計算機(jī)體系結(jié)構(gòu)實(shí)驗(yàn)報告

實(shí)驗(yàn)二結(jié)構(gòu)相關(guān)一、實(shí)驗(yàn)?zāi)康模和ㄟ^本實(shí)驗(yàn)，加深對結(jié)構(gòu)相關(guān)的理解，了解結(jié)構(gòu)相關(guān)對CPU性能的影響。二、實(shí)驗(yàn)內(nèi)容：1.用WinDLX模擬器運(yùn)行程序structure_d.s。通過模擬，找出存在結(jié)構(gòu)相關(guān)的指令對以及導(dǎo)致結(jié)構(gòu)相關(guān)的部件。記錄由結(jié)構(gòu)相關(guān)引起的暫停時鐘周期數(shù)，計算暫停時鐘周期數(shù)占總執(zhí)行周期數(shù)的百分比。論述結(jié)構(gòu)相關(guān)對CPU性能的影響，討論解決結(jié)構(gòu)相關(guān)的方法。三、實(shí)驗(yàn)程序structure_d.sLHIR2,(A>>16)&0xFFFF數(shù)據(jù)相關(guān)ADDUIR2,R2,A&0xFFFFLHIR3,(B>>16)&0xFFFFADDUIR3,R3,B&0xFFFFADDUR4,R0,R3loop:LDF0,0(R2)LDF4,0(R3)ADDDF0,F0,F4；浮點(diǎn)運(yùn)算，兩個周期，結(jié)構(gòu)相關(guān)ADDDF2,F0,F2;<-Astallisfound(anexampleofhowtoansweryourquestions)ADDIR2,R2,#8ADDIR3,R3,#8SUBR5,R4,R2BNEZR5,loop；條件跳轉(zhuǎn)TRAP#0;;Exit<-thisisacomment!!A:.double1,2,3,4,5,6,7,8,9,10B:.double1,2,3,4,5,6,7,8,9,10四、實(shí)驗(yàn)過程打開軟件，loadstructure_d.s文件，進(jìn)行單步運(yùn)行。經(jīng)過分析，此程序次循環(huán)中共有五次結(jié)構(gòu)相關(guān)。（R-stall數(shù)據(jù)相關(guān)Stall-結(jié)構(gòu)相關(guān)）1）第一個結(jié)構(gòu)相關(guān)：adddf2,,f0,f2由于前面的數(shù)據(jù)相關(guān)，導(dǎo)致上一條指令adddf0,f0,f4暫停在ID階段，所以下一條指令adddf2,,f0,f2發(fā)生結(jié)構(gòu)相關(guān)，導(dǎo)致相關(guān)的部件：譯碼部件。]ClockCycleDiagramInstructions/Cycles-IO.-9-8-6-2Ihi2,UXLIInstructions/Cycles-IO.-9-8-6-2Ihi2,UXLIdddLjii2j2,0K138IFID|intEX|MEM|WBIF|ID|iritEX|MEM|WBl「|ID|iiitEX|MUM|mIF|ID||MFM|''収addir2j2,0x8IF|ID||MFM|''収addir2j2,0x8addrii88IFIIDIintE父addrii88aHcIijr4rflr：Idf4,0x0(r3)adddf0J0J4adddf2J0J22）第二個結(jié)構(gòu)相關(guān)：ADDIR2,R2,#8，與第一個結(jié)構(gòu)相關(guān)類似。由于數(shù)據(jù)相關(guān)，上一條指令暫停在ID階段，所以導(dǎo)致下一條指令發(fā)生結(jié)構(gòu)相關(guān)?！鰘口|XInstnxdons/Cycles:10■0.7-G-5.4.3.■|口|XInstnxdons/Cycles:10■0.7-G-5.4.3.乜Ihir2.0x0adduir2j238Ihi「3』⑷adduir3..r3..0KlSBaddur4j0j3IdfO,0K0(r2)IdMJJmOIjSJadddfO..fOJ4adddf2JDJ2addii3j3.0x8ir|ID|inlEX|MEM||WB|IF||ID||ntCX]MUhl||WE：|IIFIIID|intEX|MUMIIVv;DjIFIDint：XHEMV/BIF1」intEXMLM|u/=!|IFIUR-5t^l|faddEX[l]MbMIF|呂怡II|ID|冃占tmllpddEXii|]ClockCycleDiagram3）第三、四、五個結(jié)構(gòu)相關(guān)：ADDIR2,R2,#8ADDIR3,R3,#8SUBR5,R4,R2第三個：ADDIR2,R2,#8訪存沖突，由于上一條指令處在訪存階段，所以此條指令需要暫停，為存儲器結(jié)構(gòu)相關(guān)沖突。

1ClockCycleDingImliudiincJCjdEi甜」兇2加卻d：idjM.iOj-saddi3jjXlmsdbi5j4.j2triezrSJaop1ClockCycleDingImliudiincJCjdEi甜」兇2加卻d：idjM.iOj-saddi3jjXlmsdbi5j4.j2triezrSJaopInformationaboutaddi『2』『2』addi『乙rZ.UxBAdr.:loop+Oxl0Code:0w20420008InPipeline-StagsMEMFirstCipidIs:-5LastCycle:???TdialCycles:>=6ldfD.Wad±n3j3腫田Cycles:=2(2)TerminatedsuccessfullvALU<-A+(8)(=0x140)[A=0:-:i3Ef|'1Stall(s)becauseofstructuralHazard!NuForuy-ardinq.第四個結(jié)構(gòu)相關(guān)：ADDIR3,R3,#8NuForuy-ardinq.與第一個和第二個相關(guān)類似，但此處為執(zhí)行部件沖突的結(jié)構(gòu)相關(guān)第五個：SUBR5,R4,R2與第一個類似，為譯碼部件結(jié)構(gòu)相關(guān)。Informationaboutsubr5Jr4Jir2sub「5』4』2IFIDAJr1Iood-i-Ow18Cycles:-2[2]Cycles:0(1]Code:0x00822822TernninaledsuccessJullpInPipelineInPipeline-StageIDIMAR<-PC(=Iqop+0k18]A<-R4[=0k183)IR<-M&m[IMAR](=Owaoe22822)B<-R2[=0w138)FirstCycle:-2LastCycle:???TotalCycles:>=3F匚c-F匚+4(=loop+0x1cj1Stallfs]becausecfstructuralHazard!NoStallsrequired.由于這三個相關(guān)在同一時鐘周期發(fā)生，所以算一個。4)計算暫停時鐘周期數(shù)占總執(zhí)行周期數(shù)的百分比。^WINDD{-[Statistics]|File-WindowExecuteMemoryConfigurationStati^tic&Total:139C^cle(s)execuled.IDexecutedby06Instructianfs].2Instruction[sjcurrentlyinPipeline.Haxd^waxeconflgur-ation:Menriorpsize:327E8BytesfaddEX-Stages:1,requiredCycles:2fmulEX-Stages:1requiredCycles:5fdivEX-Stages:LrequiredCycles:19Forwardingenabled.St已丄Is:FiAW軾恥；30[21.58^ofallCycle?),thereof;LDstalls:10[33.33^ofRAWstalls]Branch/Jumpstalls:10(33.33^ofRAWstalls)Floatingpointstalls;10(33.33^dRAW泅k)WAWstalls:0(0.00^ofallCycles)Structuralstalls:0(0.00^ofallCycles)Central躺b:9(SA7^ofallCycle?)Trapstalls:3(2.1E遼ofmilCjicles：]Total:425tall(sJ(30.22^ofallCycles)ConditionalBranches):Total:10(11.63^ofallInstructions].thereof:taken:9[90.00^ofallcond.Branches]nottaker:1(10.00^ofallcond.Branches)Load—-^Store—Instr-uctl口口乞：Tot^l:20(23.26^ofallInstructions],thereof:L^d;;20(100.00^ofLaad7S^re-lnstructions)Stores:□(0.00^ofLoad-/Stoie-lr$huctions]Floatingpointstageinstructions:Total:20(23.26^ofallInstructions^thereof:Additions:20(100.00^ofFloatingpointslageinst.)Multiplications;0(0.00^afFloatingpoint?tageirst]Divisiors:0(0.00^ofFloatingpointstageinst.)Ti-aps:Traps:1[1.1E嵬ofallInstructions)由上圖統(tǒng)計信息可知：此程序一共有139個周期，而程序一共循環(huán)10次，每次有5個結(jié)構(gòu)相關(guān)，但后三個在同一時鐘周期發(fā)生，一次相關(guān)暫停一個周期，所以，暫停時鐘周期數(shù)占總執(zhí)行周期數(shù)的百分比為：3*10/139=21.6%.5)論述結(jié)構(gòu)相關(guān)對CPU性能的影響，討論解決結(jié)構(gòu)相關(guān)的方法。發(fā)生結(jié)構(gòu)相關(guān)時，必然會導(dǎo)致