王爽《匯編語言》第二版習(xí)題答案

《匯編語言》第二版習(xí)題答

案

目錄

1.前言..........................................................................7

2.第1章基礎(chǔ)知識(shí)..............................................................8

2.1檢測(cè)點(diǎn)1.1.....................................................................................................................8

3.第2章寄存器................................................................9

3.1檢測(cè)點(diǎn)2.1(1)................................................................................................................9

3.2檢測(cè)點(diǎn)2.1(2)..............................................................................................................11

3.3檢測(cè)點(diǎn)2.2(1).............................................................................................................12

3.4檢測(cè)點(diǎn)2.2(2)..............................................................................................................13

3.5檢測(cè)點(diǎn)2.3...................................................................................................................14

3.6實(shí)驗(yàn)一(1)....................................................................................................................14

3.7實(shí)驗(yàn)一(2)....................................................................................................................17

3.8實(shí)驗(yàn)一(3)....................................................................................................................18

4.第3章寄存器(內(nèi)存訪問)...................................................19

4.1檢測(cè)點(diǎn)3.1(1)..............................................................................................................19

4.2檢測(cè)點(diǎn)3.1(2).............................................................................................................21

4.3檢測(cè)點(diǎn)3.2(1).............................................................................................................22

4.4檢測(cè)點(diǎn)3.2(2).............................................................................................................23

4.5實(shí)驗(yàn)2(1).....................................................................................................................23

4.6實(shí)驗(yàn)2(2).....................................................................................................................26

5.第4章第一個(gè)程序...........................................................27

1

4X編語d第二案

5.1實(shí)驗(yàn)3(1).....................................................................................................................27

5.2實(shí)驗(yàn)3(2).....................................................................................................................30

5.3實(shí)驗(yàn)3(3).....................................................................................................................32

6.第5章［BX］和loop指令......................................................32

6.1實(shí)驗(yàn)4(1).....................................................................................................................32

6.2實(shí)驗(yàn)4(2).....................................................................................................................34

6.3實(shí)驗(yàn)4(3).....................................................................................................................35

7.第6章包含多個(gè)段的程序....................................................38

7.1檢測(cè)點(diǎn)6.1(1)..............................................................................................................38

7.2檢測(cè)點(diǎn)6.1(2).............................................................................................................39

7.3實(shí)驗(yàn)5(1).....................................................................................................................40

7.4實(shí)驗(yàn)5(2).....................................................................................................................42

7.5實(shí)驗(yàn)5(3).....................................................................................................................43

7.6實(shí)驗(yàn)5(4).....................................................................................................................45

7.7實(shí)驗(yàn)5(5).....................................................................................................................45

7.8實(shí)驗(yàn)5(6).....................................................................................................................48

8.第7章更靈活的定位內(nèi)存地址的方法..........................................49

8.1實(shí)驗(yàn)6(1).....................................................................................................................49

8.2實(shí)驗(yàn)6(2).....................................................................................................................49

9.第8章數(shù)據(jù)處理的兩個(gè)基本問題...............................................50

9.1實(shí)驗(yàn)7..........................................................................................................................50

10.第9章轉(zhuǎn)移指令的原理......................................................54

2

10.1檢測(cè)點(diǎn)9.1(1)..........................................................................................................54

10.2檢測(cè)點(diǎn)9.1(2).........................................................................................................55

10.3檢測(cè)點(diǎn)9.1(3).........................................................................................................57

10.4檢測(cè)點(diǎn)9.2.................................................................................................................60

10.5檢測(cè)點(diǎn)9.3.................................................................................................................61

10.6實(shí)驗(yàn)8.......................................................................................................................62

10.7實(shí)驗(yàn)9.......................................................................................................................63

11.第10章CALL和RET指令..................................................67

11.1檢測(cè)10.1...................................................................................................................67

11.2檢測(cè)點(diǎn)10.2..............................................................................................................68

11.3檢測(cè)點(diǎn)10.3..............................................................................................................69

11.4檢測(cè)點(diǎn)10.4..............................................................................................................70

11.5檢測(cè)點(diǎn)10.5(1)........................................................................................................70

11.6檢測(cè)點(diǎn)10.5(2)........................................................................................................72

11.7課程設(shè)計(jì)一.............................................................74

11.8實(shí)驗(yàn)10⑴.................................................................................................................79

11.9實(shí)驗(yàn)10⑵.................................................................................................................81

11.10實(shí)驗(yàn)10⑶..............................................................................................................84

11.11實(shí)驗(yàn)10(4)...............................................................................................................84

12.第11章標(biāo)志寄存器.........................................................88

12.1檢測(cè)點(diǎn)11.1...............................................................................................................88

12.2檢測(cè)點(diǎn)11.2..............................................................................................................89

3

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12.3檢測(cè)點(diǎn)11.3(1).........................................................................................................91

12.4檢測(cè)點(diǎn)11.3(2).........................................................................................................92

12.5檢測(cè)點(diǎn)11.4...............................................................................................................93

12.6實(shí)驗(yàn)11.....................................................................................................................95

13.第12章內(nèi)中斷.............................................................98

13.1檢測(cè)點(diǎn)12.1(1).........................................................................................................98

13.2檢測(cè)點(diǎn)12.1(2).........................................................................................................98

13.3實(shí)驗(yàn)12.....................................................................................................................99

14.第13章int指令...........................................................100

14.1檢測(cè)點(diǎn)13.1(1)........................................................................................................100

14.2檢測(cè)點(diǎn)13.1(2).......................................................................................................101

14.3檢測(cè)點(diǎn)13.2(1).......................................................................................................103

14.4檢測(cè)點(diǎn)13.2(2).......................................................................................................103

14.5實(shí)驗(yàn)13⑴...............................................................................................................103

14.6實(shí)驗(yàn)13⑵...............................................................................................................105

14.7實(shí)驗(yàn)13⑶...............................................................................................................106

15.第14章端口..............................................................108

15.1檢測(cè)點(diǎn)14.1(1)........................................................................................................108

15.2檢測(cè)點(diǎn)14.1(2).......................................................................................................108

15.3檢測(cè)點(diǎn)14.2.............................................................................................................109

15.4實(shí)驗(yàn)14...................................................................................................................110

16.第15章外中斷............................................................114

4

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16.1檢測(cè)點(diǎn)15.1⑴.....................................................................................................114

16.2檢測(cè)點(diǎn)15.1(2).....................................................................................................114

16.3實(shí)驗(yàn)15.................................................................................................................118

17.第16章直接定址表........................................................120

17.1檢測(cè)點(diǎn)16.1.............................................................................................................120

17.2檢測(cè)點(diǎn)16.2............................................................................................................122

17.3實(shí)驗(yàn)16...................................................................................................................123

18.第17章使用BIOS進(jìn)行鍵盤輸入和磁盤讀寫.................................131

18.117.3字符串的輸入......................................................131

18.2檢測(cè)點(diǎn)17.1............................................................................................................135

18.3實(shí)驗(yàn)17...................................................................................................................136

18.4課程設(shè)計(jì)2..............................................................................................................138

19.第18章綜合研究..........................................................153

19.1研究試驗(yàn)1搭建一個(gè)精簡(jiǎn)的C語言開發(fā)環(huán)境..............................153

19.2研究實(shí)驗(yàn)2使用寄存器.................................................153

19.3研究試驗(yàn)3使用內(nèi)存空間...............................................159

20.部分章節(jié)筆記..............................................................165

20.1第11章標(biāo)志寄存器....................................................165

20.2第12章內(nèi)中斷.......................................................169

20.3第14章端口..........................................................171

20.4第15章外中斷.......................................................173

20.5第16章直接定址表...................................................176

5

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20.6第17章使用BIOS進(jìn)行鍵盤輸入和磁盤讀寫.............................177

21.附注........................................................................181

21.1《匯編語言（第2版）》勘誤...............................................181

21.2匯編測(cè)試題目（部分）....................................................183

21.3匯編測(cè)試答案（部分）....................................................183

6

4X編語d第二累

1.前言

教材：《匯編語言》（2008年4月第2版），王爽著，清華大學(xué)出版社。

長期以來，匯編語言被認(rèn)為是一門枯燥難學(xué)的課程，但王爽老師的著作《匯編語言》解決了這個(gè)問題。

本人僅將自己在學(xué)習(xí)此書時(shí)所做的答案及一些個(gè)人筆記進(jìn)行了整理并編輯成集，方便日后查閱。

后因朋友要求，發(fā)于網(wǎng)上共享。

由于本人水平有限，制作倉促，不能保證解析完全正確。

如果你在對(duì)照的過程中，發(fā)現(xiàn)了錯(cuò)誤的地方，可以寫信告知本人，在此先表示感謝。

郵箱:gray_always@qq.com;

QQ群:100354217（群主不友善,慎入）

如有其它疑問，可到專門為《匯編語言》讀者開設(shè)的論壇中尋求幫助。

網(wǎng)站地址：http:〃www.asmedu.net/bbs/forum.jsp

制作者：灰色依然

制作日期：2008年7月

如需轉(zhuǎn)載，請(qǐng)注明來源和出處，謝謝。

7

4X編語d第二案

2.第1章基礎(chǔ)知識(shí)

2.1檢測(cè)點(diǎn)1.1

檢測(cè)點(diǎn)1.1

(1)1個(gè)CPU的尋址能力為8KB,那么它的地址總線的寬度為」工位。

(2)1KB的存儲(chǔ)器有1024個(gè)存儲(chǔ)單元,存儲(chǔ)單元的編號(hào)從0到1023。

(3)1KB的存儲(chǔ)器可以存儲(chǔ)可92(2-3)個(gè)bit,1024個(gè)Byte。

(4)1GB是1073741824(2'30)個(gè)Byte、1MB是1048576(2.20)個(gè)Byte、1KB是1024(2~10)個(gè)Byte。

(5)8080、8088、80296、80386的地址總線寬度分別為16根、20根、24根、32根，則它們的尋址能力分

別為：且.(KBXJ_(MB\_16_(MB]4(GB1

(6)8080、8088、8086、80286、80386的數(shù)據(jù)總線寬度分別為8根、8根、16根、16根、32根。則它們一

次可以傳送的數(shù)據(jù)為\_1_(BX_2_(BX_2_(BX

(7)從內(nèi)存中讀取1024字節(jié)的數(shù)據(jù)，8086至少要讀512次,80386至少要讀256次。

(8)在存儲(chǔ)器中，數(shù)據(jù)和程序以二進(jìn)制形式存放。

解題過程：

(1)1KB=1O24B,8KB=1024B*8=2'N,N=13o

(2)存儲(chǔ)器的容量是以字節(jié)為最小單位來計(jì)算的，1KB=1O24B。

(3)8Bit=lByte,1024Byte=lKB(lKB=1024B=1024B*8Bit卜

(4)1GB=1O73741824B(即2-30)1MB=1O48576B(即2'20)1KB=1O24B(即2"10\

(5)一個(gè)CPU有N根地址線，則可以說這個(gè)CPU的地址總線的寬度為No這樣的CPI最多可以尋找2的N次

方個(gè)內(nèi)存單元。(一個(gè)內(nèi)存單元=lByteb

8

4X編語d第二案

(6)8根數(shù)據(jù)總線一次可以傳送8位二進(jìn)制數(shù)據(jù)(即一個(gè)字節(jié)X

(7)8086的數(shù)據(jù)總線寬度為16根(即一次傳送的數(shù)據(jù)為2B)1024B/2B=512,同理1024B/4B=256。

(8)在存儲(chǔ)器中指令和數(shù)據(jù)沒有任何區(qū)別，都是二進(jìn)制信息。

3.第2章寄存器

3.1檢測(cè)點(diǎn)2.1(1)

檢測(cè)點(diǎn)2.1

(1)寫出每條匯編指令執(zhí)行后相關(guān)寄存器中的值。

movax,62627AX=F4A3H

movah,31HAX=31A3H

moval,23HAX=3123H

addax,axAX=6246H

movbx,826CHBX=826CH

movex,axCX=6246H

movax,bxAX=826cH

addax,bxAX=04D8H

moval,bhAX=0482H

movah,blAX=6C82H

addah,ahAX=D882H

addal,6AX=D888H

addal,alAX=D810H

movax,exAX=6246H

Microsoft(R)WindowsDOS

(C)CopyrightMicrosoftCorp1990-2001.

C:\D0CUME"1\ADMINI"1>debug

-a

0ClC:0100movax,f4a3

OC1C:O1O3movah,31

OC1C:O1O5moval,23

OC1C:O1O7addax,ax

OC1C:O1O9movbx,826c

OC1C:O1OCmovex,ax

OC1C:O1OEmovax,bx

OC1C:O11Oaddax,bx

9

OC1C:O112moval,bh

OC1C:O114movah,bl

OC1C:O116addah,ah

OC1C:O118addal,6

OC1C:O11Aaddal,al

OC1C:O11Cmovax,ex

OC1C:O11E

AX=OOOOBX=OOOOCX=OOOODX=OOOOSP:FFEEBP=OOOOSI=0000DI=OOOO

DS=OC1CES=OC1CSS=OC1CCS=OC1C1P=O1OONVUPEIPLNZNAPONC

OC1C:O1OOB8A3F4MOVAX,F4A3

-t

AX=F4A3BX=OOOOCX=OOOODX=OOOOSP二FFEEBP=OOOOSI=OOOODI=OOOO

DS=OCICES=OC1CSS=OC1CCS=OC1CIP=0103NVUPEIPLNZNAPONC

OC1C:O1O3B431MOVAH,31

-t

AX=31A3BX=OOOOCX=OOOODX=OOOOSP=FFEEBP=OOOOSI=0000DI=OOOO

DS=0C1CES=OC1CSS=OC1CCS=0C1CIP=0105NVUPEIPLNZNAPONC

OC1C:O1O5B023MOVAL,23

-t

AX=3123BX=OOOOCX=OOOODX=OOOOSP=FFEEBP=OOOOSI=OOOODI=OOOO

DS=OC1CES=OC1CSS=OC1CCS=OC1CIP=0107NVUPEIPLNZNAPONC

OC1C:O1O701C0ADDAX,AX

-t

AX=6246BX=OOOOCX=OOOODX=OOOOSP二FFEEBP=OOOOSI=0000DI=OOOO

DS=OC1CES=OC1CSS=OC1CCS=OC1C1P=O1O9NVUPEIPLNZNAPONC

OC1C:O1O9BB6C82MOVBX,826C

-t

AX=6246BX=826CCX=OOOODX=OOOOSP:FFEEBP=OOOOSI=OOOODI=OOOO

DS=OCICES=OC1CSS=OC1CCS=OC1CIP=010CNVUPEIPLNZNAPONC

OC1C:O1OC89C1MOVCX,AX

-t

AX=6246BX=826CCX=6246DX=OOOOSP:FFEEBP=OOOOS1=0000DI=OOOO

DS=OCICES=OC1CSS=OC1CCS=OCICII^OIOENVUPEIPLNZNAPONC

OC1C:O1OE89D8MOVAX,BX

-t

AX=826CBX=826CCX=6246DX=OOOOSP=FFEEBP=OOOOSI-0000DI=OOOO

DS=0C1CES=OC1CSS=OC1CCS=OC1CIP=0110NVUPEIPLNZNAPONC

OC1C:O11O01D8ADDAX,BX

-t

AX=04D8BX=826CCX=6246DX=OOOOSP:FFEEBP=OOOOSI=OOOODI=OOOO

DS=OC1CES=OC1CSS=OC1CCS=OC1C1P=O112OVUPEIPLNZACPECY

OC1C:O11288F8MOVAL,BH

-t

10

AX=0482BX=826CCX=6246DX=OOOOSP二FFEEBP=OOOOS1=0000DI=OOOO

DS=OCICES=OC1CSS=OC1CCS=OC1C1P=O114OVUPEIPLNZACPECY

OC1C:O11488DCMOVAH,BL

AX=6C82BX=826CCX=6246DX=OOOOSP二FFEEBP=OOOOSI=0000DI=OOOO

DS=OC1CES=OCICSS=OC1CCS=OC1CIP=0116OVUPEIPLNZACPECY

OC1C:O11600E4ADDAH,AH

AX=D882BX=826CCX=6246DX=OOOOSP二FFEEBP=OOOOSI=0000DI=OOOO

DS=0C1CES=OC1CSS=0C1CCS=OC1CIP=0U8OVUPEINGNZACPENC

OC1C:O1180406ADDAL,06

AX=D888BX=826CCX=6246DX=OOOOSP:FFEEBP=OOOOSI=OOOODI=OOOO

DS=OCICES=0C1CSS=0C1CCS=OCICIP=0UANVUPEINGNZNAPENC

OC1C:O11AOOCOADDAL,AL

AX=D810BX=826CCX=6246DX=OOOOSP=FFEEBP=OOOOSI=0000DI=OOOO

DS=0C1CES=OC1CSS=OC1CCS=0C1CIP=01ICOVUPEIPLNZACPOCY

OC1C:O11C89C8MOVAX,CX

-t

AX=6246BX=826CCX=6246DX=OOOOSP二FFEEBP=OOOOSI=0000DI=OOOO

DS=OC1CES=OC1CSS=OC1CCS=OC1CIP=011EOVUPEIPLNZACPOCY

OC1C:O11EOBOCORCX,[SI]DS:0000=20CD

3.2檢測(cè)點(diǎn)2.1(2)

檢測(cè)點(diǎn)2.1

(2)只能使用目前學(xué)過的匯編指令，最多使用4條指令，編程計(jì)算2的4次方。

movax,2AX=2

addax,axAX=4

addax,axAX=8

addax,axAX=16

Microsoft(R)WindowsDOS

(C)CopyrightMicrosoftCorp1990-2001.

C:\DOCUME~l\ADMINI~Ddebug

-a

OC1C:O1OOmovax,2

OC1C:O1O3addax,ax

11

OC1C:O1O5addax,ax

0C1C:0107addax,ax

OC1C:O1O9

AX=OOOOBX=OOOOCX=OOOODX=OOOOSP:FFEEBP=OOOOSI=0000DI=0000

DS=OC1CES=OCICSS=OC1CCS=OC1CIP=0100NVUPEIPLNZNAPONC

OC1C:O1OOB80200MOVAX,0002

AX=0002BX=OOOOCX=OOOODX=OOOOSP二FFEEBP=OOOOSI=0000DI=OOOO

DS=0C1CES=OC1CSS=0C1CCS=OC1CIP=0103NVUPEIPLNZNAPONC

OC1C:O1O301C0ADDAX,AX

AX=0004BX=OOOOCX=OOOODX=OOOOSP:FFEEBP=OOOOSI=0000DI=OOOO

DS=OCICES=0C1CSS=0C1CCS=OCICIP=0105NVUPEIPLNZNAPONC

OC1C:O1O501C0ADDAX,AX

AX=0008BX=OOOOCX=OOOODX=OOOOSP二FFEEBP=OOOOS1=0000DI=OOOO

DS=OCICES=OC1CSS=OC1CCS=OCICIP=0107NVUPEIPLNZNAPONC

OC1C:O1O701C0ADDAX,AX

AX=OOIOBX=OOOOCX=OOOODX=OOOOSP:FFEEBP=OOOOSI=0000DI=OOOO

DS=OC1CES=OCICSS=OC1CCS=OC1CIP=0109NVUPEIPLNZACPONC

OC1C:O1O920881615AND[BX+SI+1516],CLDS:1516-00

3.3檢測(cè)點(diǎn)2.2(1)

檢測(cè)點(diǎn)2.2

(1)給定段地址為0001H,僅通過變化偏移地址尋址，CPU的尋址范圍為0010H到1000FII。

解題過程：

物理地址=5人*16+以

EA的變化范圍為Oh'ffffh

物理地址范圍為(SA*16+0h)~(SA*16+ffffh)

現(xiàn)在SA=0001h,那么尋址范圍為

(0001h*16+0h)~(0001h*16+ffffh)

=0010h'1000fh

12

4X編語言匚第二案

3.4檢測(cè)點(diǎn)2.2(2)

檢測(cè)點(diǎn)2.2

(2)有一數(shù)據(jù)存放在內(nèi)存20000H單元中，現(xiàn)給定段地址為SA,若想用偏移地址尋到此單元。則SA應(yīng)滿足的

條件是：最小為100UI,最大為200011。

當(dāng)段地址給定為1001H以下和2000H以上，CPU無論怎么變化偏移地址都無法尋到20000H單元。

解題過程：

物理地址=5人*16+￡人

20000h=SA*16+EA

SA=(20000h-EA)/16=2000h-EA/l6

EA取最大值時(shí),SA=2000h-ffffh/16=1001h,SA為最小值

EA取最小值時(shí),SA=2000h-0h/16=2000h,SA為最大值

這里的ffffH/16=fffh是通過WIN自帶計(jì)算器算的

按位移來算確實(shí)應(yīng)該為fff.fh,這里小數(shù)點(diǎn)后的f應(yīng)該是省略了

單就除法來說，應(yīng)有商和余數(shù)，但此題要求的是地址最大和最小，所以余數(shù)忽略了

如果根據(jù)位移的算法(段地址*16=16進(jìn)制左移一位)，小數(shù)點(diǎn)后應(yīng)該是不能省略的

我們可以反過來再思考下，如果SA為1000h的話，小數(shù)點(diǎn)后省略

SA=1000h,EA取最大ffffh,物理地址為Iffffh,將無法尋到20000H單元

這道題不應(yīng)看成是單純的計(jì)算題

13

4X編語d第二案

3.5檢測(cè)點(diǎn)2.3

檢測(cè)點(diǎn)2.3

下面的3條指令執(zhí)行后，cpu幾次修改IP?都是在什么時(shí)候？最后IP中的值是多少？

movax,bx

subax,ax

jmpax

答：一共修改四次

第一次：讀取movax,bx之后

第二次：讀取subax,ax之后

第三次：讀取.jmpax之后

第四次：執(zhí)行jmpax修改IP

最后IP的值為0000H，因?yàn)樽詈骯x中的值為0000H,所以IP中的值也為0000H

3.6實(shí)驗(yàn)一(1)

實(shí)驗(yàn)一查看CPU和內(nèi)存，用機(jī)器指令和匯編指令編程

2實(shí)驗(yàn)任務(wù)

(1)使用Debug,將下面的程序段寫入內(nèi)存，逐條執(zhí)行，觀察每條指令執(zhí)行后，CPU中相關(guān)寄存器中內(nèi)容的變

化。

機(jī)器碼匯編指令寄存器

b8204emovax,4E20Hax=4E20H

051614addax,1416Hax=6236H

bb0020movBX,2000Hbx=2000H

01d8addax,bxax=8236H

89c3movbx,axbx=8236H

01d8addax,bxax=046CH

b8la00movax,001AHax=001AH

bb2600movbx,0026Hbx=0026H

14

4X編語d第二案

00d8addal,blax=0040H

00deaddah,blax=2640H

00c7addbh,albx=4026H

b400movah,0ax=0040H

00d8addal,blax=0066H

049caddal,9CHax=0002H

Microsoft(R)WindowsDOS

(C)CopyrightMicrosoftCorp1990-2001.

C:\DOCUME"l\ADMINri>debug

-a

0C1C:0100movax,4e20

OC1C:O1O3addax,1416

OC1C:O1O6movbx,2000

OC1C:O1O9addax,bx

0C1C:010Bmovbx,ax

OC1C:O1ODaddax,bx

0C1C:010Fmovax,001a

OC1C:O112movbx,0026

OC1C:O115addal,bl

OC1C:O117addah,bl

OC1C:O119addbh,al

OC1C:O11Bmovah,0

OC1C:O11Daddal,bl

OC1C:O11Faddal,9c

OC1C:O121

-r

AX=0000BX=000()CX=0000DX=OOOOSP=FFEEBP=OOOOSI-0000DI=0000

DS=OC1CES=OC1CSS=OC1CCS=OC1CIP=0100NVUPEIPLNZ,PONC

OC1C:O1OOB8204EMOVAX,4E20

-t

AX-4E20BX=OOOOCX=00001)X=0000SP=FFEEBP=OOOOSI-0000DI=0000

DS=OC1CES=OC1CSS=OC1CCS=OC1CIP=0103NVUPEIPLNZ,PONC

OC1C:O1O3051614ADI)AX,1416

-t

AX=6236BX=000()CX=OOOODX-0000SP=FFEEBP=0000SI=00001)1=0000

DSRC1CES=OC1CSS=OC1CCS=OC1CIP=0106NVUPEIPLNZ,PENC

OC1C:O1O6BB0020MOVBX,2000

AX=6236BX=2000CX=0000DX-0000SP=FFEEBP=0000SI-0000[)1=0000

DS=OCICES=OC1CSS=OC1CCS=OC1CIP=0109NVUPEIPLNZ?PENC

OC1C:O1O901D8ADDAX,BX

AX=8236BX=2000CX=0000DX=0000SP=FFEEBP-0000SI-00001)1=0000

15

《匯編語言》第二版習(xí)題答案

DS=OC1CES=OC1CSS=OC1CCS=OCICIP=010BOVUPEINGNZPENC

OC1C:O1OB89C3MOVBX,AX

AX=8236BX=8236CX-00001)X=0000SP=FFEEBP=0000SI=00001)1=0000

DSRC1CES=OC1CSS=OC1CCS=OC1CIP=010DOVUPEINGNZ,PENC

OC1C:O1OD01D8ADI)

AX=046cBX=8236CX=0000DX-0000SP=FFEEBP=0000SI-0000[)1=0000

DS=OCICES=OC1CSSRC1CCS=OC1CIP=010FOVUPEIPLNZ?PECY

OC1C:O1OFB81A00MOVAX,001A

AX-001ABX=8236CX=0000DX=0000SP=FFEEBP-0000SI-0000DI=0000

DS=OC1CES=OC1C