第四章 8086指令系统

第四章

8086指令系统4.1汇编语言程序4.28068寻址方式4.38086指令系统4.4程序设计步骤内容与要求：

1.熟悉Intel8086/8088的寻址方式。

2.掌握Intel8086/8088的指令系统。

3.了解Intel80386扩充与增加的指令。4.1汇编语言程序（1）机器语言(MachineLanguage)

机器语言是一种用二进制表示指令和数据，能被机器直接识别的计算机语言。它的缺点是不直观，不易理解和记忆，因此编写、阅读和修改机器语言程序都比较繁琐。但机器语言程序是计算机惟一能够直接理解和执行的程序，具有执行速度快、占用内存少等特点。（2）汇编语言(AssemblyLanguage)

汇编语言是一种采用助记符表示的程序设计语言，即用助记符来表示指令的操作码和操作数，用标号或符号代表地址、常量或变量。助记符一般都是英文字的缩写，以方便人们书写、阅读和检查。实际上，用汇编语言编写的汇编语言源程序就是机器语言程序的符号表示，汇编语言源程序与其经过汇编所产生的目标代码程序之间有明显的一一对应关系，故也称汇编语言为符号语言。例题：1.定义问题：

已知某日的最高温度和最低温度，要求计算这一天的平均温度。2.算法设计： 最高温度加最低温度 将和除以2得到平均温度3.数据结构：

DATASEGMENTHI_TEMPDB92H;实际值可能是从传感器读取的数值LO_TEMPDB52H AV_TEMPDB? DATAENDS4.初始化表： MOVAX,DATA MOVDS,AX5.选择指令：

选择算法中每一主要动作所需的指令（按功能选择指令），并决定数据在这些指令中的存放形式，最后则按照指令的要求，用MOV或其他的指令把数据传送的正确的位置。

ADDdest,source;查指令表并阅读细节ADCdest,source;dest=dest+source+CFDIVsource;DIV02immedmodenotallowed!!6.编写程序：DATASEGMENTHI_TEMPDB92HLO_TEMPDB52HAV_TEMPDB?DATAENDSCODE SEGMENTMAIN PROC FAR

ASSUMECS:CODE,DS:DATA MOV AX, DATA MOV DS, AX MOV AL, HI_TEMP ADD AL, LO_TEMP MOV AH, 00H ADC AH, 00H MOV BL, 02 DIV BL MOV AV_TEMP,AL

MOV AH, 4CH INT 21HMAIN ENDPCODE ENDS

ENDMAINExample: PURPOSE:ADDS4WORDSOFDATA

STACK SEGMENT

DB 32DUP(?) STACK ENDS DATA SEGMENT data_in DW 234DH,1DE6H,3BC7H,566AH

ORG 10H sum DW ? DATA ENDS CODE SEGMENT MAIN PROC FAR

ASSUMECS:CODE,DS:DATA,SS:STACK MOV AX, DATA MOV DS, AX MOV CX, 04 MOV DI, OFFSETdata_in MOV BX, 00 ADD_LP: ADD BX, [DI] INC DI INC DI DEC CX JNZ ADD_LP MOV SI, OFFSETsum MOV [SI], BX MOV AH, 4CH INT 21H MAIN ENDP CODE ENDS

END MAIN ;FULLSEGMENTDEFINITION ;-----stacksegment-------- STACK SEGMENT

DB 64DUP(?) STACK ENDS ;-----datasegment-------- DATA SEGMENT ;datadefinitionsareplacedhere DATA ENDS ;-----codesegment-------- CODE SEGMENT MAIN PROCFAR

ASSUMECS:CODE,DS:DATA,SS:STACK MOVAX, DATA MOVDS, AX ------ MOVAH, 4CH INT21H MAIN ENDP CODE ENDS ENDMAIN4.28086寻址方式（Addressingmodes）

指令以不同的方式访问操作数（数据）称为寻址方式。

8086共有7种寻址方式：（1）寄存器寻址（register）（2）立即寻址（immediate）（3）直接寻址（direct）（4）寄存器间接寻址（registerindirect）（5）基址相对寻址（basedrelative）（6）变址相对寻址（indexedrelative）（7）基址变址相对寻址（basedindexedrelative）操作数在CPU内：（1）寄存器寻址（Register）： MOVBX,DX MOVES,AX ADDAL,BH

MOVCL,AX（2）立即寻址（immediate）： MOVAX,2550H MOVCX,625 MOVBL,40H MOVAX,2550H MOVDS,AX MOVDS,0123H操作数在CPU外：（3）直接寻址（direct）：

MOV

DL,[2400];movecontentsofDS:2400HintoDLPA=DS*10H+2400Example: Findthephysicaladdressofthememorylocationanditscontentsaftertheexecutionofthefollowing,assumingthatDS=1512H. MOV AL,99H MOV [3518],ALSolution: FirstALisinitializedto99H,theninlinetwo,thecontentsofALaremovedtologicaladdressDS:3518whichis1512:3518.ShiftingDSleftandaddingittotheoffsetgivesthephysicaladdressof18638H(15120H+3518H).Thatmeansaftertheexecutionofthesecondinstruction,thememorylocationwithaddress18638Hwillcontainthevalue99H.（4）寄存器间接寻址（registerindirect）：MOVAL,[BX]

;movesintoALthe;contentsofthe;memorylocation;pointedtobyDS:BX

MOVCL,[SI]

MOV

[DI],AHExample: AssumethatDS=1120,SI=2498,andAX=17FE.Showthecontentsofmemorylocationsaftertheexecution MOV [SI],AXSolution: ThecontentsofAXaremovedintomemorylocationswithlogicaladdressDS:SIandDS:SI+1;therefore,thephysicaladdressstartsatDS*10H(shiftedleft)+SI=13698.Accordingtothelittleendianconvention,lowaddress13698HcontainsFE,thelowbyte,andhighaddress13699Hwillcontain17,thehighbyte.Strcpy函数：将指针d指向的字符串复制到指针s指向的位置—寄存器间接寻址法 viod strcpy(char*s,char*d) { while((*s=*d)!=‘\0’){ s++; d++; } }（5）基址相对寻址（basedrelative）MOV

CX,[BX]+10;PA=DS*10H+BX+10

MOVAL,[BP]+5;PA=SS*10H+BP+5

MOVAL,[BP+5]

MOVAL,5[BP]（6）变址相对寻址（indexedrelative）

MOVDX,[SI]+5;PA=DS*10H+SI+5

MOVCL,[DI]+20

;PA=DS*10H+DI+20Example:

AssumethatDS=4500,SS=2000,BX=2100,SI=1486,BP=7814,andAX=2512.ShowtheexactphysicalmemorylocationwhereAXisstoredineachofthefollowing.Allvaluesarehex. (a) MOV [BX]+20,AX (b) MOV [SI]+10,AX (c) MOV [DI]+4,AX (d) MOV [BP]+12,AXSolution:

IneachcasePA=segmentregister(shiftedleft)+offsetregister+displacement. (a) DS:BX+20 location47120=(12)and47121=(25) (b) DS:SI+10 location46496=(12)and46497=(25) (c) DS:DI+4 location4D504=(12)and4D505=(25) (d) SS:BP+12 location27826=(12)and27827=(25)#include <stio.h>

#define PROFIT 15

#define MAX_PRICES 10Int cost[]={20,28,15,26,19,27,16,29,39,42};Int prices[10];Int index;Main(){ for(index=0;index<MAX_PRICES;index++) prices[index]=cost[index]+PROFIT}Strcpy函数：将指针d指向的字符串复制到指针s指向的位置—基址相对寻址法Voidstrcpy(char*s,char*d){ int i; i=0; while((s[i]=d[i]!=‘\0’) i++;}（7）基址变址相对寻址（basedindexedrelative）

MOVCH,[BX][SI]+20

;PA=DS*10H+BX+SI+20（7）基址变址相对寻址（basedindexedrelative）

MOVCL,[BX][DI]+8 ;PA=DS*10H+BX+DI+8 MOVCL,[BX+DI+8] MOVCH,[BX][SI]+20 ;PA=DS*10H+BX+SI+20

MOVAH,[BP][DI]+12

;PA=SS*10H+BP+DI+12 MOVAH,[BP][SI]+29 ;PA=SS*10H+BP+SI+29

MOVAX,[SI][DI]+displacementAddressingModeOperandDefaultSegmentRegisterRegNone(CPU)ImmediateDataNone(CPU)Direct[offset]DSRegisterindirect[BX][SI][DI]DSDSDSBasedrelative[BX]+disp[BP]+dispDSSSIndexedrelative[DI]+disp[SI]+dispDSDSBasedindexedrelative[BX][SI]+disp[BX]+[DI]+disp[BP]+[SI]+disp[BP]+[DI]+dispDSDSSSSS4.38086

指令系统指令（Instruction）和伪指令（Directives）格式：

[标号:]助记符[操作数][；注释]

[label:]mnemonic[operands][;comment]指令（Instruction）:

4.3.1数据传送类指令

4.3.2算术运算指令

4.3.3位操作指令

4.3.4串操作指令

4.3.5程序转移指令

4.3.6处理机控制指令4.3.1数据传送类指令

(1)MOV(MOVement) MOVdst,src;copysrctodst MOVCL,55H MOVDL,CL MOVAH,DL MOVAL,AH MOVBH,CL MOVCH,BH

MOVCX,468FH MOVAX,CX MOVDX,AXMOVBX,DXMOVDI,BXMOVSI,DIMOVDS,SIMOVBP,DIMOVAX,58FCHMOVDX,6678HMOVSI,924BHMOVBP,2459HMOVDS,2341HMOVCX,8876HMOVCS,3F47HMOVBH,99H指令格式中的dst表示目的操作数，src表示源操作数(下同)。指令实现的操作是将源操作数送给目的操作数。这种传送实际上是进行数据的“复制”，源操作数本身不变。这种双操作数指令在汇编语言中的表示方法，总是将目的操作数写在前面，源操作数写在后面，二者之间用一个逗号隔开。必须注意，不能用一条MOV指令实现以下传送：①存储单元之间的传送。②立即数至段寄存器的传送。③段寄存器之间的传送。(2)LEA(LoadEffectiveAddress) Format:LEAdst,src ;dst=OFFSETsrc Flags:Unchanged. Function:Loadsintothedestination(a16-bitregister)theeffectiveaddressofadirectmemoryoperand.

ORG 0100HDATA DB 34,56,87,90,76,54,13,29 … ;toaccessthesixthelement: LEA SI, DATA+5 ;SI=100H+5 MOV AL, [SI];ifBX=2000H;SI=3500HLEADX,[BX][SI]+100H;DX=2000+3500+100;=5600H(3)PUSH(PUSHwordontostack)Format:PUSHsrc;SP=SP-2Flags:Unchanged.Function:CopiesthesourcewordtothestackanddecrementsSPby2.PUSHAXPUSHDIPUSHDX(4)POP(POPwordoffstack)Format:POPdst;SP=SP+2Flags:Unchanged.Function:CopiesthewordpointedtobythestackpointertotheregisterormemorylocationindicatedbytheoperandandincrementstheSPby2.

POPDXPOPDIPOPAX(5)XCHG(eXCHanGe)Format:XCHGdst，srcFlags:Unchanged功能：将源地址与目的地址中的内容互换。MOV AX，5678H ；(AX)=5678HMOV BX，0FFFFH ；(BX)=0FFFFHXCHG AX，BX；(AX)=0FFFFH，(BX)=5678H4.3.2算术运算指令(1)加1指令INC(INCrementby1) Format:INCdst ;dst=dst+1 Flags: OF,SF,ZF,AF,PF.Unchanged:CF

INC指令是一个单操作数指令，操作数可以是寄存器或存储器操作数。 如：INCBX，即（BX）+1→BX。 加1指令可用于对计数器和地址指针进行调整。

Example:Adds5bytesofdataSTACK SEGMENT

DW 64 DUP(?)STACK ENDSDATA SEGMENTdata_in DB 25H,12H,15H,1FH,2BHsum DB ?DATA ENDSCODE SEGMENTMAIN PROC

FAR

ASSUMECS:CODE,DS:DATA,SS:STACK

MOV AX, DATA MOV DS, AX

MOV CX, 05 MOV BX, OFFSETdata_in MOV AL, 0AGAIN: ADD AL, [BX] INC BX DEC CX JNZ AGAIN MOV SUM, AL

MOV AH, 4C INT 21HMAIN ENDPCODE ENDS END MAIN

(2)ADD(ADDition)Format:ADDdst,src ;dst=dst+src

Flags: OF,SF,ZF,AF,PF,CF

MOV AL, 25H MOV BL, 34H ADD AL, BL MOVDH, 25H MOVCL, 34H ADD DH, CL MOVDH, 25H ADDDH, 34H例题：;Thisprogramcalculatesthetotalsumof5bytesofdata.

.MODELSMALL .STACK64;--------------

.DATACOUNT EQU 05;equateDATA DB 125,235,197,91,48;definebyte

ORG 0008H;originSUM DW ?;defineword;--------------

.CODEMAIN PROC FAR MOV

AX,@DATA MOV

DS,AX

MOV CX,COUNT ;CXistheloopcounter

MOVSI,OFFSETDATA;SIisthedatapointerMOV AX,00 ;AXwillholdthesum

BACK:

ADD

AL,[SI] ;addthenextbytetoALJNC

OVER ;Ifnocarry,continueINC AH ;elseaccumulatecarryinAHOVER:

INC SI ;incrementdatapointerDEC CX ;decrementloopcounterJNZ BACK ;ifnotfinished,goaddnextbyte

MOV SUM,AX ;storesum

MOV AH,4CH

INT 21H ;gobacktoDOSMAINENDP

END MAIN(3)ADC(AdditionwithCarry)Format:ADC

dst,src ;dst=dst+src+CFFlags:OF,SF,ZF,AF,PF,CF;Thisprogramcalculatesthetotalsumoffivewordsofdata.

.MODELSMALL.STACK64;--------------

.DATACOUNT EQU 05DATA DW 27345,28521,29533,30105,32375;

ORG 0010HSUM DW 2DUP(?) ;duplicate;--------------

.CODEMAIN PROC FAR MOV AX,@DATA MOV DS,AX MOV CX,COUNT ;CXistheloopcounter MOV

SI,OFFSETDATA ;SIisthedatapointer

MOV AX,00 ;AXwillholdthesum

MOV BX,AX ;BXwillholdthecarriesBACK:

ADD

AX,[SI] ;addthenextwordtoAXADC BX,0 ;addcarrytoBXINC

SI ;incrementdatapointertwice

INC SI ;topointtonextwordDEC CX ;decrementloopcounter

JNZ

BACK ;ifnotfinished,continueadding

MOV SUM,AX ;storethesumMOV

SUM+2,BX ;storethecarries

MOVAH,4CH INT

21H ;gobacktoDOSMAINENDP

ENDMAIN;Thisprogramaddsthefollowingtwomultiwordnumbersandsavesthe;result:DATA1=548FB9963CE7HandDATA2=3FCD4FA23B8DH.

.MODELSMALL

.STACK64;--------------

.DATADATA1 DQ548FB9963CE7H ;definequadword

ORG0010HDATA2 DQ3FCD4FA23B8DH

ORG0020HDATA3 DQ?;--------------

.CODEMAIN PROC FAR MOV

AX,@DATA MOV DS,AX

CLC;clearcarrybeforefirstaddition

MOV

SI,OFFSETDATA1;SIispointerforoperand1MOVDI,OFFSETDATA2;DIispointerforoperand2

MOVBX,OFFSETDATA3;BXispointerforthesum

MOVCX,04;CXistheloopcounterBACK:MOVAX,[SI];movethefirstoperandtoAXADCAX,[DI];addthesecondoperandtoAXMOV[BX],AX;storethesumINCSI;pointtonextwordofoperand1INCSIINCDI;pointtonextwordofoperand2INCDIINCBX;pointtonextwordofsumINCBXLOOPBACK;ifnotfinished,continueadding

MOVAH,4CH INT21H;gobacktoDOS MAIN ENDP

ENDMAIN(4)

DEC(Decrement)Format:DEC

dest;dest=dest-1Flags:OF,SF,ZF,AF,PF.Unchanged:CFFunction:Subtract1fromthedestinationoperand.NotethatCFisunchangedevenifavalue0000isdecrementedandbecomesFFFF.(5)SUB(Subtract)Format:SUBdest,source;dest=dest-sourceFlags:OF,SF,ZF,AF,PF,CFFunction:Subtractssourcefromdestinationandputstheresultinthedestination.Setsthecarryandzeroflagaccordingtothefollowing:CFZFdest>source00theresultispositivedest=source01theresultis0dest<source10theresultisnegativein2`scomp ThestepsforsubtractionperformedbytheinternalhardwareoftheCPUareasfollows: 1.Takesthe2`scomplementofthesource 2.Addsthistothedestination 3.Invertsthecarryandchangestheflagsaccordingly;fromthedatasegmentDATA1DB4CHDATA2DB6EHDATA3DB?;fromthecodesegment MOV DH, DATA1 SUB DH, DATA2 JNC NEXT NOT DH INC DHNEXT: MOV DATA3,DH(6)SBB(SubtractwithBorrow)Format:SBBdest,source:dest=dest-CF-sourceFlags:OF,SF,ZF,AF,PF,CFExample:;fromthedatasegmentDATA_ADD62562FAHDATA_BDD412963BHRESULTDD?;fromthecodesegment MOV AX, WORD

PTRDATA_A SUB AX, WORDPTRDATA_B MOV WORD

PTRRESULT, AX MOV AX, WORDPTRDATA_A+2 SBB AX, WORDPTRDATA_B+2 MOV WORD

PTRRESULT+2,AX(7)CMP(CompareOperands)Format:CMPdest,source;setsflagsasif“SUB”Flags:OF,SF,ZF,AF,PF,CFFunction:Comparestwooperandsofthesamesize.Thesourceanddestinationoperandsarenotaltered.

CF ZF dest>source0 0 dest=source0 1 dest<source1 0Example: DATA1 DW 235FH ….. MOV AX,0CCCCH CMP AX,DATA1 JNC OVER SUB AX,AX OVER: INC DATA1 …..Example: ….. MOV BX,7888H MOV CX,9FFFH CMP BX,CX JNC NEXT ADD BX,4000H NEXT: ADD CX,250H …..Example: ….. TEMP DB ? ….. MOV AL,TEMP CMP AL,99 JZ HOT INC BX ….. HOT : HLTExample:;Thefollowingprogramfindsthehighestamong5grades.MODELSMALL.STACK64;--------------

.DATAGRADES DB 69,87,96,45,75

ORG 0008HIGHEST DB ?;--------------

.CODEMAIN PROC FAR MOV AX, @DATA MOV

DS, AX

MOVCX,5;setuploopcounter MOV

BX,OFFSETGRADES;BXpointsto GRADEdata SUBAL,AL;ALholdshighestgradefoundsofarAGAIN:CMPAL,[BX];comparenextgradetohighestJANEXT;jumpifALstillhighestMOVAL,[BX];elseALholdsnewhighestNEXT: INCBX;pointtonextgradeLOOPAGAIN;continuesearchMOVHIGHEST,AL;storehighestgradeMOV AH,4CHINT

21H;gobacktodosMAINENDP

ENDMAINExample:ASCIIlowercasetouppercaseconversion43217650000010100111001011101110000NULDLESP0@P\p0001SOHDC1!1AQaq0010STXDC2“2BRbr0011ETXDC3#3CScs0100EOTDC4$4DTdt0101ENQNAK%5EUeu0110ACKSYN&6FVfv0111BELETB‘7GWgw1000BSCAN(8HXhx1001HTEM)9IYiy1010LFSUB*:JZjz1011VTESC+;K[k{1100FFFS,<L\l|1101CRGS-=M]m}1110SORS.>N^n~1111SIUS/?O_oDEL行0123456789101112131415列01234567.MODELSMALL.STACK64;--------------

.DATADATA1DB 'mYNAMEisjOe'

ORG 0020HDATA2 DB 14DUP(?);--------------

.CODEMAIN PROC FAR MOV AX,@DATA MOV DS,AX MOV SI,OFFSETDATA1;SIpointstooriginaldata MOV BX,OFFSETDATA2;BXpointstouppercasedata MOV CX,14;CXisloopcounterBACK: MOV AL,[SI] ;getnextcharacter CMP AL,61H ;iflessthan'a'JBOVER;thennoneedtoconvert CMP AL,7AH ;ifgreaterthan'z'JAOVER;thennoneedtoconvert AND AL,11011111B ;maskd5toconverttouppercaseOVER: MOV [BX],AL ;storeuppercasecharacter INC SI ;incrementpointertooriginal INC BX ;incrementpointertouppercasedata LOOP BACK ;continueloopingifCX>0 MOV AH,4CH INT

21H ;gobacktodosMAIN ENDP

END MAIN4）十进制数（BCD）运算指令

1．压缩BCD码调整指令（1）加法的十进制调整指令DAA（2）减法的十进制调整指令DAS

2．非压缩BCD码调整指令（1）加法的非压缩BCD码调整指令AAA（2）减法的非压缩BCD码调整指令AAS（3）乘法的非压缩BCD码调整指令AAM（4）除法的非压缩BCD码调整指令AADKeyASCII(hex)BinaryBCD(unpacked)030011000000000000131011000100000001232011001000000010333011001100000011434011010000000100535011010100000101636011011000000110737011011100000111838011100000001000939011100100001001ASCII码ASCIItounpackedBCDconversion;-----------datasegmentASC DB ‘9562481273’

ORG 0010HUNPACK DB 10DUP(?);-----------codesegment MOV CX, 5 MOV BX, OFFSETASC MOV DI, OFFSETUNPACKAGAIN: MOV AX, [BX] AND AX, 0F0FH MOV [DI], AX ADD BX, 2 ADD DI, 2 LOOP AGAIN;-----------datasegmentASC DB ‘9562481273’

ORG 0010HUNPACK DB 10DUP(?);-----------codesegment

MOV CX, 10 SUB BX, BXAGAIN: MOV AL, ASC[BX] AND AL, 0FH MOV UNPACK[BX],AL INC BX LOOP AGAINASCIItopackedBCDconversionKey

ASCII

UnpackedBCD

PackedBCD434000001007370000011101000111(47H)Example:

ORG 0010HVAL_ASC DB ‘47’VAL_BCD DB ?;------------- MOV AX,WORDPTRVAL_ASC AND AX,0F0FH XCHG AH,AL MOV CL,4 ROL AH,CL OR AL,AH MOV VAL_BCD,AL;----------------PackedBCDtoASCIIconversionPackedBCD

UnpackedBCD

ASCII29H02H&09H32H&39H0010100100000010&000010010110010&0111001Example:VAL_BCD DB 29HVAL_ASC DW ?--------------- MOV AL, VAL_BCD MOV AH, AL AND AX, 0F00FH MOV CL, 4 SHR AH, CL OR AX, 3030H XCHG AH, AL MOV VAL_ASC,AX---------------（1）十进制加法的调整指令DAADAA(DecimalAdjustafterAddition)Format:DAA;NotethatDAAworksonlyonALFlags: Affected:SF,ZF,AF,PF,CF,OFExample:----------DATA1 DB 47HDATA2 DB 25HDATA3 DB ?---------- MOV AL, DATA1 MOV BL, DATA2 ADD AL, BL DAA MOV DATA3,AL---------- Example:1.ConvertfromASCIItopackedBCD2.AddthemultibytepackedBCDandsaveit3.ConvertthepackedBCDresultintoASCII

.MODESMALL

.STACK64

.DATADATA1_ASC DB '0649147816'

ORG 0010HDATA2_ASC DB '0072687188'

ORG 0020HDATA3_BCD DB 5DUP(?)

ORG 0028HDATA4_BCD DB 5DUP(?)

ORG 0030HDATA5_ADD DB 5DUP(?)

ORG 0040HDATA6_ASC DB 10DUP(?)

.CODEMAINPROC

FAR MOVAX,@DATA MOVDS,AX

MOVBX,OFFSETDATA1_ASC MOVDI,OFFSETDATA3_BCD MOVCX,5 CALLCONV_BCD;convertASCIItoBCD MOVBX,OFFSETDATA2_ASC MOVDI,OFFSETDATA4_BCD MOVCX,5 CALLCONV_BCD;convertASCIItoBCD CALLBCD_ADD;addtheBCDoperands MOVSI,OFFSETDATA5_ADD MOVDI,OFFSETDATA6_ASC MOVCX,05 CALLCONV_ASC ;convertresulttoASCII MOV AH,4CH INT 21H;gobacktoDOSMAINENDP;---------------;THISSUBROUTINECONVERTSASCIITOPACKEDBCDCONV_BCDPROCAGAIN:MOV AX,[BX] ;BX=pointerforASCIIdata XCHG AH,AL AND AX,0F0FH ;maskASCII3s PUSH CX ;savethecounter MOV CL,4 ;shiftAHleft4bits SHL AH,CL ;togetreadyforpacking OR AL,AH ;combinetomakepackedBCD MOV[DI],AL ;DI=pointerforBCDdata ADD BX,2 ;pointtonext2ASCIIbytes INC DI ;pointtonextBCDdata POPCX ;restoreloopcounter LOOPAGAIN RETCONV_BCDENDP;THISSUBROUTINEADDSTWOMULTIBYTEPACKEDBCDBCD_ADDPROC MOVBX,OFFSETDATA3_BCD;BX=pointerforoperand1 MOVDI,OFFSETDATA4_BCD;DI=pointerforoperand2 MOVSI,OFFSETDATA5_ADD;SI=pointerforsum MOVCX,05 CLC;CLearCarryflagBACK:MOVAL,[BX]+4;getnextbyteofoperand1 ADCAL,[DI]+4;addnextbyteofoperand2 DAA ;correctforBCDaddition MOV[SI]+4,AL ;savesum DECBX ;pointtonextbyteofoperand1 DECDI ;pointtonextbyteofoperand2 DECSI ;pointtonextbyteofsum LOOPBACK RETBCD_ADDENDP;THISSUBROUTINECONVERTSFROMPACKEDBCDTOASCIICONV_ASCPROCAGAIN2:MOVAL,[SI];SI=pointerforBCDdata MOVAH,AL;duplicatetounpackANDAX,0F00FH;unpackPUSHCX;savecounterMOVCL,04;shiftright4bitstounpackSHRAH,CL;theuppernibbleORAX,3030H;makeitASCIIXCHGAH,AL;swapforASCIIstorageconventionMOV[DI],AX;storeASCIIdataINCSI;pointtonextBCDdataADDDI,2;pointtonextASCIIdataPOPCX;restoreloopcounterLOOPAGAIN2RETCONV_ASCENDP

ENDMAIN（2）加法的非压缩BCD码调整指令AAAAAA(ASCIIAdjustafterAddition)Format:AAAFlags:AffectedAFandCFFunction:ThisinstructionisusedafteranADDinstructionhasaddedtwodigitsinASCIIcode.ThismakesitpossibletoaddASCIInumberswithoutmaskingofftheuppernibble“3”.TheresultwillbeunpackedBCDinALwithcarryflagsetifneeded.ThisinstructionadjustsonlyontheALregister.AHisincrementedifthecarryflagisset.Example: MOV AL,‘5’ ;AL=35 ADD AL,‘2’ ;addtoAL32theASCIIfor2 AAA ;changes67Hto07H OR AL,30 ;ORALwith30HtogetASCIIExample: --------- SUB AH,AH ;AH=00 MOV AL,‘7’ ;AL=37H MOV BL,‘5’ ;BL=35H ADD AL,BL ;37H+35H=6CHthereforeAL=6C AAA ;changes6CHto02inALandAH=CF=1 OR AX,3030H ;AX=3132 ----------Example:

.MODELSMALL .STACK64 ;--------------

.DATA VALUE1DB'0659478127'

ORG0010H VALUE2DB'0779563678'

ORG0020H RESULT1DB10DUP(?)

ORG0030H RESULT2DB10DUP(?) ;--------------.CODEMAINPROCFAR MOVAX, @DATA MOVDS, AX

CALLASC_ADD;callASCIIadditionsubroutine CALLCONVERT;callconverttoasciisubroutine

MOVAH,4CH INT21H;gobacktoDOSMAINENDP;---------------;THISSUBROUTINEADDSTHEASCIINUMBERSANDMAKESTHE;RESULTUNPACKED.ASC_ADDPROCCLC;clearthecarryMOVCX,10;setuploopcounterMOVBX,9;pointtoLSDBACK:MOVAL,VALUE1[BX];movenextbyteofoperand1ADCAL,VALUE2[BX];addnextbyteofoperand2AAA ;adjusttomakeitunpackedBCDMOVRESULT1[BX],AL;storeBCDsumDECBX;pointtonextbyteLOOPBACKRETASC_ADDENDP;---------------;THISSUBROUTINECONVERTSUNPACKEDBCDTOASCIICONVERTPROCMOVBX,OFFSETRESULT1;BXpointstounpackedBCDdata MOVSI,OFFSETRESULT2;SIpointstoASCIIdataMOVCX,05;CXisloopcounterBACK2:MOVAX,WORDPTR