第四章 8086指令系統(tǒng)

第四章

8086指令系統(tǒng)4.1匯編語言程序4.28068尋址方式4.38086指令系統(tǒng)4.4程序設(shè)計步驟內(nèi)容與要求：

1.熟悉Intel8086/8088的尋址方式。

2.掌握Intel8086/8088的指令系統(tǒng)。

3.了解Intel80386擴充與增加的指令。4.1匯編語言程序（1）機器語言(MachineLanguage)

機器語言是一種用二進制表示指令和數(shù)據(jù)，能被機器直接識別的計算機語言。它的缺點是不直觀，不易理解和記憶，因此編寫、閱讀和修改機器語言程序都比較繁瑣。但機器語言程序是計算機惟一能夠直接理解和執(zhí)行的程序，具有執(zhí)行速度快、占用內(nèi)存少等特點。（2）匯編語言(AssemblyLanguage)

匯編語言是一種采用助記符表示的程序設(shè)計語言，即用助記符來表示指令的操作碼和操作數(shù)，用標號或符號代表地址、常量或變量。助記符一般都是英文字的縮寫，以方便人們書寫、閱讀和檢查。實際上，用匯編語言編寫的匯編語言源程序就是機器語言程序的符號表示，匯編語言源程序與其經(jīng)過匯編所產(chǎn)生的目標代碼程序之間有明顯的一一對應(yīng)關(guān)系，故也稱匯編語言為符號語言。例題：1.定義問題：

已知某日的最高溫度和最低溫度，要求計算這一天的平均溫度。2.算法設(shè)計： 最高溫度加最低溫度 將和除以2得到平均溫度3.數(shù)據(jù)結(jié)構(gòu)：

DATASEGMENTHI_TEMPDB92H;實際值可能是從傳感器讀取的數(shù)值LO_TEMPDB52H AV_TEMPDB? DATAENDS4.初始化表： MOVAX,DATA MOVDS,AX5.選擇指令：

選擇算法中每一主要動作所需的指令（按功能選擇指令），并決定數(shù)據(jù)在這些指令中的存放形式，最后則按照指令的要求，用MOV或其他的指令把數(shù)據(jù)傳送的正確的位置。

ADDdest,source;查指令表并閱讀細節(jié)ADCdest,source;dest=dest+source+CFDIVsource;DIV02immedmodenotallowed!!6.編寫程序：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尋址方式（Addressingmodes）

指令以不同的方式訪問操作數(shù)（數(shù)據(jù)）稱為尋址方式。

8086共有7種尋址方式：（1）寄存器尋址（register）（2）立即尋址（immediate）（3）直接尋址（direct）（4）寄存器間接尋址（registerindirect）（5）基址相對尋址（basedrelative）（6）變址相對尋址（indexedrelative）（7）基址變址相對尋址（basedindexedrelative）操作數(shù)在CPU內(nèi)：（1）寄存器尋址（Register）： MOVBX,DX MOVES,AX ADDAL,BH

MOVCL,AX（2）立即尋址（immediate）： MOVAX,2550H MOVCX,625 MOVBL,40H MOVAX,2550H MOVDS,AX MOVDS,0123H操作數(shù)在CPU外：（3）直接尋址（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）寄存器間接尋址（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函數(shù)：將指針d指向的字符串復(fù)制到指針s指向的位置—寄存器間接尋址法 viod strcpy(char*s,char*d) { while((*s=*d)!=‘\0’){ s++; d++; } }（5）基址相對尋址（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）變址相對尋址（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函數(shù)：將指針d指向的字符串復(fù)制到指針s指向的位置—基址相對尋址法Voidstrcpy(char*s,char*d){ int i; i=0; while((s[i]=d[i]!=‘\0’) i++;}（7）基址變址相對尋址（basedindexedrelative）

MOVCH,[BX][SI]+20

;PA=DS*10H+BX+SI+20（7）基址變址相對尋址（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

指令系統(tǒng)指令（Instruction）和偽指令（Directives）格式：

[標號:]助記符[操作數(shù)][；注釋]

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

4.3.1數(shù)據(jù)傳送類指令

4.3.2算術(shù)運算指令

4.3.3位操作指令

4.3.4串操作指令

4.3.5程序轉(zhuǎn)移指令

4.3.6處理機控制指令4.3.1數(shù)據(jù)傳送類指令

(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表示目的操作數(shù)，src表示源操作數(shù)(下同)。指令實現(xiàn)的操作是將源操作數(shù)送給目的操作數(shù)。這種傳送實際上是進行數(shù)據(jù)的“復(fù)制”，源操作數(shù)本身不變。這種雙操作數(shù)指令在匯編語言中的表示方法，總是將目的操作數(shù)寫在前面，源操作數(shù)寫在后面，二者之間用一個逗號隔開。必須注意，不能用一條MOV指令實現(xiàn)以下傳送：①存儲單元之間的傳送。②立即數(shù)至段寄存器的傳送。③段寄存器之間的傳送。(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：將源地址與目的地址中的內(nèi)容互換。MOV AX，5678H ；(AX)=5678HMOV BX，0FFFFH ；(BX)=0FFFFHXCHG AX，BX；(AX)=0FFFFH，(BX)=5678H4.3.2算術(shù)運算指令(1)加1指令I(lǐng)NC(INCrementby1) Format:INCdst ;dst=dst+1 Flags: OF,SF,ZF,AF,PF.Unchanged:CF

INC指令是一個單操作數(shù)指令，操作數(shù)可以是寄存器或存儲器操作數(shù)。 如：INCBX，即（BX）+1→BX。 加1指令可用于對計數(shù)器和地址指針進行調(diào)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例題：;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）十進制數(shù)（BCD）運算指令

1．壓縮BCD碼調(diào)整指令（1）加法的十進制調(diào)整指令DAA（2）減法的十進制調(diào)整指令DAS

2．非壓縮BCD碼調(diào)整指令（1）加法的非壓縮BCD碼調(diào)整指令A(yù)AA（2）減法的非壓縮BCD碼調(diào)整指令A(yù)AS（3）乘法的非壓縮BCD碼調(diào)整指令A(yù)AM（4）除法的非壓縮BCD碼調(diào)整指令A(yù)ADKeyASCII(hex)BinaryBCD(unpacked)030011000000000000131011000100000001232011001000000010333011001100000011434011010000000100535011010100000101636011011000000110737011011100000111838011100000001000939011100100001001ASCII碼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）十進制加法的調(diào)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）加法的非壓縮BCD碼調(diào)整指令A(yù)AAAAA(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