cs 110 computer architecture
TRANSCRIPT
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CS110ComputerArchitectureMIPSInstructionFormats
Instructor:SörenSchwertfeger
http://shtech.org/courses/ca/
School of Information Science and Technology SIST
ShanghaiTech University
1Slides based on UC Berkley's CS61C
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LevelsofRepresentation/Interpretation
lw $t0,0($2)lw $t1,4($2)sw $t1,0($2)sw $t0,4($2)
HighLevelLanguageProgram(e.g.,C)
AssemblyLanguageProgram(e.g.,MIPS)
MachineLanguageProgram(MIPS)
HardwareArchitectureDescription(e.g.,blockdiagrams)
Compiler
Assembler
MachineInterpretation
temp=v[k];v[k]=v[k+1];v[k+1]=temp;
0000 1001 1100 0110 1010 1111 0101 10001010 1111 0101 1000 0000 1001 1100 0110 1100 0110 1010 1111 0101 1000 0000 1001 0101 1000 0000 1001 1100 0110 1010 1111
ArchitectureImplementation
Anythingcanberepresentedasanumber,
i.e.,dataorinstructions
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LogicCircuitDescription(CircuitSchematicDiagrams)
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Review:AllocatingSpaceonStack• Chastwostorageclasses:automaticandstatic– Automatic variablesarelocaltofunctionanddiscardedwhenfunctionexits
– Staticvariablesexistacrossexitsfromandentriestoprocedures
• Usestackforautomatic(local)variablesthatdon’tfitinregisters
• Procedureframeor activationrecord:segmentofstackwithsavedregistersandlocalvariables
• SomeMIPScompilersuseaframepointer($fp)topointtofirstwordofframe
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StackBefore,During,AfterCall
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RecursiveFunctionFactorialint fact (int n)
{
if (n < 1) return (1);
else return (n * fact(n-1));
}
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RecursiveFunctionFactorialFact:
# adjust stack for 2 itemsaddi $sp,$sp,-8# save return addresssw $ra, 4($sp) # save argument nsw $a0, 0($sp) # test for n < 1slti $t0,$a0,1 # if n >= 1, go to L1beq $t0,$zero,L1# Then part (n==1) return 1addi $v0,$zero,1 # pop 2 items off stackaddi $sp,$sp,8 # return to callerjr $ra
L1: # Else part (n >= 1) # arg. gets (n – 1)addi $a0,$a0,-1 # call fact with (n – 1)jal Fact # return from jal: restore nlw $a0, 0($sp)# restore return addresslw $ra, 4($sp) # adjust sp to pop 2 itemsaddi $sp, $sp,8 # return n * fact (n – 1)mul $v0,$a0,$v0# return to the callerjr $ra
6mul isapseudoinstruction
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LevelsofRepresentation/Interpretation
lw $t0,0($2)lw $t1,4($2)sw $t1,0($2)sw $t0,4($2)
HighLevelLanguageProgram(e.g.,C)
AssemblyLanguageProgram(e.g.,MIPS)
MachineLanguageProgram(MIPS)
HardwareArchitectureDescription(e.g.,blockdiagrams)
Compiler
Assembler
MachineInterpretation
temp=v[k];v[k]=v[k+1];v[k+1]=temp;
0000 1001 1100 0110 1010 1111 0101 10001010 1111 0101 1000 0000 1001 1100 0110 1100 0110 1010 1111 0101 1000 0000 1001 0101 1000 0000 1001 1100 0110 1010 1111
ArchitectureImplementation
Anythingcanberepresentedasanumber,
i.e.,dataorinstructions
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LogicCircuitDescription(CircuitSchematicDiagrams)
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ENIAC(U.Penn.,1946)FirstElectronicGeneral-PurposeComputer
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• Blazinglyfast(multiplyin2.8ms!)– 10decimaldigitsx10decimaldigits
• Butneeded2-3daystosetupnewprogram,asprogrammedwithpatchcordsandswitches
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BigIdea:Stored-Program
Computer
– Instructionsarerepresentedasbitpatterns- canthinkoftheseasnumbers
– Therefore,entireprogramscanbestoredinmemorytobereadorwrittenjustlikedata
– Canreprogramquickly(seconds),don’thavetorewirecomputer(days)
– Knownasthe“vonNeumann”computersafterwidelydistributedtechreportonEDVACproject• Wrote-updiscussionsofEckertandMauchly• AnticipatedearlierbyTuringandZuse
FirstDraftofaReportontheEDVACby
JohnvonNeumannContractNo.W–670–ORD–4926
BetweentheUnitedStatesArmyOrdnanceDepartmentandthe
UniversityofPennsylvaniaMooreSchoolofElectricalEngineering
UniversityofPennsylvania
June 30,1945
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Consequence#1:EverythingAddressed
• Sinceallinstructionsanddataarestoredinmemory,everythinghasamemoryaddress:instructions,datawords– bothbranchesandjumpsusethese
• Cpointersarejustmemoryaddresses:theycanpointtoanythinginmemory– Unconstraineduseofaddressescanleadtonastybugs;upto
youinC;limitedinJavabylanguagedesign• Oneregisterkeepsaddressofinstructionbeingexecuted:
“ProgramCounter”(PC)– Basicallyapointertomemory:IntelcallsitInstructionPointer(a
bettername)
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Consequence#2:BinaryCompatibility
• Programsaredistributedinbinaryform– Programsboundtospecificinstructionset– DifferentversionforMacintoshesandPCs
• Newmachineswanttorunoldprograms(“binaries”)aswellasprogramscompiledtonewinstructions
• Leadsto“backward-compatible”instructionsetevolvingovertime
• SelectionofIntel8086in1981for1st IBMPCismajorreasonlatestPCsstilluse80x86instructionset;couldstillrunprogramfrom1981PCtoday
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InstructionsasNumbers(1/2)
• Currentlyalldataweworkwithisinwords(32-bitchunks):– Eachregisterisaword.– lw andsw bothaccessmemoryonewordatatime.
• Sohowdowerepresentinstructions?– Remember:Computeronlyunderstands1sand0s,so“add $t0,$0,$0”ismeaningless.
– MIPS/RISCseekssimplicity:sincedataisinwords,makeinstructionsbefixed-size32-bitwordsalso
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InstructionsasNumbers(2/2)
• Onewordis32bits,sodivideinstructionwordinto“fields”.
• Eachfieldtellsprocessorsomethingaboutinstruction.
• Wecoulddefinedifferentfieldsforeachinstruction,butMIPSseekssimplicity,sodefine3basictypesofinstructionformats:– R-format– I-format– J-format
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InstructionFormats
• I-format:usedforinstructionswithimmediates,lw andsw (sinceoffsetcountsasanimmediate),andbranches(beq andbne)– (butnottheshiftinstructions;later)
• J-format:usedforj andjal• R-format:usedforallotherinstructions• Itwillsoonbecomeclearwhytheinstructionshavebeenpartitionedinthisway
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R-FormatInstructions(1/5)
• Define“fields”ofthefollowingnumberofbitseach:6+5+5+5+5+6=32
• Forsimplicity,eachfieldhasaname:
• Important:Ontheseslidesandinbook,eachfieldisviewedasa5- or6-bitunsignedinteger,notaspartofa32-bitinteger– Consequence:5-bitfieldscanrepresentanynumber0-31,while
6-bitfieldscanrepresentanynumber0-63
6 5 5 5 65
opcode rs rt rd functshamt
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R-FormatInstructions(2/5)
• Whatdothesefieldintegervaluestellus?– opcode:partiallyspecifieswhatinstructionitis• Note:Thisnumberisequalto0 forallR-Formatinstructions
– funct:combinedwithopcode,thisnumberexactlyspecifiestheinstruction
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• Question:Whyaren’topcode andfunct asingle12-bitfield?–We’llanswerthislater
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R-FormatInstructions(3/5)
• Morefields:– rs (SourceRegister):usually usedtospecifyregistercontainingfirstoperand
– rt (TargetRegister):usually usedtospecifyregistercontainingsecondoperand(notethatnameismisleading)
– rd (DestinationRegister):usually usedtospecifyregisterwhichwillreceiveresultofcomputation
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R-FormatInstructions(4/5)
• Notesaboutregisterfields:– Eachregisterfieldisexactly5bits,whichmeansthatitcanspecifyanyunsignedintegerintherange0-31.Eachofthesefieldsspecifiesoneofthe32registersbynumber.
– Theword“usually”wasusedbecausethereareexceptionsthatwe’llseelater
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R-FormatInstructions(5/5)• Finalfield:– shamt:Thisfieldcontainstheamountashiftinstructionwillshiftby.Shiftinga32-bitwordbymorethan31isuseless,sothisfieldisonly5bits(soitcanrepresentthenumbers0-31)
– Thisfieldissetto0 inallbuttheshiftinstructions
• Foradetaileddescriptionoffieldusageforeachinstruction,seegreeninsertinCOD(Youcanbringwithyoutoallexams)
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R-FormatExample(1/2)• MIPSInstruction:
add $8,$9,$10
opcode =0(lookupintableinbook)funct =32(lookupintableinbook)rd =8(destination)rs =9(firstoperand)rt =10(secondoperand)shamt =0(notashift)
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R-FormatExample(2/2)• MIPSInstruction:
add $8,$9,$10Decimalnumberperfieldrepresentation:
Binarynumberperfieldrepresentation:
hexrepresentation: 012A 4020hex
CalledaMachineLanguageInstruction
0 9 10 8 320
000000 01001 01010 01000 10000000000hex
21
opcode rs rt rd functshamt
= 19,546,144ten
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I-FormatInstructions(1/4)
• Whataboutinstructionswithimmediates?– 5-bitfieldonlyrepresentsnumbersuptothevalue31:immediatesmaybemuchlargerthanthis
– Ideally,MIPSwouldhaveonlyoneinstructionformat(forsimplicity):unfortunately,weneedtocompromise
• DefinenewinstructionformatthatispartiallyconsistentwithR-format:– Firstnoticethat,ifinstructionhasimmediate,thenitusesatmost2registers.
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I-FormatInstructions(2/4)• Define“fields”ofthefollowingnumberofbitseach:6+5+5+16=32bits
– Again,eachfieldhasaname:
– KeyConcept:OnlyonefieldisinconsistentwithR-format.Mostimportantly,opcode isstillinsamelocation.
6 5 5 16
opcode rs rt immediate
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I-FormatInstructions(3/4)• Whatdothesefieldsmean?– opcode:sameasbeforeexceptthat,sincethere’snofunct field,opcode uniquelyspecifiesaninstructioninI-format
– ThisalsoanswersquestionofwhyR-formathastwo6-bitfieldstoidentifyinstructioninsteadofasingle12-bitfield:inordertobeconsistentaspossiblewithotherformatswhileleavingasmuchspaceaspossibleforimmediatefield.
– rs:specifiesaregisteroperand(ifthereisone)– rt:specifiesregisterwhichwillreceiveresultofcomputation(thisiswhyit’scalledthetarget register“rt”)orotheroperandforsomeinstructions.
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I-FormatInstructions(4/4)• TheImmediateField:– addi,slti,sltiu,theimmediateissign-extended to32bits.Thus,it’streatedasasignedinteger.
– 16bitsè canbeusedtorepresentimmediateupto216 differentvalues
– Thisislargeenoughtohandletheoffsetinatypicallw orsw,plusavastmajorityofvaluesthatwillbeusedintheslti instruction.
– Later,we’llseewhattodowhenavalueistoobigfor16bits
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I-FormatExample(1/2)• MIPSInstruction:
addi $21,$22,-50
opcode =8(lookupintableinbook)rs =22(registercontainingoperand)rt =21(targetregister)immediate =-50(bydefault,thisisdecimalinassemblycode)
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I-FormatExample(2/2)• MIPSInstruction:
addi $21,$22,-50
8 22 21 -50
001000 10110 10101 1111111111001110
Decimal/field representation:
Binary/field representation:
hexadecimal representation: 22D5 FFCEhex
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QuestionWhichinstructionhassamerepresentationasinteger35ten?
a) add$0,$0,$0b) subu $s0,$s0,$s0c) lw $0,0($0)d) addi $0,$0,35e) subu $0,$0,$0
Registersnumbersandnames:0:$0,..8:$t0,9:$t1,..15:$t7,16:$s0,17:$s1,..23:$s7
Opcodesandfunctionfields:add:opcode =0,funct =32subu:opcode =0,funct =35addi:opcode =8lw:opcode =35
opcode rs rt offset
rd functshamtopcode rs rt
opcode rs rt immediate
rd functshamtopcode rs rt
rd functshamtopcode rs rt
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DealingWithLargeImmediates
• Howdowedealwith32-bitimmediates?– Sometimeswanttouseimmediates >± 215 withaddi,lw,sw andslti
– Bitwiselogicoperationswith32-bitimmediates
• Solution:Don’tmesswithinstructionformats,justaddanewinstruction
• LoadUpperImmediate (lui)– lui reg,imm–Moves16-bitimm intoupperhalf(bits16-31)ofreg andzerosthelowerhalf(bits0-15)
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lui Example
• Want:addiu $t0,$t0,0xABABCDCD– Thisisapseudo-instruction!
• Translatesinto:lui $at,0xABAB # upper 16ori $at,$at,0xCDCD # lower 16addu $t0,$t0,$at # move
• Nowwecanhandleeverythingwitha16-bitimmediate!
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Onlytheassemblergetstouse$at($1)
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BranchingInstructions
• beq andbne– Needtospecifyatargetaddressifbranchtaken– Alsospecifytworegisterstocompare
• UseI-Format:
– opcode specifiesbeq (4)vs.bne (5)– rs andrt specifyregisters– Howtobestuseimmediate tospecifyaddresses?
31
opcode rs rt immediate31 0
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BranchingInstructionUsage
• Branchestypicallyusedforloops(if-else,while,for)– Loopsaregenerallysmall(<50instructions)– Functioncallsandunconditionaljumpshandledwithjumpinstructions(J-Format)
• Recall: Instructionsstoredinalocalizedareaofmemory(Code/Text)– Largestbranchdistancelimitedbysizeofcode– Addressofcurrentinstructionstoredintheprogramcounter(PC)
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PC-RelativeAddressing
• PC-RelativeAddressing: Usetheimmediatefieldasatwo’scomplementoffsettoPC– BranchesgenerallychangethePCbyasmallamount
– Canspecify± 215 addressesfromthePC
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BranchCalculation
• Ifwedon’t takethebranch:– PC = PC + 4 = nextinstruction
• Ifwedo takethebranch:– PC = (PC+4) + (immediate*4)
• Observations:– immediate isnumberofinstructionstojump(remember,specifieswords)eitherforward(+)orbackwards(–)
– BranchfromPC+4 forhardwarereasons;willbeclearwhylaterinthecourse
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BranchExample(1/2)
• MIPSCode:Loop: beq $9,$0,End
addu $8,$8,$10addiu $9,$9,-1j Loop
End:
• I-Formatfields:opcode =4 (lookuponGreenSheet)rs =9 (firstoperand)rt =0 (secondoperand)immediate =???
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StartcountingfrominstructionAFTERthebranch
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BranchExample(2/2)
• MIPSCode:Loop: beq $9,$0,End
addu $8,$8,$10addiu $9,$9,-1j Loop
End:
Fieldrepresentation(decimal):
Fieldrepresentation(binary):
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4 9 0 331 0
000100 01001 00000 000000000000001131 0
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QuestionsonPC-addressing
• Doesthevalueinbranchimmediatefieldchangeifwemovethecode?– Ifmovingindividuallinesofcode,thenyes– Ifmovingallofcode,thenno
• Whatdowedoifdestinationis>215instructionsawayfrombranch?– Otherinstructionssaveus– beq $s0,$0,far bne $s0,$0,next# next instr à j far
next: # next instr37
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J-FormatInstructions(1/4)
• Forbranches,weassumedthatwewon’twanttobranchtoofar,sowecanspecifyachangeinthePC
• Forgeneraljumps(j andjal),wemayjumptoanywhere inmemory– Ideally,wewouldspecifya32-bitmemoryaddresstojumpto
– Unfortunately,wecan’tfitbotha6-bitopcodeanda32-bitaddressintoasingle32-bitword
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J-FormatInstructions(2/4)
• Definetwo“fields”ofthesebitwidths:
• Asusual,eachfieldhasaname:
• KeyConcepts:– Keepopcode fieldidenticaltoR-FormatandI-Formatforconsistency
– Collapseallotherfieldstomakeroomforlargetargetaddress 39
6 2631 0
opcode target address31 0
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J-FormatInstructions(3/4)
• Wecanspecify226 addresses– Stillgoingtoword-alignedinstructions,soadd00 aslasttwobits(multiplyby4)
– Thisbringsusto28bitsofa32-bitaddress
• Takethe4highestorderbitsfromthePC– Cannotreacheverywhere,butadequatealmostallofthetime,sinceprogramsaren’tthatlong
– Onlyproblematicifcodestraddlesa256MBboundary• Ifnecessary,use2jumpsorjr (R-Format)instead
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J-FormatInstructions(4/4)
• Jumpinstruction:– NewPC={ (PC+4)[31..28],targetaddress,00}
• Notes:– {,,}meansconcatenation{4bits,26bits,2bits}=32bitaddress• Bookuses||instead
– Arrayindexing:[31..28]meanshighest4bits– Forhardwarereasons,usePC+4insteadofPC
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AssemblerPseudo-Instructions• CertainCstatementsareimplementedunintuitivelyinMIPS– e.g.assignment(a=b)viaadd$zero
• MIPShasasetof“pseudo-instructions”tomakeprogrammingeasier– Moreintuitivetoread,butgettranslatedintoactualinstructionslater
• Example:move dst,src
translatedintoaddi dst,src,0
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AssemblerPseudo-Instructions
• Listofpseudo-instructions:http://en.wikipedia.org/wiki/MIPS_architecture#Pseudo_instructions
– Listalsoincludesinstructiontranslation• LoadAddress(la)– la dst,label– Loadsaddressofspecifiedlabelintodst
• LoadImmediate (li)– li dst,imm– Loads32-bitimmediateintodst
• MARShasadditionalpseudo-instructions– SeeHelp(F1)forfulllist
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AssemblerRegister
• Problem:–Whenbreakingupapseudo-instruction,theassemblermayneedtouseanextraregister
– Ifitusesaregularregister,it’lloverwritewhatevertheprogramhasputintoit
• Solution:– Reservearegister($1 or$at for“assemblertemporary”)thatassemblerwillusetobreakuppseudo-instructions
– Sincetheassemblermayusethisatanytime,it’snotsafetocodewithit
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MultiplyandDivide• Examplepseudo-instruction:
mul $rd,$rs,$rt– Consistsofmult whichstorestheoutputinspecialhiandloregisters,andamovefromtheseregistersto$rd
mult $rs,$rtmflo $rd
• mult anddiv havenothingimportantintherd fieldsincethedestinationregistersarehi andlo
• mfhi andmflo havenothingimportantinthers andrt fieldssincethesourceisdeterminedbytheinstruction(seeCOD)
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IntegerMultiplication(1/3)
• Paperandpencilexample(unsigned):Multiplicand 1000 8 Multiplier x1001 9
10000000
0000+100001001000
• m bitsx n bits=m +n bitproduct
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IntegerMultiplication(2/3)
• InMIPS,wemultiplyregisters,so:– 32-bitvaluex 32-bitvalue=64-bitvalue
• SyntaxofMultiplication(signed):– mult register1,register2–Multiplies32-bitvaluesinthoseregisters&puts64-bitproductinspecialresultregs:• putsproductupperhalfinhi,lowerhalfin lo
– hi andlo are2registersseparatefromthe32generalpurposeregisters
– Usemfhi register&mflo register tomovefromhi,lo toanotherregister
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IntegerMultiplication(3/3)• Example:– inC: a = b * c;– inMIPS:
• letb be$s2;letc be$s3;andleta be$s0 and$s1 (sinceitmaybeupto64bits)
mult $s2,$s3 # b*cmfhi $s0 # upper half of
# product into $s0mflo $s1 # lower half of
# product into $s1•Note:Often,weonlycareaboutthelowerhalfoftheproduct.•Pseudo-inst.mul expandstomult/mflo.
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IntegerDivision(1/2)
• Paperandpencilexample(unsigned):1001 Quotient Divisor
1000|1001010 Dividend-1000
101011010
-100010 Remainder
(or Modulo result)• Dividend=Quotientx Divisor+Remainder
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• SyntaxofDivision(signed):– div register1,register2
– Divides32-bitregister1by32-bitregister2:
– putsremainderofdivisioninhi,quotientinlo
• ImplementsCdivision(/)andmodulo(%)
• ExampleinC: a = c / d; b = c % d;
• inMIPS:a«$s0;b«$s1;c«$s2;d«$s3
div $s2,$s3 # lo=c/d, hi=c%dmflo $s0 # get quotientmfhi $s1 # get remainder
IntegerDivision(2/2)
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MALvs.TAL
• TrueAssemblyLanguage(TAL)– Theinstructionsacomputerunderstandsandexecutes
• MIPSAssemblyLanguage(MAL)– Instructionstheassemblyprogrammercanuse(includespseudo-instructions)
– EachMALinstructionbecomes1ormoreTALinstruction
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Question
WhichofthefollowingplacetheaddressofLOOPin$v0?1) la $t1, LOOP
lw $v0, 0($t1)
2) jal LOOPLOOP: addu $v0, $ra, $zero
3) la $v0, LOOP
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1 2 3A)T, T, TB)T, T, FC)F, T, TD)F, T, FE)F, F, T
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Summary• I-Format: instructionswithimmediates,lw/sw (offsetisimmediate),andbeq/bne– Butnottheshiftinstructions– BranchesusePC-relativeaddressing
• J-Format: j andjal (butnotjr)– Jumpsuseabsoluteaddressing
• R-Format: allotherinstructions
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opcode rs rt immediateI:
opcode target addressJ:
opcode functrs rt rd shamtR: