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语言学课后习题答案汇编

Chapter 1

Introduction

1. De?ne the following terms brie?y.

(1) linguistics: the scienti?c or systematic study of language.

(2) language: a system of arbitrary vocal symbols used for human communication.

(3) arbitrariness: the absence of similarity between the form of a linguistic sign and what it

relates to in reality, e.g. the word dog does not look like a dog.

(4) duality: the way meaningless elements of language at one level (sounds and letters)

combine to form meaningful units (words) at another level.

(5) competence: knowledge of the grammar of a language as a formal abstraction and

distinct from the behavior of actual language use, i.e. performance.

(6) performance: Chomsky’s te rm for actual language behavior as distinct from the

knowledge that underlies it, or competence.

(7) stylistics: the study of how literary effects can be related to linguistic features.

(8) phatic communion: Language is used to establish an atmosphere or maintain

social contact between the speaker and the hearer.

(9) functionalism: the study of the forms of language in reference to their social function in

communication.

(10) formalism: the study of the abstract forms of language and their internal relations.

(11) synchronic linguistics: the study of language and speech as they are used at a given

moment and not in terms of how they have evolved over time.

(12) diachronic linguistics: the study of linguistic change over time in contrast to looking at

language as it is used at a given moment.

2. No, language is human-speci?c. Human language has seven design features, including arbitrariness, duality, productivity, interchangeability, displacement, specialization and cultural transmission. These features are found utterly lacking in do gs’ or pigs’ noises and thus set human language apart from animal cry systems.

3. Arbitrariness refers to the fact that there is no logical or intrinsic connection bet ween a particular sound and the meaning it is associated with. For example, for the same animal dog, in English we call it /d0g/, in Chinese as “gou”, but “yilu” in Japanese; it barks wow wow in English but wang wang in Chinese. Of course, onomatopoetic words such as “quack-quack” and “bang” are exceptions, but words like these are re latively few compared with the total number of words in a language.

4. A human baby does not speak any language at birth. What language the baby is going to speak is determined by the culture he is born into. A Chinese baby born and brought up in London by an English family will speak English, while an English child brought up in Beijing by a Chinese aunt will speak Chinese. That is to say, language cannot be transmitted through heredity. It is culturally transmitted.

5. Firstly, linguistics describes languages and does not lay down rules of correctness while traditional grammar emphasizes correctness. Secondly, linguistics regards the spoken language as primary, while traditional grammar emphasizes the priority of the written language. Thirdly, traditional grammar is based on Latin and it tries to impose the Latin categories and structures on other languages, while linguistics describes each language on its own merits.

6. A descriptive approach attempts to tell what is in the language while the prescriptive approach tells people what should be in the language. Most modern linguistics is descriptive, whereas traditional grammars are prescriptive.

7. Synchronic linguistics studies language at one particular time while diachronic linguistics studies language developments through time. Synchronic linguistics focuses on the state of language at any point in history while diachronic linguistics focuses on the differences in two or more than two states of language over decades or centuries.

8. No, human language has the design feature of specialization. It refers to the fact that man does not have a total physical involvement in the act of communication. For example, a mother can tell a story to her child while slicing up a cake. However, wolves can only respond to a stimulus and is totally involved physically in the communication process. Thus, a wolf cannot have a language similar to man’s, even though it could express a thousand different emotions. Besides, the aspect of productivity also distinguishes human language from wolf ’s postures.

Chapter 2

The Sounds of Language

1. De?ne each of the following terms brie?y.

(1) articulator: the tongue, lips, and velum, which change the shape of the vocal tract to

produce different speech sounds.

(2) assimilation: a phonological process whereby a sound becomes phonetically similar (or

identical) to a neighboring sound, e.g. a vowel becomes [+nasal] when followed by a

[+nasal] consonant.

(3) consonant: a speech sound produced by partial or complete closure of part of the vocal

tract, thus obstructing the air?ow and creating audible friction. Consonants are

described in terms of voicing, place of articulation, and manner of articulation.

(4) elision: the leaving out of a sound or sounds in speech.

(5) intonation: the variation in pitch and stress which gives beat and rhythm to the tune the

voice plays in ordinary speech.

(6) phoneme: the abstract element of a sound, ident i?ed as being distinctive in a particular

language.

(7) phonetics: the study of linguistic speech sounds, how they are produced,

how they are perceived, and their physical properties.

(8) phonology: the study of the abstract systems underlying the sounds of language.

(9) stress: the prominence given to certain sounds in speech.

(10) voicing: the vibration in the larynx caused by air from the lungs passing

through the vocal cords when they are partly closed; speech sounds are said to be either

voiced or voiceless.

(11) voiceless: When the vocal cords are spread apart, the airstream from the

lungs is not obstructed at the space between vocal cords and passes freely.

The sounds produced in this way are described as voiceless sounds.

(12) vowel: a speech sound produced without signi?cant constriction of the air ?owing

through the oral cavity.

2. Minimal pair test or substitution test.

Minimal pair test or substitution test is to see whether substituting one sound for 4another results in a different word. If it does, the two sounds represent different phonemes. For example, as to the English word bear, if we substitute p for b, we get the word pear, the two are different words. Then /b/ and /p/ represent different phonemes. Other examples are chunk/junk, ban/bin, bet/beat, ?ne/vine, side/site, etc.

3. Take the word big for example, in the word big /big/, /b/ is the onset, /i/ is the nucleus and /g/ is the coda. The difference between open syllables and closed syllables is whether the words have codas. If there are codas, they are closed syllables, such as pig, hat and at; if not, they are open syllables, such as do, I, tea and key.

4. (1) stop, consonant

(2) back, rounded, vowel

5. (1) voiceless / voiced

(2) bilabial / labiodental

(3) close / semi-open

(4) stop / nasal

(5) alveolar / palatal

(6) alveolar / dental

(1) kit/git, bucker/bugger, bag/back

(2) mark/nark, smack/snack, sum/sun

(3) best/vest, ober/over, lib/live

(4) bore/more, abate/amate, mob/mom

(5) pat/fat, apt/aft (AmE), harp (BrE)/half

7. (1) The stresses are placed on the second syllable except for “promise”. We may easily conclude that the verbs usually are stressed on the second syllable.

(2) Syllable representations of the words:

collide [k2#laid] elect [i#lekt] consider [k2n#sid2]

Chapter 3

Morphology

1. De?ne the following terms brie?y.

(1) morphology: the study of the structure of words.5

(2) morpheme: the smallest unit of language that carries meaning or serves a

grammatical function.

(3) free morpheme: a morpheme that can stand alone as a word.

(4) bound morpheme: a morpheme that can not stand alone as a word, e.g.

ment (as in establishment), and -er (as in painter).

(5) morph: the smallest meaningful phonetic segments of an utterance on the level

of parole.

(6) allomorph: a phonetic form in which a morpheme is realized, e.g. -s, -es, and en are all

allomorphs (in writing) of the plural morpheme.

(7) derivation: the formation of new words by adding af?xes to other words or morphemes in

morphology and word formation.

(8) clipping: the process by which parts of a word of more than one syllable have been cut

off, and reduced to a shorter form.

(9) acronym: words which are composed of the ?rst letter of a series of words and are

pronounced as single words. Examples: NATO, radar and yuppy.

(10) initialism: Some new words are composed of the ?rst letters of a series of words

and pronounced by saying each letter in them. Such words are called initialism.

(11) blending: A single new word can be formed by combining two separate forms.

Typically, blending is ?nished by taking only the beginning of one word and

joining it to the end of another word. For example, brunch is formed by the

shortened forms of breakfast and lunch.

(12) root: the morpheme that remains when all af?xes are stripped from a complex word, e.g.

system from un- + system + atic + ally.

(13) stem: the base to which one or more af?xes are attached to create a more complex

form that may be another stem or a word. For example, book is the stem of bookish. (14) pre?x: Af?xes can be joined to the beginning of the root or stem, in which case

t hey are called pre?xes.

(15) suf?x: Af?xes can be joined to the end of the root or stem, in which case they are called

suf?xes.

2. (3), (5), (7)

汇编语言试题及参考答案

汇编语言试题及参考答案一,填空题 1.ZF标志位是标志结果是否为零的,若结果,ZF为( 1 ),否则ZF为( 0 ).当ZF为1时,SF为( 0 ) 2.标号可以有两种类型属性.它们是( )和( ) 3.8位无符号整数的表示范围为0--255,写成16进制形式为( ),8位有符号整数的表示范围为-128--+127,写成16进制形式为( ) 4.伪指令DB,GROUP 和NAME 三个标号名字域必须有名字的是( ),不得有名字的是( ),可有可无名字的是( ). 5.循环程序通常由( )( )( )和循环结果外理四部分构成 6.在数值不达式中,各种运算符可可混合使用,其优先次序规则*,/( )于+,-;XOR,OR( )于AND,LT( )于GT 7. 宏指令定义必须由伪指令( )开始,伪指令( )结束,两者之间的语句称为( ) 8.调用程序与子程序之间的参数传递方法有四种,即堆栈法( )( )( ) 9.分别用一条语句实现下述指明的功能 (1)栈顶内容弹出送字变量AYW( ) (2)双字变量AYD存放的地址指针送ES和SI( ) (3)变量AY2类型属性送AH( ) (4)不允许中断即关中断( ) (5)将字符串'HOW ARE YOU!'存入变量AYB( ) (6)子程序返回调用程序( ) (7)地址表达式AYY[4]的偏移地址送字变量ADDR( ) (8)AX的内容加1,要求不影响CF( ) (9)BX的内容加1,要求影响所有标志位( ) (10)若操作结果为零转向短号GOON( ) 二,单选题 1.IBM PC微机中,有符号数是用( )表示的 1.原码 2.补码 3.反码 4.BCD码 2.把汇编源程序变成代码程序的过程是( ) 1.编译 2.汇编 3.编辑 4.链接

汇编语言课后习题答案

2.1已知DS＝2000H、BX＝0100H、SI＝0002H，存储单元[20100H]～[20103H]依次存放12345678H，[21200H]～[21203H]依次存放2A4C B765H，说明下列每条指令执行完后AX寄存器的内容。（1）mov ax,1200h （2）mov ax,bx （3）mov ax,[1200h] （4）mov ax,[bx] （5）mov ax,[bx+1100h] （6）mov ax,[bx+si] （7）mov ax,[bx][si+1100h] 〔解答〕（1）AX＝1200H （2）AX＝0100H （3）AX＝4C2AH;偏移地址=bx=0100h （4）AX＝3412H;偏移地址=bx=0100h （5）AX＝4C2AH;偏移地址=bx+1100h=1200h （6）AX＝7856H;偏移地址=bx+si=0100h+0002h=0102h （7）AX＝65B7H;偏移地址=bx+si+1100h=0100h+0002h+1100h=1202h 〔习题2.2〕指出下列指令的错误（1）mov cx,dl （2）mov ip,ax （3）mov es,1234h （4）mov es,ds （5）mov al,300 （6）mov[sp],ax （7）mov ax,bx+di

（8）mov20h,ah 〔解答〕（1）两操作数类型不匹配（2）IP指令指针禁止用户访问（3）立即数不允许传给段寄存器（4）段寄存器之间不允许传送（5）两操作数类型不匹配（6）目的操作数应为[SI] （7）源操作数应为[BX+DI] （8）立即数不能作目的操作数〔习题2.5〕已知SS=2200H、SP=00B0H，画图说明执行下面指令序列时，堆栈区和SP的内容如何变化？mov ax,8057h push ax mov ax,0f79h push ax pop bx pop[bx] 〔解答〕 mov ax,8057h push ax mov ax,0f79h push ax pop bx;bx=0f79h pop[bx];DS:[0f79h]=8057h

《新编简明英语语言学教程》第二版课后练习题答案

《新编简明英语语言学教程》第二版练习题参考答案 Chapter 1 Introduction 1. How do you interpret the following definition of linguistics: Linguistics is the scientific study of language. 答：Linguistics is based on the systematic investigation of linguistic data, conducted with reference to some general theory of language structure. In order to discover the nature and rules of the underlying language system, the linguists has to collect and observe language facts first, which are found to display some similarities, and generalizations are made about them; then he formulates some hypotheses about the language structure. The hypotheses thus formed have to be checked repeatedly against the observed facts to fully prove their validity. In linguistics, as in any other discipline, data and theory stand in a dialectical complementation, that is, a theory without the support of data can hardly claim validity, and data without being explained by some theory remain a muddled mass of things. 2. What are the major branches of linguistics? What does each of them study? 答：The major branches of linguistics are: (1) phonetics: it studies the sounds used in linguistic communication; (2) phonology: it studies how sounds are put together and used to convey meaning in communication; (3) morphology: it studies the way in which linguistic symbols representing sounds are arranged and combined to form words; (4) syntax: it studies the rules which govern how words are combined to form grammatically permissible sentences in languages;

真题汇编(一★)参考答案及解析

真题汇编（一）参考答案及解析一、单项选择题 1.【答案】A。解析：人类历史上最早、最完整论述教育问题的著作是我国的《学记》。 2.【答案】C。解析：亚里士多德的和谐教育思想以灵魂论为基础、以培养优良的公民为目标、以德智体和谐发展为内容、以音乐陶冶为途径。其最早根据儿童身心发展的特点提出按年龄划分教育阶段的主张。 3.【答案】A。解析：“用一把钥匙开一把锁”意为用不同的方法解决不同的问题，在教学中教师要因材施教，有针对性地对学生进行个别指导。 4.【答案】C。解析：奥苏贝尔指出，有意义学习过程的实质是将符号所代表的知识与学习者认知结构中已有知识的适当观念建立非人为的和实质性的联系。 5.【答案】B。解析：“君子欲化民成俗，其必由学乎”出自《礼记?学记》，意思是君子如果要教化人民，形成良好的风俗习惯，一定要从教育入手。表达了这样一种治学精神，即教育担负的最大责任不是传承知识，而是移风易俗。“古之王者，建国君民，教学为先”意为古代的君王建立国家，治理民众，都把教育当作首要的事情。都体现了社会本位论的教育目的观。 6.【答案】C。解析：教师劳动的创造性主要表现在三个方面：因材施教、教育方法的不断更新和教育机智。 7.【答案】D。解析：美国教育学家布鲁纳提出了结构教学理论和发现式学习法。他认为学习是一个主动形成和发展认知结构的过程，是在内在动机的推动下，学习者主动对新知识加以选择、转换、储存和应用的过程。 8.【答案】D。解析：刚能引起感觉的最小刺激量称为绝对感觉阈限，与之相应的感觉能力称为绝对感受性，二者在数值上成反比关系。能觉察的刺激物的最小差异量称为差别感觉阈限或最小可觉差，与之相应的感受能力称为差别感受性，它与差别阈限在数值上成反比。 9.【答案】A。解析：“入芝兰之室，久而不闻其香”在心理学中把这种现象称为感觉的适应。适应现象指的是在同一刺激持续作用于同一感受器而产生的感受性提高或降低的变化。 10.【答案】A。解析：有利于教育的原则是班级组织建构的首要原则。 11.【答案】B。解析：心理咨询的根本目标是使来访学生在以后遇到类似的问题时可以自己独立处理，而不是动辄就寻找咨询者的帮助。直接目标是咨询者向咨询对象提供心理上的帮助，以使后者克服当前的心理问题，并提高自身应付挫折的能力。

汇编语言课后习题解答

第1章基础知识检测点1.1(第9页) （1）1个CPU的寻址能力为8KB，那么它的地址总线的宽度为13位。（2）1KB的存储器有1024个存储单元，存储单元的编号从0到1023。（3）1KB的存储器可以存储8192（2^13）个bit，1024个Byte。（4）1GB是1073741824（2^30）个Byte、1MB是1048576（2^20）个Byte、1KB是1024（2^10）个Byte。（5）8080、8088、80296、80386的地址总线宽度分别为16根、20根、24根、32根，则它们的寻址能力分别为: 64（KB）、1（MB）、16（MB）、4（GB）。（6）8080、8088、8086、80286、80386的数据总线宽度分别为8根、8根、16根、16根、32根。则它们一次可以传送的数据为: 1（B）、1（B）、2（B）、2（B）、4（B）。（7）从内存中读取1024字节的数据，8086至少要读512次，80386至少要读256次。（8）在存储器中，数据和程序以二进制形式存放。

（1）1KB=1024B，8KB=1024B*8=2^N，N=13。（2）存储器的容量是以字节为最小单位来计算的，1KB=1024B。（3）8Bit=1Byte，1024Byte=1KB（1KB=1024B=1024B*8Bit）。（4）1GB=1073741824B（即2^30）1MB=1048576B（即2^20）1KB=1024B（即2^10）。（5）一个CPU有N根地址线，则可以说这个CPU的地址总线的宽度为N。这样的CPU 最多可以寻找2的N次方个内存单元。（一个内存单元=1Byte）。（6）8根数据总线一次可以传送8位二进制数据（即一个字节）。（7）8086的数据总线宽度为16根（即一次传送的数据为2B）1024B/2B=512，同理1024B/4B=256。（8）在存储器中指令和数据没有任何区别，都是二进制信息。

2016《汇编语言》练习题库参考答案分析

华中师范大学网络教育学院《汇编语言程序设计》练习测试题库参考答案一、选择题答案： 1.B 2.C 3.B 4.C 5.C 6.C 7.A 8.A 9.D 10.C 11.B 12.A 13.C 14.B 15.A 16.D 17.A 18.A 19.D 20.A 21.A 22.D 23.C 24.B 25.C 26.D 27.B 28.D 29.A 30.A 31.D 32.A 33.D 34.D 35.D 36.D 37.B 38.C 39.A

41.A 42.C 43.B 44.B 45.C 46.D 47.D 48.D 49.A 50.A 51.D 52.B 53.C 54.A 55.B 56.C 57.A 58.C 59.D 60.C 61.A 62.D 63.C 64.C 65.D 66.A 67.B 68.C 69.B 70.D 71.D 72.A 73.D 74.C 75.D 76.C 77.C 78.D 79.D 80.B 二、判断题答案： 1. ×

3. √ 4. √ 5. × 6. √ 7. × 8. × 9. × 10. × 11. √ 12．√ 13. × 14. √ 15. √ 16. √ 17. × 18. × 19. √ 20. √ 21. √ 22. √ 23. √ 24. × 25. √ 26. × 27. √ 28 √ 29. √ 30. √ 31. √ 32. √ 33. √ 34. √ 35. √ 36. √ 37. √ 38. √ 39. × 40. √ 41. × 42. √ 三、填空题答案：

汇编语言王爽第二版课后答案

第一章基础知识检测点(第8页) ---------------------- (1) 13 (2) 1024,0,1023 (3) 8192,1024 (4) 2^30,2^20,2^10 (5) 64,1,16,4 (6) 1,1,2,2,4 (7) 512,256 (8) 二进制注意: 1.第4题中的符号'^'指求幂运算(如: 2^30指2的30次方) 第二章寄存器(CPU工作原理) 检测点(第18页) ---------------------- (1)写出每条汇编指令执行后相关寄存器中的值。第一空：F4A3H 第二空：31A3H 第三空：3123H 第四空：6246H 第五空：826CH 第六空：6246H 第七空：826CH 第八空：04D8H 第九空：0482H 第十空：6C82H 第十一空：D882H 第十二空：D888H 第十三空：D810H 第十四空：6246H (2)只能使用目前学过的汇编指令，最多使用4条指令，编程计算2的4次方。解答如下： mov ax,2 add ax,ax add ax,ax add ax,ax 检测点(第23页) ---------------------- (1)00010H,1000FH (2)1001H,2000H 第2题说明：因为段的起始地址要为16的倍数。所以当段地址小于1001H或大于2000H时CPU都无法寻到。检测点(第33页) ----------------------答：CPU修改了4次IP的值。情况如下：第1次：执行完mov ax,bx后第2次：执行完sub ax,ax后第3次：读入jmp ax后第4次：执行完jmp ax后最后IP的值为0 实验1 查看CPU和内存，用机器指令和汇编指令编程(第33页) ----------------------------------------------------- 1.预备知识：Debug的使用 <此部分略> 2.实验任务(第43页) (1) <此部分略> (2) <此部分略> (3) 通过DEBUG中的D命令查看到主板的生产日期[以月、日、年，分隔符为'/'的格式]存储在内存ffff:0005~ffff:000C(共8个字节单元中)处。此生产日期不能被改变，因为其具有‘只读’属性。 (4) 通过向内存中的显存写入数据，使计算机根据写入的数据进行ASCII转换，并将转换后且可打印的字符输出到屏幕上。<注：关于显存的详细讨论不在此题范围> 第三章寄存器(内存访问) 检测点(第52页) ---------------------- (1)(题目：略) 第一空：2662H 第二空：E626H 第三空：E626H 第四空：2662H 第五空：D6E6H 第六空：FD48H 第七空：2C14H 第八空：0000H 第九空：00E6H 第十空：0000H 第十一空：0026H 第十二空：000CH 提示：此题可在DEBUG中利用E命令在本机上按照题目中所给出的内存单元及其数据进行相应地修改，然后再用A命令进行写入(题目中所给出的)相应的汇编指令，最后再进行T命令进行逐步执行，以查看相应结果。

汇编语言程序设计(钱晓捷)课后答案

汇编语言程序设计(第二版) 钱晓捷习题答案第二章（０１） 2.1 （1）AX＝1200h （2）AX＝0100h （3）AX＝4C2Ah （4）AX＝3412h （5）AX＝4C2Ah （6）AX＝7856h （7）AX＝65B7h 2.2(1) 两操作数类型不匹配 (2) IP指令指针禁止用户访问 (3) 立即数不允许传给段寄存器 (4) 段寄存器之间不允许传送 (5) 两操作数类型不匹配 (6) 目的操作数应为[ BP ] (7) 源操作数应为[BX+DI] (8) 立即数不能作目的操作数 2.3 lea bx,table ；获取table的首地址，BX＝200H mov al,8 ；传送欲转换的数字，AL＝8 xlat ；转换为格雷码，AL＝12H 2.4 堆栈是一种按“先进后出”原则存取数据的存储区域。堆栈的两种基本操作是压栈和出栈，对应的指令是PUSH和POP。 2.5 mov ax,8057h push ax mov ax,0f79h push ax pop bx ;bx=0f79h pop [bx] ;DS:[0f79h]=8057h 2.6 AL=89h CF ZF SF OF PF AL=12h 1 0 0 1 1 AL=0afh 0 0 1 0 1 AL=0afh 1 0 1 0 1 AL=00h 0 1 0 0 1 AL=0ffh 0 0 1 0 1 AL=00h 0 1 0 0 1 2.7 W=X+Y+24-Z

2.8 （1）ADD DX,BX （2）ADD AL,[BX+SI] （3）ADD [BX+0B2H],CX （4）ADD WORD PTR [0520H],3412H （5）ADD AL,0A0H 2.9；为了避免与操作数地址混淆，将题中X,Y,Z,V 字操作数改为A,B,C,D mov ax,X ；ax=A imul Y ；dx,ax = A*B (将操作数看作符号数，以下同） mov cx,ax mov bx,dx ；bx,ax <-- dx,ax =A*B mov ax,Z ；ax = C cwd ；dx,ax =C （扩展符号后为双字） add cx,ax adc bx,dx ；bx,cx <-- bx,cx+dx,ax=A*B+C sub cx,540 sbb bx,0 ；bx,cx<-- A*B+C-540 mov ax, V ；ax= D cwd ；dx,ax= D （扩展符号后为双字） sub ax, cx sbb dx, bx ；dx,ax = dx,ax - bx,cx = D-(A*B+C-540) idiv X ；运算结果：[D-(A*B+C-540h)]/A ；ax存商，dx存余数 2.10;（1）xchg的操作数不能是立即数（2不能对CS直接赋值（3）两个操作数不能都是存储单元（4）堆栈的操作数不能是字节量（5）adc的操作数不能是段寄存器（6）没有确定是字节还是字操作（7）in不支持超过FFH的直接寻址（8）out只能以AL/AX为源操作数第二章(02) 2.11; 指令AX的值CF OF SF ZF PF Mov ax,1407h1470h----- And ax,ax1470h00000 Or ax,ax1470h00000 Xor ax,ax000011 Not ax0ffffh----- Test ax,0f0f0h0ffffh00101 注意: 1. mov, not指令不影响标志位 2. 其他逻辑指令使CF=OF=0, 根据结果影响其他标志位。

汇编语言课后习题答案郑晓薇

习题一 1 分别将下列二进制数作为无符号数和带符号数转换为十进制和十六进制数 11010011，01110111，10000011，00101111 ，10101010 查看正确答案无符号数: 11010011=211=D3H，01110111=119=77H，10000011=131=83H，00101111=47=2FH，10101010=170=AAH 带符号数：11010011= -45=D3H，01110111=+119=77H，10000011= -125=83H，00101111=+47=2FH，10101010= -86=AAH 2 十六进制运算 1A52H+4438H ，3967H－2D81H，37H×12H ，1250H×4H 查看正确答案 5E8AH，0BE6H，3DEH，4940H 3 将十进制数变为8位补码，做运算（结果用二进制、十六进制、十进制表示） 29+53，73－24，-66+82 ，-102－15 查看正确答案 00011101+00110101=01010010=52H=82 01001001+11101000=00110001=31H=49 10111110+01010010=00010000=10H=16 10011010+11110001=10001011=8BH= -117 4 用压缩BCD码计算（结果用二进制、BCD码、十进制表示） 29+53，73－24，66+18 ，132+75 查看正确答案 00101001+01010011=01111100+00000110=10000010=82H=82 01110011－00100100=01001111－00000110=01001001=49H=49 01100110+00011000=01111110+00000110=10000100=84H=84 0000000100110010+01110101=0000000110100111+00000110 =0000001000000111=0207H=207 5 符号位扩展（字节扩展为字，字扩展为双字） 20A3H，94H ，3456H ，7FH ，EC00H 查看正确答案 000020A3H，FF94H，00003456H，007FH，FFFFEC00H 6 若机器字长为16位，其无符号数表示范围是多少？带符号数表示范围是多少？分别用十进制和十六进制表示。查看正确答案无符号数：0～65535，0000H～FFFFH；带符号数：-32768～+32767，8000H～7FFFH 7 写出下列十六进制数所能代表的数值或编码：（1）38H （2）FFH （3）5AH （4）0DH

汇编语言课后习题答案___郑晓薇(整理后的)

习题一 1 无符号数: 11010011=211=D3H，01110111=119=77H， 10000011=131=83H， 00101111=47=2FH， 10101010=170=AAH 带符号数：11010011= -45=D3H，01110111=+119=77H，10000011= -125=83H， 00101111=+47=2FH，10101010= -86=AAH 2 5E8AH，0BE6H，3DEH，4940H 3 00011101+00110101=01010010=5 2H=82 01001001+11101000=00110001=3 1H=49 10111110+01010010=00010000=1 0H=16 10011010+11110001=10001011=8 BH= -117 4 00101001+01010011=01111100+0 0000110=10000010=82H=82 01110011－00100100=01001111－00000110=01001001=49H=49 01100110+00011000=01111110+0 0000110=10000100=84H=84 0000000100110010+01110101=00 00000110100111+00000110 =0000001000000111=0207H=207 5 000020A3H，FF94H，00003456H，007FH，FFFFEC00H 6 无符号数：0～65535，0000H～FFFFH；带符号数：-32768～+32767，8000H～7FFFH 7 (1) 38H等于十进制数56，是十进制数56的补码，数字8的ASCII码，十进制数38的压缩BCD码 (2) FFH等于十进制数-1，是带符号数-1的补码，无符号数为255 (3) 5AH等于十进制数90，是十进制数90的补码，大写字母Z的ASCII码 (4) 0DH等于十进制数13，是十进制数13的补码，回车符的ASCII码 8 (1) 108=01101100B=6CH，补码01101100B，压缩BCD码0000000100001000，ASCII码313038H (2) 46=00101110B=2EH，补码00101110B，压缩BCD码01000110，ASCII码3436H (3) –15=11110001B=F1H，补码11110001B，ASCII码2D3135H (4) 254=0000000011111110B=00FEH，补码0000000011111110B，压缩BCD 码0000001001010100，ASCII码323534H 9 (1) 56+63=01110111B，CF=0，SF=0，ZF=0，OF=0 (2) 83-45=00100110B，CF=0，SF=0，ZF=0，OF=0 (3) -74+29=11010011B，CF=0， SF=1，ZF=0，OF=0 (4) -92－37=01111111B，CF=1， SF=0，ZF=0，OF=1 10 回车、换行、响铃、ESC键、空格键、@、P、p 习题二 9 最少划分为16个逻辑段，最多划分为65536个逻辑段（每段16个字节） 10 CS:IP，DS:BX、SI、DI，ES:BX、SI、DI，SS:SP、BP 11 字节单元保存8位数，字单元保存16位数。根据源操作数的属性确定要访问的是字节单元还是字单元。 12 对于字节单元来说，偶地址和奇地址一样；对于字单元而言，最好用偶地址保存，可减少CPU的访存次数。 13 首单元的物理地址=38AE8H，末单元的物理地址 =38AE8H+7FFFH=40AE7H

英语语言学第1-3章课后练习题答案

《新编简明英语语言学教程》第二版第1-3章练习题参考答案 Chapter 1 Introduction P13 1. How do you interpret the following definition of linguistics: Linguistics is the scientific study of language？答：Linguistics is based on the systematic investigation of linguistic data, conducted with reference to some general theory of language structure. In order to discover the nature and rules of the underlying language system, the linguists has to collect and observe language facts first, which are found to display some similarities, and generalizations are made about them; then he formulates some hypotheses about the language structure. The hypotheses thus formed have to be checked repeatedly against the observed facts to fully prove their validity. In linguistics, as in any other discipline, data and theory stand in a dialectical complementation, that is, a theory without the support of data can hardly claim validity, and data without being explained by some theory remain a muddled mass of things. 2. What are the major branches of linguistics What does each of them study 答：The major branches of linguistics are: (1) phonetics: it studies the sounds used in linguistic communication; (2) phonology: it studies how sounds are put together and used to convey meaning in communication; (3) morphology: it studies the way in which linguistic symbols representing sounds are arranged and combined to form words; (4) syntax: it studies the rules which govern how words are combined to form grammatically permissible sentences in languages; (5) semantics: it studies meaning conveyed by language; (6) pragmatics: it studies the meaning in the context of language use. 3. In what basic ways does modern linguistics differ from traditional grammar 答：The general approach thus traditionally formed to the study of language over the years is roughly referred to as “t raditional gramma r.” Modern linguistics differs from traditional g rammar in several basic ways. Firstly, linguistics is descriptive while traditional grammar is prescriptive. Second, modem linguistics regards the spoken language as primary, not the written. Traditional grammarians, on the other hand, tended to emphasize, maybe over-emphasize, the importance of the written word, partly because of its permanence. Then, modem linguistics differs from traditional grammar also in that it does not force languages into a Latin-based framework. 4. Is modern linguistics mainly synchronic or diachronic Why 答：In modem linguistics, a synchronic approach seems to enjoy priority over a diachronic one. Because people believed that unless the various states of a language in different historical periods are successfully studied, it would be difficult to describe the changes that have taken place in its historical development. 5. For what reasons does modern linguistics give priority to speech rather than to writing 答：Speech and writing are the two major media of linguistic communication. Modem linguistics regards the spoken language as the natural or the primary medium of human language for some obvious reasons. From the point of view of linguistic evolution, speech is prior to writing. The writing system of any langu age is always “invented” by its users to record spee ch when the need arises. Even in today's world there are still many languages that can only be spoken but not written. Then in everyday communication, speech plays a greater role than writing in terms of the amount of information conveyed. And also, speech is always the way in which every native speaker acquires his mother tongue, and writing is learned and taught later when he goes to school. For modern linguists, spoken language reveals many true features of human speech while written language is only the “revised” record of spe ech. Thus their data for investigation and analysis are mostly drawn from everyday speech, which they regard as authentic.

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