计算机组成与结构：chapter9 Computer Arithmetic

ComputerOrganization&ArchitectureChapter9

ComputerArithmetic9.1Arithmetic&LogicUnitDoesthecalculationsEverythingelseinthecomputeristheretoservethisunitController,registers,memory,I/OHandlesintegers,floatingpoint(real)numbersMaybeseparateFPU(mathsco-processor)MaybeonchipseparateFPU(486DX+)ALUInputsandOutputs4.2IntegerRepresentationOnlyhave0&1torepresenteverythingPositivenumbersstoredinbinarye.g.41=00101001NominussignNoperiod(.)n-bitbinarydigitsan-1an-2….a1a0denotesanunsignedintegerASign-MagnitudeRepresentationLeftmostbitissignbit0meanspositive1meansnegative+18=00010010-18=10010010ProblemsNeedtoconsiderbothsignandmagnitudeinarithmeticTworepresentationsofzero(+0and-0)RarelyusedTwo’sComplementRepresentation+3=00000011+2=00000010+1=000000010=00000000-1=11111111-2=11111110-3=11111101Complementofpositivenumber=unsignednumbercomplementofnegativenumber=bitwisenotunsignednumber+1GeometricDepictionofTwosComplementIntegersBenefitsOnerepresentationofzeroArithmeticworkseasily(seelater)Negatingisfairlyeasy3=00000011Booleancomplementgives 11111100Add1toLSBNegationSpecialCase10=00000000Bitwisenot11111111Add1to+1Result100000000Overflowisignored,so:0=0NegationSpecialCase2-128=10000000bitwisenot01111111Add1toLSB+1Result10000000So:-(-128)=-128XMonitorMSB(signbit)ItshouldchangeduringnegationRangeofNumbers8bit2’scomplement+127=01111111=27-1-128=10000000=-2716bit2scomplement+32767=01111111111111111=215-1-32768=10000000000000000=-215ConversionBetweenLengthsPositivenumberpackwithleadingzeros+18=00010010+18=0000000000010010Negativenumberspackwithleadingones-18=11101110-18=1111111111101110Extensionruleofcomplement:leftpacksignbitProofForabinarynumber:A,an-1an-2….a1a0

IfAispositive,an-1=0,then：IfAisnegative,an-1=1,then:9.3IntegerArithmeticNegationSign-magnitude:invertthesignbitComplement:bitwisenot+1E.g.18=0001001011101101+111101110-18=1110111000010001+100010010An-bitbinary:[-2n-1，2n-1-1]proofAddition&SubtractionCommentonAdditionandSubtractionNormalbinaryadditionMonitorsignbitforoverflowOverflowRule:iftwopositivenumbersortwonegativenumbersareadded,iffthesignofresultbecomesinversion,overflowoccurs.

SubtractionRule:Takecomplementofsubtrahendandaddtominuendi.e.a-b=a+(-b)NosubtractorinacomputerSoweonlyneedadditionandcomplementcircuitsHardwareforAdditionandSubtractionMultiplicationUnsignedintegersbyPencil-and-paper1011Multiplicand(11dec)x1101Multiplier(13dec) 1011Partialproducts0000Note:ifmultiplierbitis1copy1011 multiplicand(placevalue)1011 otherwisezero10001111Product(143dec)Note:needdoublelengthresultBycomputerMoreefficientAdditiononpartialproductratherthanwaitinguntiltheendNoneedforstorageallthepartialproducts,fewerregistersareneededSavesometimeForeach1onthemultiplier,anadditionandashiftoperationarerequired,butforeach0,onlyashiftisrequiredBlockDiagramofMultiplierSpecification3n-bitregisters:Q,M,A;one1-bitregister:C;initiallysetA,Cto0ControllogicreadsthebitsofthemultiplieroneatatimeIfQ0=1,thenthemultiplicandisaddedtotheAandtheresultsisstoredintheAThen,allofthebitsofC,AandQareshiftedrightonebit,Q0islostIfQ0=0,justrightshiftTheresulting2n-bitproductisintheAandQExecutionofExampleFlowchartforUnsignedBinaryMultiplicationMultiplyingNegativeNumbersStraightforwardmultiplicationdoesnotwork!1001(-7)x0011(3)1001100111011(27)(-5)XSolution1ConverttopositiveifrequiredMultiplyasaboveIfsignsweredifferent,negateanswerSolution2Booth’salgorithmBooth’sAlgorithmExampleofBooth’sAlgorithmExampleofBooth’sAlgorithm0000001101001InitialValues0011100111001ShiftRight0001110011001ShiftRight

1010110011001A=A+M1101011001001ShiftRight

0111001101001A=A-M

AQQ-1M1110101101001ShiftRight

ExampleofBooth’sAlgorithm(3)DivisionMorecomplexthanmultiplicationNegativenumbersarereallybad!BasedonlongdivisionDivisionofUnsignedBinaryIntegersFlowchartforUnsignedBinaryDivisionNegativeNumbersDivisionThemethodgiveninourbookistootroublesomeAsimpleschemeConverttopositiveifrequireddivisionasaboveIfsignsweredifferent,negateanswerSignIntegerdivision9.4FloatingPointRepresentationRealNumbersNumberswithfractionsCouldbedoneinpurebinary1001.1010=24+20+2-1+2-3=9.625Whereisthebinarypoint?Fixed?VerylimitedMoving?Howdoyoushowwhereitis?Principles±SB±E±:SignS:significandE:exponent,incomputer,usuallybiasedrepresentation±SBE‘B:baseisimplicit,neednotbestoredTypically,thebias=2k-1-1k:numberofbitsofexponentE.g.8-bitfield,thebiasis127Trueexponentvalues:[-127,128]Exponentvalue=trueexponent+biasThus,storedexponentvalueisunsignedTypical32-bitFloat-pointFormatSignsforExponentExponentisinexcessorbiasednotatione.g.Excess(bias)127means8bitexponentfieldPurevaluerange0-255Subtract127togetcorrectvalueRange-127to+128NormalizationFPnumbersareusuallynormalizedi.e.exponentisadjustedsothatleadingbit(MSB)ofmantissais1Sinceitisalways1thereisnoneedtostoreitc.f.Scientificnotationwherenumbersarenormalizedtogiveasingledigitbeforethedecimalpointe.g.3.123x103FPRangesFPRangesFora32bitnumber8bitexponent+/-21283.4x1038AccuracyTheeffectofchanginglsbofmantissa23bitmantissa2-231.2x10-7About6decimalplacesRelation:whenbitsofEareincreased,weexpandtherangeofexpressibledata,butresultinreductionofprecision(Elarger,mantissasmaller,scopelarger,precisionsmaller)Trade-offbetweenrangeandprecision

DensityofFloatingPointNumbers9.5Floating-pointarithmeticForadditionandsubtraction,wemustensurethatbothoperandshavethesameexponentvalue,whilemultiplicationanddivisionaremoresimple.X=XSBXe,Y=YSBYeX+Y=(XSBXe-Ye+YS)

BYeX-Y=(XSBXe-Ye-YS)

BYeXY=(XSYS)

BXe+Ye

X/Y=(XSYS)

BXe-YeFloating-pointoperationmaycausesomeproblems:ExponentoverflowExponentunderflowSignificantunderflowSignificantoverflowFPArithmetic+/-CheckforzerosAlignsignificands(adjustingexponents)ShiftingtheradixpointonthesmalleroperandOncetheoverflowoccurs,theinfluenceissmaller

AddorsubtractsignificandsNormalizeresultFPAdd&SubFPArit