输出B中断CDMAD总线控制50.指令系统中采用不同寻址方式的目的是_。A提高从内存获取数据的速度B提高从外存获取数据的速度C降低操作码的译码难度D扩大寻址空间并提高编程灵活性第20页 共20页第 20 页 共 20 页第 20 页 共 20 页第 20 页 共 20 页第 20 页 共 20 页第 20 页 共 20 页第 20 页 共 20 页第 20 页 共 20 页第 20 页 共 20 页第 20 页 共 20 页第 20 页 共 20 页