計算機組成原理與接口技術(shù)課件-第8次課

Chapter1

The80x86MicroprocessorTHE80x86IBMPCANDCOMPATIBLECOMPUTERSCHAPTER1MainContent

Registersofthe80x86MOVandADDinstructionsCodesegment,Datasegment,StacksegmentandExtrasegmentLogicaladdressandPhysicaladdressPUSHandPOPinstructionsFlagregisterAddressingmodesofthe8086

Sec.1.2Insidethe8088/8086CHAPTER1Figure1-1InternalBlockDiagramofthe8088/86CPU(ReprintedbypermissionofIntelCorporation,

CopyrightIntelCorp.1989)Sec.1.2Insidethe8088/8086RegistersCHAPTER1

Differentregistersinthe8088/86areusedfordifferentfunctions.Thefirstletterofeachregisterindicatesitsuse.AXisusedfortheaccumulator.BXisabaseaddressingregister.CXisacounterinloopoperations.DXpointstodatainI/Ooperations.Sec.1.2Insidethe8088/8086RegistersCHAPTER1

Differentregistersinthe8088/86areusedfordifferentfunctions.Sec.1.3IntroductiontoassemblyprogrammingCHAPTER1

Aprogramthatconsistsof0sand1siscalledmachinelanguage.Assemblylanguagesprovidesmnemonicsforthemachinecodeinstructions,plusotherfeaturesthatmadeprogrammingfasterandlesspronetoerror.Assemblylanguageprogramsmustbetranslatedintomachinecodebyaprogramcalledanassembler.Assemblylanguageisreferredtoasalow-levellanguagebecauseitdealsdirectlywiththeinternalstructureoftheCPU.

CHAPTER1

OneofthemostcommonlyusedassemblersisMASMbyMicrosoft.Othertools:LinkDebug

Debug.execanbefoundinC:\windows\system32\OS(XP)

Sec.1.3IntroductiontoassemblyprogrammingSec.1.3IntroductiontoassemblyprogrammingMOVinstructionCHAPTER1

Simplystated,theMOVinstructioncopiesdatafromonelocationtoanother.Ithasthefollowingformat:MOVdestination,source;copysourceoperandtodestinationThefollowingprogramfirstloadsCLwithvalue55H,thenmovesthisvaluearoundtovariousregistersinsidetheCPU.MOVCL,55HMOVDL,CLMOVAH,DLTheuseof16-bitregistersisdemonstratedbelow.MOVCX,468FHMOVAX,CXMOVDX,AXSec.1.3IntroductiontoassemblyprogrammingMOVinstructionCHAPTER1Inthe8086CPU,datacanbemovedamongalltheregistersshowninTable1-2(excepttheflagregister)aslongasthesourceanddestinationregistersmatchinsize.Codesuchas“MOVAL,DX”willcauseanerror.Theexceptionoftheflagregistermeansthatthereisnosuchinstructionas“MOVFR,AX”.Datacanbemoveddirectlyintononsegmentregistersonly,usingtheMOVinstruction.Forexample,MOVAX,58FCH;LEGALMOVDX,6678H;LEGALMOVSI,924BH;LEGALMOVBP,2459H;LEGALMOVDS,2341H;ILLEGALMOVCX,8876H;LEGALMOVCS,3F47H;ILLEGALMOVBH,99H;LEGALSec.1.3IntroductiontoassemblyprogrammingMOVinstructionCHAPTER1Fromthediscussionabove,notethefollowingthreepoints:1.Valuescannotbeloadeddirectlyintoanysegmentregister(CS,DS,ES,orSS).MOVAX,2345H;load2345HintoAXMOVDS,AX;thenloadthevalueofAXintoDS

MOVDI,1400H;load1400HintoDIMOVES,DI;thenmoveitintoES,nowES=DI=14002.IfavaluelessthanFFHismovedintoa16-bitregister,therestofthebitsareassumedtobeallzeros.Forexample,in“MOVBX,5”theresultwillbeBX=0005;thatisBH=00andBL=05.3.Movingavaluethatistoolargeintoaregisterwillcauseanerror.MOVBL,7F2H;ILLEGAL:7F2Hislargerthan8bitsMOVAX,2FE456h;ILLEGALSec.1.3IntroductiontoassemblyprogrammingMOVinstructionCHAPTER1Inthe8086CPU,datacanbemovedamongalltheregisters,aslongasthesourceanddestinationregistersmatchinsize(Excepttheflagregister.)Thereisnosuchinstructionas"MOVFR,AX".Codesuchas"MOVAL,DX"willcauseanerror.Onecannotmovethecontentsofa16-bitregister

intoan8-bitregister.Sec.1.3IntroductiontoassemblyprogrammingADDinstructionCHAPTER1

TheADDinstructionhasthefollowingformat:ADDdestination,source;ADDthesourceoperandtothedestinationToaddtwonumberssuchas25Hand34H,eachcanbemovedtoaregisterandthenaddedtogether:MOVAL,25H;move25intoALMOVBL,34H;move34intoBLADDAL,BL;AL=AL+BLTheprogramabovecanbewritteninmanyways,dependingontheregistersused.Anotherwaymightbe:MOVDH,25H;move25intoDHMOVCL,34H;move34intoCLADDDH,CL;addCLtoDH:DH=DH+CLSec.1.3IntroductiontoassemblyprogrammingADDinstructionCHAPTER1

Itisnotnecessarytomovebothdataitemsintoregistersbeforeaddingthemtogether.MOVDH,25H;loadoneoperandintoDHADDDH,34H;addthesecondoperandtoDHTheprogramabovecanbewritteninmanyways,dependingontheregistersused.Anotherwaymightbe:MOVDH,25H;move25intoDHMOVCL,34H;move34intoCLADDDH,CL;addCLtoDH:DH=DH+CLThelargestnumberthatan8-bitregistercanholdisFFH.MOVAX,34EH;move34EHintoAXMOVDX,6A5H;move6A5HintoDXADDDX,AX;addAXtoDX:DX=DX+AXAgain,any16-bitnonsegmentregisterscouldhavebeenusedtoperformtheactionabove.E.g.MOVCX,34EH;ADDCX,6A5HSec.1.4IntroductiontoprogramsegmentsCHAPTER1

AtypicalAssemblylanguageprogramconsistsatleastthreesegments:acodesegment,adatasegment,andastacksegment.ThecodesegmentcontainstheAssemblylanguageinstructionsthatperformthetasksthattheprogramwasdesignedtoaccomplish.Thedatasegmentisusedtostoreinformation(data)thatneedstobeprocessedbytheinstructionsinthecodesegment.Thestackisusedtostoreinformationtemporarily.

Sec.1.4IntroductiontoprogramsegmentsCHAPTER1Asegmentisanareaofmemorythatincludesupto64Kbytesandbeginsonanaddressevenlydivisibleby16(Suchanaddressendsin0H).The8088/86canonlyhandleamaximumof64Kbytesofcodeand64Kbytesofdataand64Kbytesofstackatanygiventime,althoughithasarangeof1megabyteofmemorybecauseofits20addresspins(220=1megabyte).

Howtomovethiswindowof64Kbytestocoverall1megabyteofmemory?Sec.1.4IntroductiontoprogramsegmentsLogicaladdressandphysicaladdressCHAPTER1

InIntelliteratureconcerningthe8086,therearethreetypesofaddressesmentionedfrequently:thephysicaladdress,theoffsetaddress,andthelogicaladdress.Thephysicaladdressisthe20-bitaddressthatisactuallyputontheaddresspinsofthe8086microprocessorsanddecodedbythememoryinterfacingcircuitry.Thisaddresscanhavearangeof00000HtoFFFFFHforthe8086andreal-mode286,386,and486CPUs.ThisisanactualphysicallocationinRAMorROMwithinthe1megabytememoryrange.Theoffsetaddressisalocationwitha64K-bytesegmentrange.There,anoffsetaddresscanrangefrom0000HtoFFFFH.Thelogicaladdressconsistsofasegmentvalueandanoffsetaddress.Sec.1.4IntroductiontoprogramsegmentsCodesegmentCHAPTER1

ThelogicaladdressofaninstructionalwaysconsistsofaCS(codesegment)andanIP(instructionpointer),showninCS:IPformat.

CSIP00523F59:StartwithCS.ShiftleftCS.AddIP.0052000523F5E2Sec.1.4IntroductiontoprogramsegmentsCodesegmentCHAPTER1Example1-1

IfCS=24F6HandIP=634AH,show:ThelogicaladdressTheoffsetaddressAndcalculate:(c)Thephysicaladdress(d)Thelowerrange(e)TheupperrangeofthecodesegmentSolution:24F6:634A(b)634A(c)2B2AA(24F60+634A)(d)24F60(24F60+0000)(e)34F5F(24F60+FFFF)Sec.1.4IntroductiontoprogramsegmentsLogicaladdressvs.physicaladdressinthecodesegment

ThethreecolumnsshowthelogicaladdressofCS:IP,themachinecodestoredatthataddressandthecorrespondingAssemblylanguagecode.ADDAX,1F35H05351F1132:0112ADDCX,BX01D91132:0110ADDAX,DX01D01132:010EMOVAH,20HB4201132:010CMOVBL,9FHB39F1132:010AMOVBH,AL88C71132:0108MOVCX,DX89D11132:0106MOVDL,72HB2721132:0104MOVDH,86HB6861132:0102MOVAL,57HB0571132:0100AssemblylanguagemnemonicsandoperandMachinelanguageOpcodeandoperandLogicaladdressCS:IPSec.1.4IntroductiontoprogramsegmentsLogicaladdressvs.physicaladdressinthecodesegment

ThethreecolumnsshowthelogicaladdressofCS:IP,themachinecodestoredatthataddressandthecorrespondingAssemblylanguagecode.MOVAL,57HB0571132:0100AssemblylanguagemnemonicsandoperandMachinelanguageOpcodeandoperandLogicaladdressCS:IPInstruction"MOVAL,57"hasamachinecodeofB057.B0istheopcodeand57istheoperand.Thebyteataddress1132:0100containsB0,theopcodeformovingavalueintoregisterAL.Address1132:0101containstheoperandtobemovedtoAL.Sec.1.4IntroductiontoprogramsegmentsDatasegmentCHAPTER1

Assumethataprogramisbeingwrittentoadd5bytesofdata,suchas25H,12H,15H,1FH,and2BH,whereeachbyterepresentsaperson’sdailyovertimepay.Onewaytoaddthemisasfollows.MOVAL,00HADDAL,25HADDAL,12HADDAL,15HADDAL,1FHADDAL,2BHIntheprogramabove,thedataandcodearemixedtogetherintheinstructions.Theproblemwithwritingtheprogramthiswayisthatifthedatachanges,thecodemustbesearchedforeveryplacethedataisincluded,andthedataretyped.Sec.1.4IntroductiontoprogramsegmentsDatasegmentCHAPTER1Thefollowingdemonstrateshowdatacanbestoredinthedatasegmentandtheprogramrewrittensothatitcanbeusedforanysetofdata.Assumethattheoffsetforthedatasegmentbeginsat200H.Thedataisplacedinmemorylocations:DS:0200=25H;DS:0201=12H;DS:0202=15H;DS:0203=1FH;DS:0204=2BHAndtheprogramcanberewrittenasfollows:MOVAL,00HADDAL,[0200]ADDAL,[0201]ADDAL,[0202]

ADDAL,[0203]ADDAL,[0204]Sec.1.4IntroductiontoprogramsegmentsDatasegmentCHAPTER1The8086/88allowsonlytheuseofregistersBX,SI,andDIasoffsetregistersforthedatasegment.Inotherwords,whileCSusesonlytheIPregisterasanoffset,DSusesonlyBX,DI,andSItoholdtheoffsetaddressofthedata.Thetermpointerisoftenusedforaregisterholdinganoffsetaddress.Inthefollowingexample,BXisusedasapointer:MOVAL,00HMOVBX,0200HADDAL,[BX]INCBXADDAL,[BX]INCBXADDAL,[BX]

Sec.1.4IntroductiontoprogramsegmentsLogicaladdressandphysicaladdressinthedatasegmentCHAPTER1

Thephysicaladdressfordataiscalculatedusingthesamerulesasforthecodesegment.Example1-2

AssumetheDSis5000andoffsetis1950.Calculatethephysicaladdressofthebyte.Solution:DS:offset00050591:StartwithDS.ShiftDS

left.Addtheoffset.00050000505915Sec.1.4IntroductiontoprogramsegmentsLogicaladdressandphysicaladdressinthedatasegmentCHAPTER1Example1-3

IfDS=7FA2Handtheoffsetis438EH,Calculatethephysicaladdress.(b)Calculatethelowerrange.(c)Calculatetheupperrangeofthedatasegment.(d)Showthelogicaladdress.Solution:83DAE(7FA20+438E)(b)7FA20(7FA20+0000)(c)8FA1F(7FA20+FFFF)(d)7FA2:438EExample1-4AssumethattheDSregisteris578C.Toaccessagivenbyteofdataatphysicalmemorylocation67F66,doesthedatasegmentcovertherangewherethedataislocated?Ifnot,whatchangesneedtobemade?Solution:No,sincetherangeis578C0to678BF,location67F66isnotincludedinthisrange.Toaccessthatbyte,DSmustbechangedsothatitsrangewillincludethatbyte.Sec.1.4IntroductiontoprogramsegmentsLittleendianconventionCHAPTER1

Previousexamplesused8-bitor1-bytedata.Inthiscasethebytesarestoredoneafteranotherinmemory.Whathappenswhen16-bitdataisused?MOVAX,35F3HMOV[1500],AXIncaseslikethis,thelowbytegoestothelowmemorylocationandthehighbytesgoestothehighmemoryaddress.DS:1500=F3DS:1501=35Example1-5Assumememorylocationswiththefollowingcontents:DS:6826=48andDS:6827=22.ShowthecontentsofregisterBXintheinstructionMOVBX,[6826]Solution:DS:6826=48DS:6827=22BHBL4822Sec.1.5Moreaboutsegmentsinthe80x86Whatisastack?CHAPTER1

Thestackisasectionofread/writememory(RAM)usedbytheCPUtostoreinformationtemporarily.TheCPUneedsthisstorageareasincethereareonlyalimitednumberofregisters.SincethestackisinRAM,ittakesmuchlongertoaccesscomparedtoaccesstimeofregisters.Sec.1.5Moreaboutsegmentsinthe80x86HowstacksareaccessedCHAPTER1

ThetwomainregistersusedtoaccessthestackaretheSS(stacksegment)andSP(stackpointer)register.Theseregistersmustbeloadedbeforeanyinstructionsaccessingthestackareused.ThestoringofaCPUregisterinthestackiscalled

apush

,andloadingthecontentsofthestackintotheCPUregisteriscalled

apop.

Theprincipleis“Lastin,Firstout”.Sec.1.5Moreaboutsegmentsinthe80x86PushontothestackCHAPTER1

WhenPUSHisexecuted,thecontentsoftheregisteraresavedonthestackandSPisdecrementedby2.Foreverybyteofdatasavedonthestack,SPisdecrementedonce,andsincepushissavingthecontentsofa16-bitregister,itisdecrementedtwice,

PoppingthecontentsofthestackbackintotheCPUregisteristheoppositeprocessofpushing.Sec.1.5Moreaboutsegmentsinthe80x86MoreaboutPushandPOPCHAPTER1

PushIP;ILLEGALPOPIP;ILLEGAL

Push1234H;ILLEGALin8086legalin80286

PUSHAandPOPAarelegalinstructionsin80286,PUSHAcopiesAX,CX,DX,BX,SP,BP,DI,andSItothestack.POPAremovesthewordcontentsforthefollowingregistersfromthestack:SI,DI,BP,SP,BX,DX,CX,AX.Sec.1.5Moreaboutsegmentsinthe80x86PushingontothestackCHAPTER1Sec.1.5Moreaboutsegmentsinthe80x86PoppingthestackCHAPTER1Sec.1.5Moreaboutsegmentsinthe80x86Logicaddressvs.physicaladdressforthestackCHAPTER1

Theexactphysicallocationofthestackdependsonthevalueofthestacksegment(SS)andSP,thestackpointer.ThelogicaddressisSS:SP.Tocomputerthephysicaladdressforthestack,thesameprincipleisappliedaswasusedforthecode(CS:IP)anddatasegment(DS:BX,DS:DIorDS:SI).

Sec.1.5Moreaboutsegmentsinthe80x86Afewmorewordsaboutsegmentsinthe80x86CHAPTER1Asinglephysicaladdresscanbelongtomanydifferentlogicaladdresses.E.g.Logicaladdress(hex)

Physicaladdress(hex)1000:5020150201500:0020150201502:0000150201400:1020150201302:200015020Sec.1.5Moreaboutsegmentsinthe80x86Afewmorewordsaboutsegmentsinthe80x86CHAPTER1WhenaddingtheoffsettotheshiftedsegmentregisterresultsinanaddressbeyondthemaximumallowedrangeofFFFFFH,wrap-aroundwilloccur.Sec.1.5Moreaboutsegmentsinthe80x86OverlappingCHAPTER1Incalculatingthephysicaladdress,itispossiblethattwosegmentscanoverlap,whichisdesirableinsomecircumstances.Sec.1.5Moreaboutsegmentsinthe80x86FlagregisterCHAPTER1Sixflags,calledconditionalflags,indicatesomeconditionresultingafteraninstructionexecutes.Thesesixare

CF,PF,AF,ZF,SF,and

OF.Theremainingthree,oftencalledcontrolflags,control

theoperationofinstructionsbeforetheyareexecuted.Sec.1.5Moreaboutsegmentsinthe80x86CodesfortheFlagRegisterCHAPTER1NC(nocarry)CY(carry)CFcarryflagPO(parityodd)PE(parityeven)PFparityflagNA(noauxiliarycarry)AC(auxiliarycarry)AFauxiliarycarryflagNZ(notzero)ZR(zero)ZFzeroflagPL(plus,orpositive)NG(negative)SFsignflagDI(disableinterrupt)EI(enableinterrupt)IFinterruptflagUP(up)DN(down)DFdirectionflagNV(nooverflow)OV(overflow)OFoverflowflagCodeWhenReset(=0)CodeWhenSet(=1)FlagSec.1.5Moreaboutsegmentsinthe80x86FlagregisterandADDinstructionCHAPTER1TheoverflowflagwillbecoveredinChapter6,sinceitrelatesonlytosignednumberarithmetic.Example1-10

Showhowtheflagregisterisaffectedbytheadditionof38Hand2FH.Solution:MOVBH,38H;BH=38HADDBH,2FH;add2FtoBH,nowBH=67H3800111000+2F

001011116701100111CF=0sincethereisnocarrybeyondd7PF=0sincethereisanoddnumberof1sintheresultAF=1sincethereisacarryfromd3tod4ZF=0sincetheresultisnotzeroSF=0sinced7oftheresultiszeroSec.1.5Moreaboutsegmentsinthe80x86FlagregisterandADDinstructionCHAPTER1

Thesameconceptsapplyfor16-bitaddition,asshowninExample1-12.Example1-13

ShowhowtheflagregisterisaffectedbyMOVAX,AAAAH;AX=AAAAHADDAX,5556H;nowAX=0000HSolution:AAAA10101010

1010

1010+5556

0101010101010110000000000000

00000000CF=1sincethereisacarrybeyond

d15PF=1sincethereisanevennumberof1s

inthelowerbyteAF=1sincethereisacarryfromd3tod4ZF=1sincetheresultiszeroSF=0sinced15oftheresultiszeroSec.1.5Moreaboutsegmentsinthe80x86UseofthezeroflagforloopingCHAPTER1

Oneofthemostwidelyusedapplicationsoftheflagregisteristheuseofzeroflagtoimplementprogramloops.MOVCX,05H;CXholdstheloopcountMOVBX,0200H;BXholdstheoffsetdataaddressMOVAL,00H;initializeALADD_LP:ADDAL,[BX];addthenextbytetoALINCBX;incrementthedatapointerDECCX;decrementtheloopcounterJNZADD_LP;jumptonextiterationifcounternotzeroSec.1.680x86addressingmodesCHAPTER1

TheCPUcanaccessoperands(data)invariousways,calledaddressingmodes.Thenumberofaddressingmodesisdeterminedwhenmicroprocessorisdesignedandcannotbechanged.The80x86providesatotalofsevendistinctaddressingmodes:1.Register2.Immediate3.Direct4.Registerindirect5.Basedrelative6.Indexedrelative7.BasedindexedrelativeSec.1.680x86addressingmodesRegisteraddressingmodeCHAPTER1

Theregisteraddressingmodeinvolvestheuseofregisterstoholdthedatatobemanipulated.

Memoryisnotaccessed

whenthisaddressingmodeisexecuted;therefore,itisrelativelyfast.MOVBX,DXMOVES,AXADDAL,BH

Itshouldbenotedthatthesourceanddestinationregistersmustmatchinsize.Sec.1.680x86addressingmodesImmediateaddressingmodeCHAPTER1

Intheimmediateaddressingmode,thesourceoperandisaconstant.Immediateaddressingmodecanbeusedtoloadinformationintoanyoftheregisters

exceptthesegmentregistersandflagregisters.Examples:MOVAX,2550H;move2550HintoAXMOVCX,625;loadthedecimalvalue625intoCXMOVBL,40H;load40HintoBLTomoveinformationtothesegmentregisters,thedatamustfirstbemovedtoageneral-purposeregisterandthentothesegmentregister.Example:MOVAX,2550HMOVDS,AXInotherwords,thefollowingwouldproduceanerror:MOVDS,0123H;illegal!!!Sec.1.680x86addressingmodesDirectaddressingmodeCHAPTER1

ThisaddressistheoffsetaddressandonecancalculatethephysicaladdressbyshiftinglefttheDSregisterandaddingittotheoffsetasfollows:MOVDL,[2400];movecontentsofDS:2400HintoDLExample1-15

Findthephysicaladdressofthememorylocationanditscontentsaftertheexecutionofthefollowing,assumingthatDS=1512H.MOVAL,99HMOV[3518H],ALSolution:FirstALisinitializedto99H,theninlinetwo,thecontentsofALaremovedto

logicaladdress

DS:3518H

whichis

1512:3518.ShiftingDSleftandaddingittotheoffsetgivesthephysicaladdressof

18638H(15120H+3518H=18638H).

Thatmeansaftertheexecutionofthesecondinstruction,thememorylocationwithaddress18638Hwillcontainthevalue99H.Sec.1.680x86addressingmodesRegisterindirectaddressingmodeCHAPTER1

Theregistersusedforthispurposeare

SI,DI,andBX.Iftheseregistersareusedaspointers,thatis,iftheyholdtheoffsetofthememorylocation,theymustbecombinedwith

DSinordertogeneratethe20-bitphysicaladdress.MOVAL,[BX];movesintoALthecontentsofthememorylocation;pointed

tobyDS:BXMOVCL,[SI];movecontentsofDS:SIintoCLMOV[DI],AH;movecontentsofAHintoDS:DIExample1-16AssumethatDS=1120,SI=2498,andAX=17FE.ShowthecontentsofmemorylocationsaftertheexecutionofMOV[SI],AXSolution:ThecontentsofAXaremovedintomemorylocationswithlogicaladdressDS:SIandDS:SI+1.Lowaddress13698HcontainFE,andhighaddress13699Hcontains17H.Sec.1.680x86addressingmodesBasedrelativeaddressingmodeCHAPTER1

Inthebasedrelativeaddressingmode,baseregistersare

BXandBP.Thedefaultsegmentsare

DSforBXandSSforBP.Forexample:MOVCX,[BX]+10;moveDS:BX+10andDS:BX+10+1intoCX;PA=DS(shiftedleft)+BX+10

AlternativecodingsareMOVCX,[BX+10]orMOVCX,10[BX]InthecaseoftheBPregister,MOVAL,[BP]+5;PA=SS(shiftedleft)+BP+5Again,alternativecodingsareMOVAL,[BP+5]orMOVAL,5[BP]Sec.1.680x86addressingmodesIndexedrelativeaddressingmodeCHAPTER1

Theindexedrelativeaddressingmodeworksthesameasthebasedrelativeaddressingmode,exceptthatregistersDIandSIholdtheoffsetaddress.Examples:MOVDX,[SI]+5;PA=DS(shiftedleft)+SI+5MOVCL,[DI]+20;PA=DS(shiftedleft)+DI+20Example1-17AssumethatDS=4500,SS=2000,BX=2100,SI=1486,DI=8500,BP=7814,andAX=2512.ShowtheexactphysicalmemorylocationwhereAXisstoredineachofthefollowing.Allvaluesareinhex.MOV[BX]+20,AX(b)MOV[SI]+10,AX(c)MOV[DI]+4,AX(d)MOV[BP]+12,AXSolution:DS:BX+20location47120=(12)and47121=(25)DS:SI+10

location46496=(12)and46497=(25)DS:DI+4

location4D504=(12)and4D505=(25)SS:BP+12

location27826=(12)and27827=(25)Sec.1.680x86addressingmodesBasedindexedaddressingmodeCHAPTER1

Bycombiningbasedandindexedaddressingmodes,anewaddressingmodeisderivedcalledthebasedindexedaddressingmode.MOVCL,[BX][DI]+8;PA=DS(shiftedleft)+BX+DI+8MOVCH,[BX][SI]+20;PA=DS(shiftedleft)+BX+SI+20MOVAH,[BP][DI]+12;PA=

SS

(shiftedleft)+BP+DI+12MOVAH,[BP][SI]+29;PA=

SS

(shiftedleft)+BP+SI+29Thecodingoftheinstructionsabovecanvary;forexample,thelastexamplecouldhavebeenwrittenMOVAH,[BP+SI+29]OrMOVAH,[SI+BP+29]NotethatMOVAX,“[SI][DI]+displacement”is

illegal.Sec.1.680x86addressingmodesSegmentoverridesCHAPTER1Table1-3:OffsetRegistersforVariousSegmentsSP,BPSI,DI,BXSI,DI,BXIPOffsetregister(s):SSESDSCSSegmentregister:Table1-4:SampleSegmentOverridesDS:BX+DI+32SS:BX+DI+32MOVSS:[BX][DI]+32,AXDS:BX+12ES:BX+12MOVCX,ES:[BX]+12SS:BPDS:BPMOVAX,DS:[BP]DS:SISS:SIMOVDX,SS:[SI]SS:BPCS:BPMOVAX,CS:[BP]DefaultSegmentSegmentUsedInstructionSec.1.680x86addressingmodesSummaryCHAPTER1Table1-5:Summaryof80x86addressingmodesDS,DSSS,SS[BX][SI]+disp,[BX][DI]+disp[BP][SI]+disp,[BP][DI]+dispBasedindexedrelativeDS,DS[DI]+disp,[SI]+dispIndexedrelativeDS,SS[BX]+disp,[BP]+dispBasedrelativeDS,DS,DS[BX],[SI],[DI]Registerindirect