MemoryUseinAssemblyLanguage

CS64:ComputerOrganizationandDesignLogicLecture#6Fall2018

ZiadMatni,Ph.D.

Dept.ofComputerScience,UCSB

Administrative

•  ReminderthatyourmidtermexamisonOctober31st–  1weekfromWednesday–  Sametime/placeasregularlecture–  DSPstudents:makearrangementsASAP

•  Lab#3isdueTODAYby11:59pm•  Lab#4inlabtomorrowanddueonFriday(asusual)

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LectureOutline

•  MoreBranchingExamples

•  AccessingDatainMemory

•  MemoryAddressing

•  InstructionRepresentation

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MoreBranchingExamplesinty;if(x==5){

y=8;}elseif(x<7){

y=x+x;}else{

y=-1;}print(y)

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.textmain:#t0:xandt1:y

li$t0,5 #exampleli$t2,5 #what’sthis?beq$t0,$t2,equal_5

#checkiflessthan7 li$t2,7 slt$t3,$t0,$t2bne$t3,$zero,less_than_7#fallthroughtofinalelse li$t1,-1 jafter_branchesequal_5: li$t1,8 jafter_branches

less_than_7: add$t1,$t0,$t0#couldjumptoafter_branches,#butthisiswhatwewillfall#throughtoanywaysafter_branches:#printoutthevalueiny($t1) li$v0,1move$a0,$t1syscall#exittheprogramli$v0,10syscall

PopQuiz!•  Youhave5minutestofillinthemissingcode.•  Fillinthe4blankspaces:

main: #assume$t0hasbeendeclaredearlier(nothere) li$t1,0 li________________ blt_________________________ li$t1,1exit: ________________________ ________________________

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InC++,thecodewouldbe:if(t0>=5)t1=1;

elset1=0;

PopQuizAnswers!•  Youhave5minutestofillinthemissingcode.•  Fillinthe4blankspaces:

main: #assume$t0hasbeendeclaredearlier(nothere) li$t1,0 li________________ blt_________________________ li$t1,1;exit: ________________________ ________________________

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InC++,thecodewouldbe:if(t0>=5)t1=1;

elset1=0;

$t2,5#somethingtocompare!$t0,$t2,exit

li$v0,10syscall

LargerDataStructures

•  Recall:registersvs.memory– Wherewoulddatastructures,arrays,etc.go?– Whichisfastertoaccess?Why?

•  Somedatastructureshavetobestoredinmemory– Soweneedinstructionsthat“shuttle”datato/fromtheCPUandcomputermemory(RAM)

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AccessingMemory•  Twobaseinstructions:–  load-word(lw)frommemorytoregisters–  store-word(sw)fromregisterstomemory

•  MIPSlacksinstructionsthatdomorewithmemorythanaccessit(e.g.,retrievesomethingfrommemoryandthenadd)–  Operationsaredonestep-by-step– MarkofRISCarchitecture

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MemoryRs

lw

sw

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.datanum1:.word42num2:.word7num3:.space1.textmain:

lw$t0,num1lw$t1,num2add$t2,$t0,$t1sw$t2,num3li$v0,1lw$a0,num3syscall

li$v0,10syscall

Example4Whatdoesthisdo?

MemoryRs

lw

sw

AddressingMemory•  Ifyou’renotusingthe.datadeclarations,thenyouneedstartingaddressesofthedatainmemorywithlwandswinstructions

Example: lw$t0,0x0000400A (ßnotarealaddress)Example: lw$t0,0x0000400A($s0)(ßnotarealaddress)

•  1word=32bits(inMIPS)–  So,ina32-bitunitofmemory,that’s4bytes–  Representedwith8hexadecimals 8x4bits=32bits…checksout…

•  MIPSaddressessequentialmemoryaddresses,butnotin“words”–  AddressesareinBytesinstead–  MIPSwordsmuststartataddressesthataremultiplesof4–  Calledanalignmentrestriction

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ThisisfoundonyourMIPSReferenceCard

NOTE:Notallmemoryaddressescanbeaccessedbytheprogrammer.Althoughtheaddressspaceis32bits,thetopaddressesfrom0x80000000to0xFFFFFFFFarenotavailabletouserprograms.TheyareusedmostlybytheOS.

Howmuchmemorydoesaprogrammergettodirectly

useinMIPS?

MemoryAllocationMap

MappingMIPSMemory(saythat10timesfast!)

•  Imaginecomputermemorylikeabigarrayofwords•  Sizeofcomputermemoryis:

232=4Gbits,or512MBytes(MB)–  Weonlygettouse2Gbits,or256MB–  That’s(256MB/groupsof4B)=64millionwords

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8bits 8bits 8bits 8bitsword

MIPSComputerMemoryAddressingConventions

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1A 80 C5 29

52 00 37 EE

B1 11 1A A5

0x00000x00010x00020x00030x00040x00050x00060x00070x00080x00090x000A0x000B

Aàà

MIPSComputerMemoryAddressingConventions

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1A 80 C5 29

52 00 37 EE

B1 11 1A A5

0x00030x00020x00010x00000x00070x00060x00050x00040x000B0x000A0x00090x0008

Bßß

or...

ATaleof2Conventions…

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BIGEND(MSByte)getsaddressedfirst

LITTLEEND(LSByte)getsaddressedfirst

TheUseofBigEndianvs.LittleEndian

Origin:JonathanSwift(author)in“Gulliver'sTravels”.Somepeoplepreferredtoeattheirhardboiledeggsfromthe“littleend”first(thus,littleendians),whileothersprefertoeatfrom

the“bigend”(i.e.bigendians).

•  MIPSuserstypicallygowithBigEndianconvention–  MIPSallowsyoutoprogram“endian-ness”

•  MostIntelprocessorsgowithLittleEndian…

•  It’sjustaconvention–itmakesnodifferencetoaCPU!

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GlobalVariablesRecall:•  Typically,globalvariablesareplaceddirectlyinmemory,notregisters

•  lwandswforloadwordandsaveword–  lwisNOTthesameaslaisNOTthesameasmove!!!–  Syntax:

lwregister_destination,N(register_with_address)WhereN=offsetofaddressinbytes

•  Let’stakealookat:access_global.asm

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access_global.asmLoadAddress(la)andLoadWord(lw).datamyVariable:.word42.textmain:la$t0,myVariableßWHAT’SIN$t0??lw$t1,0($t0) ßWHATDIDWEDOHERE??

li$v0,1move$a0,$t1syscall ßWHATSHOULDWESEEHERE??

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$t0=&myVariable

access_global.asmStoreWord(sw)(…continuingfromlastpage…)li$t1,5sw$t1,0($t0) ßWHAT’SIN$t0AGAIN??li$t1,0lw$t1,0($t0) ßWHATDIDWEDOHERE??li$v0,1move$a0,$t1syscall ßWHATSHOULDWESEEHERE??

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Arrays

•  Question:Asfarasmemoryisconcerned,whatisthemajordifferencebetweenanarrayandaglobalvariable?– Arrayscontainmultipleelements

•  Let’stakealookat:–  print_array1.asm–  print_array2.asm–  print_array3.asm

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print_array1.asmintmyArray[] ={5,32,87,95,286,386};

intmyArrayLength=6;intx;for(x=0;x<myArrayLength;x++){print(myArray[x]);print("\n");

}

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FlowChartforprint_array1

#Ccode:#intmyArray[]=#{5,32,87,95,286,386}#intmyArrayLength=6#for(x=0;x<myArrayLength;x++){#print(myArray[x])#print("\n")}.datanewline:.asciiz"\n"myArray:.word5328795286386myArrayLength:.word6.textmain:

#t0:x#initializexli$t0,0

loop:#getmyArrayLength,putresultin$t2#$t1=&myArrayLengthla$t1,myArrayLengthlw$t2,0($t1)

#seeifx<myArrayLength#putresultin$t3slt$t3,$t0,$t2#jumpoutifnottruebeq$t3,$zero,end_main

#getthebaseofmyArrayla$t4,myArray

#figureoutwhereinthearrayweneed#toreadfrom.Thisisgoingtobethearray#address+(index<<2).Theshiftisa #multiplicationbyfourtoindexbytes#asopposedtowords.#Ultimately,theresultisputin$t7sll$t5,$t0,2add$t6,$t5,$t4lw$t7,0($t6)

#printitout,withanewlineli$v0,1move$a0,$t7syscallli$v0,4la$a0,newlinesyscall

#incrementindexaddi$t0,$t0,1

#restartloopjloop

end_main:

#exittheprogramli$v0,10syscall

print_array2.asm

•  Sameasprint_array1.asm,exceptthatintheassemblycode,weliftredundantcomputationoutoftheloop.

•  Thisisthesortofthingadecentcompiler(clangorgccorg++,forexample)willdowithaHLLprogram

•  Yourhomework:Gothroughthisassemblycode!

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print_array3.asmintmyArray[]={5,32,87,95,286,386};

intmyArrayLength=6;int*p;for(p=myArray;p<myArray+myArrayLength;p++){print(*p);print("\n");

}

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Yourhomework:Gothroughthisassemblycode!

YOURTO-DOs

•  ReviewALLthedemocode– Availableviatheclasswebsite

•  Assignment#4– DueFriday

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