NumberSystems

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NumberRepresentation

BinaryHexadecimalDecimal

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TopicstobeDiscussed

• Howarenumericdataitemsactuallystoredincomputermemory?

• Howmuchspace(memorylocations)isallocatedforeachtypeofdata?– int,float,char,double,etc.

• Howarecharactersandstringsstoredinmemory?– Alreadydiscussed.

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NumberSystem::TheBasics

• Weareaccustomedtousingtheso-calleddecimalnumbersystem.– Tendigits::0,1,2,3,4,5,6,7,8,9– Everydigitpositionhasaweightwhichisapowerof10.– Base orradix is10.

• Example:

234 =2x102 +3x101 +4x100

250.67=2x102 +5x101 +0x100 +6x10-1 +7x10-2

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BinaryNumberSystem

• Twodigits:– 0and1.– Everydigitpositionhasaweightwhichisapowerof2.– Base orradix is2.

• Example:

110=1x22 +1x21 +0x20

101.01=1x22 +0x21 +1x20 +0x2-1 +1x2-2

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Binary-to-DecimalConversion

• Eachdigitpositionofabinarynumberhasaweight.– Somepowerof2.

• Abinarynumber:B=bn-1 bn-2 …..b1 b0 .b-1 b-2 …..b-m

Correspondingvalueindecimal:

D=Σ bi 2ii=-m

n-1

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Examples

1. 101011à 1x25 +0x24 +1x23 +0x22 +1x21 +1x20

=43(101011)2 =(43)10

2. .0101à 0x2-1 +1x2-2 +0x2-3 +1x2-4

=.3125(.0101)2 =(.3125)10

3. 101.11à 1x22 +0x21 +1x20 +1x2-1 +1x2-2

5.75(101.11)2 =(5.75)10

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Decimal-to-BinaryConversion

• Considertheintegerandfractionalpartsseparately.• Fortheintegerpart,

– Repeatedlydividethegivennumberby2,goonaccumulating theremainders,untilthenumberbecomeszero.

– Arrangetheremaindersinreverseorder.

• Forthefractionalpart,– Repeatedlymultiplythegivenfractionby2.

• Accumulatetheintegerpart(0or1).• Iftheintegerpartis1,chopitoff.

– Arrangetheintegerpartsintheorder theyareobtained.

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Example1::239

2 2392119--- 12 59--- 1229--- 12 14--- 127--- 02 3--- 121--- 120--- 1

(239)10 =(11101111)2

2 64232--- 02 16--- 028--- 02 4--- 022--- 02 1--- 020--- 1

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Example2::64

(64)10 =(1000000)2

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Example3::.634

.634x2=1.268

.268x2=0.536

.536x2=1.072

.072x2=0.144

.144x2=0.288::

(.634)10 =(.10100……)2

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Example4::37.0625

(37)10 =(100101)2

(.0625)10 =(.0001)2

∴ (37.0625)10 =(100101.0001)2

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HexadecimalNumberSystem

• Acompactwayofrepresentingbinarynumbers.

• 16differentsymbols(radix=16).0à 0000 8à 10001à 0001 9à 10012à 0010 Aà 10103à 0011 Bà 1011

4à 0100 Cà 11005à 0101 Dà 11016à 0110 Eà 1110

7à 0111 Fà 1111

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Binary-to-HexadecimalConversion

• Fortheintegerpart,– Scanthebinarynumberfromrighttoleft.– Translateeachgroupoffourbitsintothecorrespondinghexadecimaldigit.• Addleading zerosifnecessary.

• Forthefractionalpart,– Scanthebinarynumberfromlefttoright.– Translateeachgroupoffourbitsintothecorrespondinghexadecimaldigit.• Addtrailing zerosifnecessary.

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Examples

1. (1011 0100 0011)2 =(B43)16

2. (10 1010 0001)2 =(2A1)16

3. (.1000 010)2 =(.84)16

4. (101 .0101 111)2 =(5.5E)16

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Hexadecimal-to-BinaryConversion

• Translateeveryhexadecimaldigitintoits4-bitbinaryequivalent.– Discardleadingandtrailingzerosifdesired.

• Examples:(3A5)16 =(001110100101)2(12.3D)16 =(00010010.00111101)2(1.8)16 =(0001.1000)2

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NumberRepresentation

UnsignedandSignednumbers

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UnsignedBinaryNumbers

• Ann-bitbinarynumberB=bn-1bn-2 ….B2b1b0

(2n distinctcombinationsarepossible,0to2n−1)

• Forn=3,thereare8distinctcombinations.– 000,001,010,011,100,101,110,111

• Rangeofnumbersthatcanberepresentedn=8 è 0to28−1(255)n=16 è 0to216−1(65535)n=32 è 0to232−1(4294967295)

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SignedIntegerRepresentation

• Manyofthenumericaldataitemsthatareusedinaprogramaresigned(positiveornegative).– Question::Howtorepresentsign?

• Threepossibleapproaches:a)Sign-magnituderepresentationb)One’scomplementrepresentation

c) Two’scomplementrepresentation

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Sign-magnitudeRepresentation

• Forann-bitnumberrepresentation– Themostsignificantbit(MSB)indicatessign

0à positive1à negative

– Theremainingn-1bitsrepresentmagnitude.

b0b1bn-2bn-1

MagnitudeSign

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Example::n=4

0000à +00001à +10010à +20011à +30100à +40101à +50110à +60111à +7

1000à -01001à -11010à -21011à -31100à -41101à -51110à -61111à -7

15distinctnumberscanberepresented

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Contd.

• Rangeofnumbersthatcanberepresented:Maximum::+(2n-1 – 1)Minimum::– (2n-1 – 1)

• Aproblem:Twodifferentrepresentationsofzero.

+0 à 0 000….0–0 à 1 000….0

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One’sComplementRepresentation

• Basicidea:– Positivenumbersarerepresentedexactlyasinsign-magnitudeform.

– Negativenumbersarerepresentedin1’scomplementform.

• Howtocomputethe1’scomplementofanumber?– Complementeverybitofthenumber(1à0and0à1).– MSBwillindicatethesignofthenumber.

0à positive

1à negative

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Example::n=4

0000à +00001à +10010à +20011à +30100à +40101à +50110à +60111à +7

1000à -71001à -61010à -51011à -41100à -31101à -21110à -11111à -0

Tofindtherepresentationof,say,-4,firstnotethat

+4=0100

−4=1’scomplementof0100=1011

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Contd.• Rangeofnumbersthatcanberepresented:

Maximum::+(2n-1 – 1)Minimum::– (2n-1 – 1)

• Aproblem:Twodifferentrepresentationsofzero.

+0à 0000….0−0à 1111….1

• Advantageof1’scomplementrepresentation– Subtractioncanbedoneusingaddition.– Leadstosubstantial savingincircuitry.– Signextensionispossibletoincreasenumberofbitstorepresent.

-3=1100(in4bits)=11111100(in8bits)+3=0011(in4bits)=00000011(in8bits)

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Two’sComplementRepresentation

• Basicidea:– Positivenumbersarerepresentedexactlyasinsign-magnitudeform.

– Negativenumbersarerepresentedin2’scomplementform.

• Howtocomputethe2’scomplementofanumber?– Complementeverybitofthenumber(1à0and0à1),andthenadd1totheresultingnumber.

– MSBwillindicatethesignofthenumber.0à positive

1à negative

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Example::n=4

0000à +00001à +10010à +20011à +30100à +40101à +50110à +60111à +7

1000à -81001à -71010à -61011à -51100à -41101à -31110à -21111à -1

Tofindtherepresentationof,say,-4,firstnotethat

+4=0100

−4=2’scomplementof0100=1011+1=1100

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Contd.• Rangeofnumbersthatcanberepresented:

Maximum::+(2n-1 – 1)Minimum::– 2n-1

• Advantage:– Uniquerepresentationofzero.– Subtractioncanbedoneusingaddition.– Leadstosubstantial savingincircuitry.– Signextensionispossibletoincreasenumberofbitstorepresent.

-3=1101(in4bits)=11111101(in8bits)+3=0011(in4bits)=00000011(in8bits)

• Almostallcomputerstodayusethe2’scomplementrepresentationforstoringnegativenumbers.

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Contd.

• InC(typicalvalues):– shortint

• 16bitsè +(215–1)to–215

– int• 32bitsè +(231–1)to–231

– longint• 64bitsè +(263–1)to–263

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Binaryoperations

AdditionSubtractionusingaddition

Binaryaddition

• Rulesforaddingtwobits:0+0=00+1=11+0=11+1=10,i.e.,0withcarryof1

• Additionexamplesforunsignednumbers:0 1 0 1 0 1 1 1 1 0 0 10 0 0 1 0 0 1 1 1 0 1 0======= ======= =========0 1 1 0 1 0 1 0 1 0 0 1 1

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Carry

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SubtractionUsingAddition::1’sComplement

• HowtocomputeA– B?– Computethe1’scomplementofB(say,B1).

– ComputeR=A+B1

– Ifthecarryobtainedafteradditionis‘1’• AddthecarrybacktoR(calledend-aroundcarry).

• Thatis,R=R+1.

• Theresultisapositivenumber.

Else• Theresultisnegative,andisin1’scomplementform.

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Example1::6– 2

1’scomplementof2=1101

6 :: 0 1 1 0-2 :: 1 1 0 1

1 0 0 1 11

0 1 0 0 à +4

End-aroundcarry

Assume4-bitrepresentations.

Sincethereisacarry,itisaddedbacktotheresult.

Theresultispositive.

RB1A

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Example2::3– 5

1’scomplementof5=1010

3 :: 0 0 1 1-5 :: 1 0 1 0

1 1 0 1 Assume4-bitrepresentations.

Sincethereisnocarry,theresultisnegative.

1101isthe1’scomplementof0010,thatis,itrepresents–2.

AB1

R

-2

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SubtractionUsingAddition::2’sComplement

• HowtocomputeA– B?– Computethe2’scomplementofB(say,B2).

– ComputeR=A+B2

– Ifthecarryobtainedafteradditionis‘1’• Ignorethecarry.

• Theresultisapositivenumber.

Else• Theresultisnegative,andisin2’scomplementform.

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Example1::6– 2

2’scomplementof2=1101+1=1110

6 :: 0 1 1 0-2 :: 1 1 1 0

1 0 1 0 0

Assume4-bitrepresentations.

Presenceofcarryindicatesthattheresultispositive.

Noneedtoaddtheend-aroundcarrylikein1’scomplement.

AB2

R

Ignorecarry +4

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Example2::3– 5

2’scomplementof5=1010+1=1011

3 :: 0 0 1 1-5 :: 1 0 1 1

1 1 1 0 Assume4-bitrepresentations.

Sincethereisnocarry,theresultisnegative.

1110isthe2’scomplementof0010,thatis,itrepresents–2.

AB2

R

-2

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Floating-pointnumberrepresentation

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Floating-pointNumbers• Therepresentationsdiscussedsofarappliesonlyto

integers.– Cannotrepresentnumberswithfractionalparts.

• Wecanassumeadecimalpointbeforea2’scomplementnumber.– Inthatcase,purefractions(withoutintegerparts)canbe

represented.

• Wecanalsoassumethedecimalpointsomewhereinbetween.– Thislacksflexibility.– Verylargeandverysmallnumberscannotberepresented.

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RepresentationofFloating-PointNumbers• Afloating-pointnumberFisrepresentedbyadoublet<M,E>:

F=MxBE

• B: exponentbase(usually2)• M: mantissa• E:exponent

– Misusuallyrepresentedin2’scomplementform,withanimplieddecimalpointbeforeit.

• Forexample,Indecimal,

0.235x106

Inbinary,0.101011x20110

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Example::32-bitrepresentation

– Mrepresentsa2’scomplementfraction1>M>−1

– Erepresentstheexponent(in2’scomplementform)127>E>−128

• Pointstonote:– Thenumberofsignificantdigits dependsonthenumberofbits

inM.• 6significantdigitsfor24-bitmantissa.

– Therange ofthenumberdependsonthenumberofbitsinE.• 1038 to10−38 for8-bitexponent.

M E

24 8

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AWarning

• Therepresentationforfloating-pointnumbersasshownisjustforillustration.

• Theactualrepresentationisalittlemorecomplex.

• InC:– float::32-bitrepresentation– double::64-bitrepresentation

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RepresentationofCharactersandStrings

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RepresentationofCharacters

• Manyapplications havetodealwithnon-numericaldata– Charactersandstrings.– Theremustbeastandardmechanismtorepresentalphanumericand

othercharactersinmemory.

• Threestandards inuse:– ExtendedBinaryCodedDecimalInterchangeCode(EBCDIC)

• UsedinolderIBMmachines.

– AmericanStandardCodeforInformationInterchange(ASCII)• Mostwidelyusedtoday.

– UNICODE• Usedtorepresentallinternationalcharacters.• UsedbyJava.

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ASCIICode

• Eachindividualcharacterisnumericallyencodedintoaunique7-bitbinarycode.– Atotalof27 or128differentcharacters.

– Acharacterisnormallyencodedinabyte(8bits),withtheMSBnotbeenused.

• Thebinaryencodingofthecharactersfollowaregularordering.– Digitsareorderedconsecutivelyintheirpropernumericalsequence

(0to9).– Letters(uppercaseandlowercase)arearrangedconsecutivelyin

theirproperalphabeticorder.

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SomeCommonASCIICodes

‘A’::41(H) 65(D)‘B’::42(H) 66(D)………..‘Z’::5A(H) 90(D)

‘a’::61(H) 97(D)‘b’::62(H) 98(D)………..‘z’::7A(H) 122(D)

‘0’::30(H) 48(D)‘1’::31(H) 49(D)………..‘9’::39(H) 57(D)

‘(‘::28(H) 40(D)‘+’::2B(H) 43(D)‘?’::3F(H) 63(D)‘\n’::0A(H) 10(D)‘\0’::00(H) 00(D)

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CharacterStrings

• Twowaysofrepresentingasequenceofcharactersinmemory.– Thefirstlocationcontainsthenumberofcharactersinthestring,followedbytheactualcharacters.

– Thecharactersfollowoneanother,andisterminatedbyaspecialdelimiter.

oeH5 ll

⊥leH ol

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StringRepresentationinC

• InC,thesecondapproachisused.– The‘\0’characterisusedasthestringdelimiter.

• Example:“Hello”à

• Anullstring“”occupiesonebyteinmemory.– Onlythe‘\0’character.

‘\0’leH ol