test10

一、简答题(16分，每小题4分，共4题)

1. 在调试程序DEBUG提示符“-”下，输入命令R，显示结果为：

AX=1234 BX=0100 CX=1357 DX=2468 SP=0110 BP=0120 SI=0002 DI=0130

DS=1000 ES=1010 SS=1100 CS=2000 IP=0120 NV UP EI PL NZ NA PO NC 输入d 1000:0，显示结果为：

1000:0000 40 27 AA C3 E8 AD FE 3C-0A 75 E0 C3 BB 67 48 E8 @'.....<.u...gH.

1000:0010 E1 03 E8 AD 03 B0 2C AA-E8 99 FE 98 8B D0 8A E0 ......,.........

1000:0020 B0 2B 0A E4 79 04 B0 2D-F6 DC AA 8A C4 EB BC E8 .+..y..-........

1000:0030 48 00 B0 2C AA A0 A8 56-BE 84 3C 80 3E A6 56 01 H..,...V..<.>.V.

输入d 1000:100，显示结果为：

1000:0100 75 03 BE 94 3C 98 03 F0-03 F0 A5 C3 D0 E8 D0 E8 u...<...........

1000:0110 D0 E8 24 03 BE A4 3C EB-EC 24 07 EB E5 C6 06 A6 ..$...<..$......

1000:0120 56 01 E8 15 00 B0 2C AA-A0 A8 56 EB E5 E8 0D FF V.....,...V.....

1000:0130 E8 DF FF C6 06 A6 56 01-EB 10 E8 00 FF EB 0E C6 ......V.........

输入d 1100:0，显示结果为：

1100:0000 AA E8 D8 FE EB EE 8B 1E-0B 4A B8 42 58 EB BC 03 .........J.BX...

1100:0010 1E 15 4A B8 53 49 EB CD-E8 99 FD 98 03 06 83 56 ..J.SI.........V

1100:0020 92 8A C6 E8 C5 FE 8A C2-E9 C0 FE E8 86 FD 8A D0 ................

1100:0030 E8 81 FD 8A F0 03 16 83-56 EB E6 24 07 E8 02 FF ........V..$....

输入d 1100:100，显示结果为：

1100:0100 B0 2C A3 32 C0 E9 FA FE-32 C0 E8 EB FE B0 2C A5 .,.2....2.....,.

1100:0110 B7 5B AA 33 DB C6 06 AA-56 03 EB A5 E8 F1 FF B0 .[.3....V.......

1100:0120 2C AA 32 C0 E9 D1 FE C6-06 A6 56 00 EB 05 C6 06 ,.2.......V.....

1100:0130 A6 56 01 24 07 E9 39 FE-C6 05 33 47 C3 E8 A3 01 .V.$..9...3G....

试说明下列各指令执行完后AX寄存器的内容。

指令 AX的内容（十六进制）

① POP AX ①

② MOV AX,BX ②

③ MOV AX,[BP][SI] ③

④ MOV AX,[DI] ④

⑤ MOV AX,[BX+10H] ⑤

⑥ MOV AX,[BX][SI] ⑥

⑦ MOV AX,[BX+SI+10H] ⑦

⑧ MOV AX,[BP] ⑧

2. 下面程序完成什么功能？（对寄存器AX、DX的值发生何种变化）

MOV CX,5

NEXT: SHL AX,1

RCL DX,1

ADC AX,0

LOOP NEXT

3. 数据段有变量定义语句：

NUM DB 130

现要计算NUM÷7，并将商保存在ANS中。有学生编写下面程序段，请仔细阅读，如果发现有错误，将其改正过来，如果认为没有错误，请注明“无错误”。（假设段寄存器DS存储数据段的段地址）

MOV AX,NUM

DIV 7

MOV ANS,AL

4. 不使用串指令，写出实现下列程序段相同功能的程序段。

MOV AX,DATA

MOV DS,AX

MOV ES,AX

LEA SI,X

LEA DI,Y

MOV CX,100H

CLD

REPE CMPSB

JCXZ L2

L3: ┆

二、程序填空题（共16分，每空2分，共8个空）

下面程序输入16位以内的二进制数（输入时回车作为输入结束），然后以有符号十进制数形式显示出来。将空格处指令补充完整，使得程序能完整实现所需要的功能。

CODE SEGMENT

ORG 100H

ASSUME CS:CODE,DS:CODE,ES:CODE,SS:CODE

MAIN PROC NEAR

CALL I2

CALL D10

( )……………①

INT 21H

MAIN ENDP

I2 PROC NEAR

PUSH BX

PUSH CX

PUSH DX

PUSH SI

PUSH DI

XOR BX,BX

MOV CX,16

L20: MOV AH,1

INT 21H

( )……………②

JE L21

CMP AL,30H

JB L102

CMP AL,31H

AND AX,01H

SHL BX,1

ADD BX,AX

( )……………③L21: MOV AX,BX

POP DI

POP SI

POP DX

POP CX

( )……………④

RET

I2 ENDP

D10 PROC NEAR

PUSH AX

PUSH BX

PUSH CX

PUSH DX

PUSH SI

PUSH DI

( )……………⑤

MOV AH,9

LEA DX,CRLF

INT 21H

OR BX,BX

JNS L100

MOV AH,2

( )……………⑥

INT 21H

NEG BX

L100: MOV AX,BX

XOR CX,CX

MOV SI,10

L101: XOR DX,DX

( )……………⑦

PUSH DX

INC CX

CMP AX,0

JNZ L101

L102: POP DX

MOV AH,2

( )……………⑧

INT 21H

LOOP L102

POP DI

POP SI

POP DX

POP CX

POP AX

RET

CRLF DB 0DH,0AH,24H

D10 ENDP

CODE ENDS

END MAIN

三、阅读程序（共23分，第1、2小题各8分，第3小题7分）

1.仔细阅读下面程序，说明该程序的功能。

DSEG SEGMENT

FILE1 DB "8086ASM.EXE",0

FILE2 DB "MASM.EXE",0

HANDLE1 DW ?

HANDLE2 DW ?

BUFFER DB 512 DUP(?)

ERRMSG DB "File Handle Error!",0DH,0AH,24H

DSEG ENDS

SSEG SEGMENT STACK

DW 100H DUP(?)

TOP LABEL WORD

SSEG ENDS

CSEG SEGMENT

ASSUME CS:CSEG, DS:DSEG, SS:SSEG START PROC FAR

MOV AX,SSEG

MOV SS,AX

LEA SP,TOP

MOV AX,DSEG

MOV DS,AX

LEA DX,FILE2

MOV AH,3DH

MOV AL,0

INT 21H

JC ERROR

MOV HANDLE2,AX

LEA DX,FILE1

MOV CX,20H

MOV AH,3CH

INT 21H

JC ERROR

MOV HANDLE1,AX

NEXT: MOV AH,3FH

MOV BX,HANDLE2

MOV CX,512

LEA DX,BUFFER

INT 21H

CMP AX,0

JE CLOSEFILES

MOV CX,AX

MOV AH,40H

MOV BX,HANDLE1

LEA DX,BUFFER

INT 21H

JC ERROR

JMP NEXT

CLOSEFILES:

MOV AH,3EH

MOV BX,HANDLE2

INT 21H

JC ERROR

MOV AH,3EH

MOV BX,HANDLE1

INT 21H

JC ERROR

JMP EXIT

ERROR: MOV AH,9

LEA DX,ERRMSG

INT 21H

EXIT: MOV AH,4CH

INT 21H

START ENDP

CSEG ENDS

END START

2. 仔细阅读下面程序，说明该程序的功能。

CODE SEGMENT

ORG 100H

ASSUME CS:CODE,DS:CODE,ES:CODE,SS:CODE P22 PROC NEAR

MOV CX,N

LEA SI,NUM

LP1: NOT BYTE PTR[SI]

INC SI

LOOP LP1

MOV CX,N

LEA SI,NUM

STC

LP2: ADC BYTE PTR[SI],0

INC SI

LOOP LP2

MOV AH,4CH

INT 21H

DB 67H,89H,0DEH,13H,24H,35H,46H,57H

N EQU $-NUM

P22 ENDP

CODE ENDS

END P22

3.下面是某程序的汇编列表文件，请将程序执行过程中堆栈最满时每个单元的内容填入下表。假设程序执行时PSP（Program Segment Prefix）段地址为13CBH，STACK段的段地址为13DBH，CODE1段的段地址为13DFH，CODE2的段地址为13E1H。(不考虑INT 21H体内对栈的使用)

地址机器码汇编语言指令

0000 STACK SEGMENT STACK

0000 0030[0] DW 30H DUP(0)

0060 TOP LABEL WORD

0060 STACK ENDS

0000 CODE1 SEGMENT

ASSUME CS:CODE1,SS:STACK

0000 P1 PROC FAR

0000 B8 ---- R MOV AX,STACK

0003 8E D0 MOV SS,AX

0005 8D 26 0060 R LEA SP,TOP

0009 B8 0100 MOV AX,0100H

000C 9A 000C ---- R CALL FAR PTR P4

0011 2E: A3 0019 R MOV CS:[ANS],AX

0015 B4 4C MOV AH,4CH

0017 CD 21 INT 21H

0019 ???? ANS DW ?

001B P1 ENDP

001B P2 PROC FAR

001B 05 0020 ADD AX,20H

001E CB RET

001F P2 ENDP

001F CODE1 ENDS

0000 CODE2 SEGMENT

ASSUME CS:CODE2

0000 P3 PROC NEAR

0000 05 0030 ADD AX,30H

0003 9A 001B ---- R CALL FAR PTR P2

0008 05 0003 ADD AX,3H

000B C3 RET

000C P3 ENDP

000C P4 PROC FAR

000C 05 0040 ADD AX,40H

000F E8 0000 R CALL NEAR PTR P3

0015 CB RET

0016 P4 ENDP

0016 CODE2 ENDS

END P1

偏移地址

( )

( )

( )

( )

( )

SP

四、程序设计题（45分，5小题中任选3题，每题15分，写清楚题号，编写超过3题时，必须注明哪3题作为评分题，否则按得分最少的3题作为评分题）

1. 编一个程序，实现从键盘输入一个长度不超过300个字符的串（以回车键作为输入结束），然后在下一行以倒序输出所输入的字符。

2. 编写一个程序，从键盘输入一个0～65535之间的10进制无符号数，然后以16进制形式显示出所输入的数。

3. 编写一个程序，将一个包含有30个字数据的数组M分成两个数组：奇数数组ODD和偶数数组EVEN，并把这两个数组中元素的个数以二进制形式显示出来(不能使用DIV/IDIV指令)。

4. 已知数组A包含20个互不相等的字整数，数组B包含30个互不相等的字整数，试编写一个程序，将既在A数组中又在B数组中出现的数存放在数组C中。

5. 已知在首地址为DATA的字数组中存放一系列有符号数（首元素为数据个数），试编写一个程序求出它们的平均值放在变量AVER中，并求出数组中有多少个数大于该平均值，将大于平均值的元素个数保存在变量COUNT中。（注意，这些数据的累加和可能超出-32768～32767之间）

五、附加题（15分，只有在前四题总分达到60分或以上时，本附加题的分数才生效，且加上附加题后的总分不会超100分）

输入一个数，判断该数否属于数列1、2、4、5、7、9、10、12、14、16、17、19、21、23、25、······，如果属于该数列，输出“YES”，否则输出“NO”。（可能有用的算法：用减奇数次数的方法，求一个数的近似平方根，这个平方根是一个整数。如求17的平方根，可以用17相继减去奇数1、3、5、7、…，当结果为负数或0时停止，即：17-1-3-5-7-9＜0，可以看出，17在减去5次奇数后结果变为负数，可以近似认为17的平方根在4与5之间。25-1-3-5-7-9=0，25的平方根为5）

附录1：部分字符ASCII(16进制数)

字符响铃回车换行空格 $ + - 0 1 2 (9)

ASCII 07 0D 0A 20 24 2B 2D 30 31 32 (39)

字符 A B C … Y Z a b c … y z

ASCII 41 42 43 … 595A 61 62 63 … 79 7A

附录2：部分DOS功能调用参数（功能号为16进制数）

功能号功能描述调用参数返回参数

2 显示输出 DL=输出字符ASCII 无

5 打印机输出 DL=输出字符ASCII 无

9 显示字符串 DS:DX=串首址,以$结无

束字符串

0A 键盘输入到缓冲区 DS:DX=缓冲区首地址 (DS:DX+1)=实际输入的字符个数

首字节保存缓冲区容量 (DS:DX+2)=输入的字符串开始地址

3C 建立文件 DS:DX=文件名ASCIIZ CF=0时，AX=文件HANDLE

CX=文件属性(一般为20H) CF=1时，操作失败

3D 打开文件 DS:DX=文件名ASCIIZ CF=0时，AX=文件HANDLE

AL=方式（0:读，1:写） CF=1时，操作失败

3E 关闭文件 BX=文件HANDLE CF=1时，操作失败

3F 读文件 DS:DX=缓冲区首地址 CF=0时，AX=读取字符个数

BX=文件HANDLE 当AX=0时表示文件结束

CX=读字符个数 CF=1时，操作失败

40 写文件 DS:DX=缓冲区首地址 CF=1时，操作失败

BX=文件HANDLE

CX=写字符个数

42 移动文件指针 DXCX=32位的以移动字节数 CF=0时，DXAX=32位的新指针位置 AL=方式码 CF=1时，操作失败

0:从文件头往后移，正数

1:从当前位置往前后移

2:从文件尾往前移，正数

BX=文件HANDLE

4C 程序结束 AL=返回码

附录3：部分设备文件的的文件代号（HANDLE）

0:标准输入设备（即键盘）

1:标准输出设备（即显示器）

4:标准打印设备（即打印机）

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?经济上，发展中国家的劳动力被剥削，资源被取用，而产品却不能够进人发达国家，国内产业受到外商和外企的冲击；?社会上，外国的产品流人一个国家，会改变人们的生活习惯和购买习惯；年轻人比较喜欢外同产品，比如说食品、衣服、电子产品和娱乐方式； ?文化上，本土文化会被削弱，本土语言会受到威胁，文化多样性会逐步损失； ?环境上，工业化生产和交通量的加大会增加温室气体的排放；游人的增多会对自然环境构成威胁。 练习5 Topic 1: Increasing travels between countries enable people to learn different cultures or to increase tension between people from different countries? 外来人口的增加有利于文化的交流： ?游客或者移民都具备经济利益和价值（commercial interest and economic value)；而外同人的停留时间是和东道国的合作密切相关的（The length of stay depends on the cooperation of the host society.);很多当地人因此表现得非常好客和友好，这有助于文化的交流（Many local people are very friendly and hospitable, which promote the cultural communications.)； ?对其他的文化和人民更为了解（a better understanding of other cultures and other peoples),改变人们对其他民族和文化的态度（alter one's attitudes towards another people or culture);人们因此可以接受不同的文化和价值观（embrace different values and cultures),可以和平共处（create motivation to coexist peacefully), 并促进对"多元文化的理解(promote multi-cultural understanding)； ?促进文化的融合（promote integration),消除文化障碍（remove cultural barriers), 外来人口的增加导致冲突：?违反当地的习俗会激怒当地人（breach of local customs can irritate the locals); ?不同文化的人看待同一事物总有不同的角度（harbour different perceptions),并给予不同的解释(different interpretations)，比如对手势、衣服、言行举止有不同的理解；举止不当会引起当地人意想不到的反应(provoke unanticipated responses); ?外国游客增多会对当地环境造成压力，甚至破坏环境，而移民增多可能导致工作机会减少，致使竞争更加激烈，这些都有可能引起当地人的反感（cause resentment in local communities). Globalization is a catch-all term that refers to any activity that involves more than one country, for example, travel from one country to another. The dramatic increase in transnational travel in recent years has sparked controversy over the potential impacts of this trend on individual countriesf especially those new member states of globalization. Some people are concerned that the upsurge in new arrivals will prompt local hostility against visitors instead of promoting their understanding on mutual cultural background. This notion should be rejected as one can see many facts in favor of this development between countries. The first reason why international travels would never bring conflict is rooted in the fact that both visitors and locals are economically motivated. International travel opens up opportunities for business development throughout the world. Entrepreneurs are interested not only in the domestic market but also in the oversea market. Foreigners should learn the culture of a country before winning over the local people. In turn, locals should show their hospitality to visitors in exchange for their trust. They share a view that acceptance of each other's cultural background is a necessary condition for cooperation.

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It is inevitable that traditional cultures will be lost as technology develops. Technology and traditional cultures are incompatible. To what extent do you agree or disagree with this view? Some people believe that technological developments lead to the loss of traditional cultures. While this may be true in the case of some societies, others seem to be unaffected by technology and the modern world. On the one hand, the advances in technology that have driven industrialisation in developed countries have certainly contributed to the disappearance of traditional ways of life. For example, in pre-industrial Britain, generations of families grew up in the same small village communities. These communities had a strong sense of identity, due to their shared customs and beliefs. However, developments in transport, communications and manufacturing led to the dispersal of families and village communities as people moved to the cities in search of work. Nowadays most British villages are inhabited by commuters, many of whom do not know their closest neighbours. On the other hand, in some parts of the world traditional cultures still thrive. There are tribes in the Amazon Rainforest, for example, that have been completely untouched by the technological developments of the developed world. These tribal communities continue to hunt and gather food from the forest, and traditional skills are passed on to children by parents and elders. Other traditional cultures, such as farming communities in parts of Africa, are embracing communications technologies. Mobile phones give farmers access to information, from weather predictions to market prices, which helps them to prosper and therefore supports their culture. In conclusion, many traditional ways of life have been lost as a result of advances in technology, but other traditional communities have survived and even flourished. (260 words, band 9)