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2015下半年软件设计师考试真题及答案

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A、modern B、conventional C、different D、formal

【题目74】

In a world where it seems we already have too much to do,and too many things to think about, it seems the last thing we need is something new that we have to learn.But use cases do solve a problem with requirements:with (71) declarative requirements it's hard to describe steps and sequences of events.Use cases,stated simply, allow description of sequences of events that, taken together, lead to a system doing something useful.As simple as this sounds,this is important. When confronted only with a pile of requiements, it's often(72)to make sense of what the authors of the requirements really wanted the system to do.In the preceding example, use cases reduce the ambiguity of the requirements by specifying exactly when and under what conditions certain behavior occurs;as such, the sequence of the behaviors can be regarded as a requirement. Use cases are particularly well suited to capture approaches. Although this may sound simple, the fact is that (73) requirement capture approaches, with their emphasis on declarative requirements and \statements,completely fail to capture fail to capture the (74) of the system's behavior. Use cases are a simple yet powerful way to express the behavior of the system in way that all stakeholders can easily understand.But,like anything, use cases come with their own problems, and as useful as they are,they can be (75). The result is something that is as bad, if not worse, that the original problem.Therein it's important to utilize use cases effectively without creating a greater problem than the one you started with. A、statics B、nature C、dynamics

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D、originals

【题目75】

In a world where it seems we already have too much to do,and too many things to think about, it seems the last thing we need is something new that we have to learn.But use cases do solve a problem with requirements:with (71) declarative requirements it's hard to describe steps and sequences of events.Use cases,stated simply, allow description of sequences of events that, taken together, lead to a system doing something useful.As simple as this sounds,this is important. When confronted only with a pile of requiements, it's often(72)to make sense of what the authors of the requirements really wanted the system to do.In the preceding example, use cases reduce the ambiguity of the requirements by specifying exactly when and under what conditions certain behavior occurs;as such, the sequence of the behaviors can be regarded as a requirement. Use cases are particularly well suited to capture approaches. Although this may sound simple, the fact is that (73) requirement capture approaches, with their emphasis on declarative requirements and \statements,completely fail to capture fail to capture the (74) of the system's behavior. Use cases are a simple yet powerful way to express the behavior of the system in way that all stakeholders can easily understand.But,like anything, use cases come with their own problems, and as useful as they are,they can be (75). The result is something that is as bad, if not worse, that the original problem.Therein it's important to utilize use cases effectively without creating a greater problem than the one you started with. A、misapplied B、applied C、used D、powerful

-----------------答案及解析-----------

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【答案1】: 答案:D 【解析】

本题考查计算机组成基础知识。DMA控制器在需要的时候代替CPU作为总线主设备,在不受CPU干预的情况下,控制I/O设备与系统主存之间的直接数据传输。DMA操作占用的资源是系统总线,而CPU并非在整个指令执行期间即指令周期内都会使用总线,故DMA请求的检测点设置在每个机器周期也即总线周期结束时执行,这样使得总线利用率最高。

【答案2】: 答案:A 【解析】

本题考查计算机组成基础知识。计算机中不同容量、不同速度、不同访问形式、不同用途的各种存储器形成的是一种层次结构的存储系统。所有的存储器设备按照一定的层次逻辑关系通过软硬件连接起来,并进行有效的管理,就形成了存储体系。不同层次上的存储器发挥着不同的作用。-般计算机系统中主要有两种存储体系Cache存储体系由Cache和主存储器构成,主要目的是提高存储器速度,对系统程序员以上均透明;虚拟存储体系由主存储器和在线磁盘存储器等辅存构成,主要目的是扩大存储器容量,对应用程序员透明。

【答案3】: 答案:B 【解析】

本题考查计算机组成基础知识。在计算机中使用了类似于十进制科学计数法的方法来表示二进制实数,因其表示不同的数时小数点位置的浮动不固定而取名浮点数表示法。浮点数编码由两部分组成阶码(即指数,为带符号定点整数,常用移码表示,也有用补码的)和尾数(是定点纯小数,常用补码表示,或原码表示)。因此可以知道,浮点数的精度由尾数的位数决定,表示范围的大小则主要由阶码的位数决定。

【答案4】:

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答案:C 【解析】

本题考查计算机组成基础知识。随着主存增加,指令本身很难保证直接反映操作数的值或其地址,必须通过某种映射方式实现对所需操作数的获取。指令系统中将这种映射方式称为寻址方式,即指令按什么方式寻找(或访问)到所需的操作数或信息(例如转移地址信息等)。可以被指令访问到的数据和信息包括通用寄存器、主存、堆栈及外设端口寄存器等。指令中地址码字段直接给出操作数本身,而不是其访存地址,不需要访问任何地址的寻址方式被称为立即寻址。

【答案5】: 答案:B 【解析】

本题考查计算机组成基础知识。直接计算16进制地址包含的存储单元个数即可。DABFFH-B3000H+1=27C00H=162816=159k,按字节编址,故此区域的存储容量为159kB。 【答案6】: 答案:A 【解析】

本题考查计算机组成与结构基础知识。计算机技术发展使得机器性能提高,随着高级语言的发展,程序员需要更强大的命令,指令集往往结合应用需要不断扩展,推动了指令集越来越复杂,形成了CISC,即Complex Instruction Set Computer,就是使用复杂指令集系统的计算机。与其对应的是RISC,即Reduced Instruction Set Computer,精简指令集系统的计算机。

【答案7】: 答案:A 【解析】

本题考查的是网络攻击的基础知识。网络攻击有主动攻击和被动攻击两类。其中主动攻击是

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指通过一系列的方法,主动向被攻击对象实施破坏的一种攻击方式,例如重放攻击、IP地址欺骗、拒绝服务攻击等均属于攻击者主动向攻击对象发起破坏性攻击的方式。流量分析攻击是通过持续检测现有网络中的流量变化或者变化趋势,而得到相应信息的一种被动攻击方式。

【答案8】: 答案:B 【解析】

本题考查的是防火墙基础知识。防火墙是一种放置在网络边界上,用于保护内部网络安全的网络设备。它通过对流经数据流进行分析和检查,可实现对数据包的过滤、保存用户访问网络的记录和服务器代理功能。防火墙不具备检査病毒的功能。

【答案9】: 答案:C 【解析】

本题考查网管命令netstat-n的含义以及端口的作用。从netstat -n的输出信息中可以看出,本地主机192.168.0.200使用的端口号2011、2038、2052都不是公共端口号。 根据状态提示信息,其中己经与主机128.105.129.30的80端口建立了普通连接, 与主机100.29.200.110正在等待建立连接, 与主机202.100.112.12的443端口建立连接,由于443端口主要用于HTTPS服务,是提供加密和通过安全端口传输的另一种HTTP协议,所以是建立了安全连接。

【答案10】: 答案:C 【解析】

我国著作权法在第10条对权利内容作了较为详尽而具体的规定,指明著作权的内容包括人身权利和财产权利。著作人身权是指作者享有的与其作品有关的以人格利益为内容的权利,也称为精神权利,包括发表权、署名权、修改权和保护作品完整权。著作人身权与作者的身份紧密联系,永远属于作者本人,即使作者死亡,其他任何人不能再拥有它。所以,我国著

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2015下半年软件设计师考试真题及答案

A、modernB、conventionalC、differentD、formal【题目74】Inaworldwhereitseemswealreadyhavetoomuchtodo,andtoomanythingstothinkabout,itseemsthelastthingweneedisso
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