Integers II, Floating Point I - University of Washington

CSE351, Summer 2020L05: Integers II, Floating Point I

IntegersII,FloatingPointICSE351Summer2020

Instructor: TeachingAssistants:PorterJones AmyXu

CallumWalkerSamWolfsonTimMandzyuk

http://xkcd.com/1953/


Administrivia

v Questionsdoc:https://tinyurl.com/CSE351-7-1v hw4andhw5dueMonday7/6– 10:30amv hw6andhw7dueFriday7/10– 10:30am

§ WillpostMonday’sslideslatertodaysoyoucangetstarted

v Lab1adueMonday(7/6)(trytofinishbyFriday!)§ Submitpointer.c andlab1Areflect.txt toGradescope

v Lab1breleasedtomorrow,due7/10§ Bitmanipulationproblemsusingcustomdatatype§ Today’sbonusslideshavehelpfulexamples,omorrow’ssectionwillhavehelpfulexamplestoo

2


Gradescope LabTurnin

v MakesureyoupasstheFileandCompilationCheck!

v Doesn’tindicateifyoupassedalltests,justindicatesthatallthecorrectfileswerefoundandtherewerenocompilationorruntimeerrors.

v Usethetestingprogramsweprovidetocheckyoursolutionforcorrectness(onattu ortheVM)

3


QuickAside:CMacros

v Lab1bwillhaveyouusesomeCmacrosforbitmasksv Syntaxisoftheform:

#define NAME expressionv Cannowuse“NAME”insteadof“expression”incodev Usefultohelpwithreadability/factoringincode

§ Especiallyusefulfordefiningconstantssuchasbitmasks!

v AreNOTexactlythesameasaconstantinJava§ Doesnaïvecopyandreplacebefore compilation.§ Everywherethecharacters“NAME”appearinthecode,thecharacters“expression”willnowappearinstead.

v SeeLecture4(IntegersI)slidesforexampleusages4


Integers

v Binaryrepresentationofintegers§ Unsignedandsigned

v Shiftingandarithmeticoperations– usefulforLab1av InC:Signed,UnsignedandCastingv Consequencesoffinitewidthrepresentations

§ Overflow,signextension

5


Two’sComplementArithmetic

v Thesameadditionprocedureworksforbothunsignedandtwo’scomplementintegers§ Simplifieshardware: onlyonealgorithmforaddition§ Algorithm: simpleaddition,discardthehighestcarrybit

• Calledmodularaddition:resultissummodulo 2!

v 4-bitExamples:

6

HW. TC.

0100+0011

4+3

= 0111 = 7

HW. TC. HW. TC.

1100+0011

-4+3

0100+1101

4-3

= 1111 =-1 =10001 = 1


WhyDoesTwo’sComplementWork?

v Forallrepresentablepositiveintegers𝑥,wewant:

§ Whatarethe8-bitnegativeencodingsforthefollowing?

7

bitrepresentationof –𝑥+ bitrepresentationof –𝑥

0 (ignoringthecarry-outbit)

00000001+ ????????00000000

00000010+ ????????00000000

11000011+ ????????00000000


WhyDoesTwo’sComplementWork?

v Forallrepresentablepositiveintegers𝑥,wewant:

§ Whatarethe8-bitnegativeencodingsforthefollowing?

8

bitrepresentationof –𝑥+ bitrepresentationof –𝑥

0 (ignoringthecarry-outbit)

00000001+ 11111111100000000

00000010+ 11111110100000000

11000011+ 00111101100000000

Thesearethebitwisecomplementplus1!-x == ~x + 1


UMax – 1

0

TMax

TMin

–1–2

0/UMin

UMax

TMaxTMax +1

2’sComplementRange

UnsignedRange

Signed/UnsignedConversionVisualizedv Two’sComplement→ Unsigned

§ OrderingInversion§ Negative→ BigPositive

9


ValuesToRememberv UnsignedValues

§ UMin = 0b00…0= 0

§ UMax = 0b11…1= 2! − 1

v Example: Valuesfor𝑤 = 64

10

v Two’sComplementValues§ TMin = 0b10…0

= −2!"#

§ TMax = 0b01…1= 2!"# − 1

§ −1 = 0b11…1

Decimal HexUMax 18,446,744,073,709,551,615 FF FF FF FF FF FF FF FF

TMax 9,223,372,036,854,775,807 7F FF FF FF FF FF FF FF

TMin -9,223,372,036,854,775,808 80 00 00 00 00 00 00 00

-1 -1 FF FF FF FF FF FF FF FF

0 0 00 00 00 00 00 00 00 00


Integers


v Shiftingandarithmeticoperations– usefulforLab1av InC:Signed,UnsignedandCastingv Consequencesoffinitewidthrepresentations

§ Overflow,signextension

11


InC:Signedvs.Unsigned

v Casting§ Bitsareunchanged,justinterpreteddifferently!

• int tx,ty;• unsigned int ux,uy;

§ Explicit casting• tx =(int)ux;• uy =(unsigned int)ty;

§ Implicit castingcanoccurduringassignmentsorfunctioncalls• tx =ux;• uy =ty;

12


CastingSurprises

v Integerliterals(constants)§ Bydefault,integerconstantsareconsideredsigned integers

• Hexconstantsalreadyhaveanexplicitbinaryrepresentation

§ Use“U”(or“u”)suffixtoexplicitlyforceunsigned• Examples: 0U,4294967259u

v ExpressionEvaluation§ Whenyoumixedunsignedandsignedinasingleexpression,thensignedvaluesareimplicitlycasttounsigned

§ Includingcomparisonoperators<,>,==,<=,>=

13

!!!


CastingSurprises

v 32-bitexamples:§ TMin =-2,147,483,648,TMax =2,147,483,647

14

!!!

LeftConstant Order RightConstant Interpretation000000000000000000000000000000000

0U00000000000000000000000000000000

-111111111111111111111111111111111

000000000000000000000000000000000

-111111111111111111111111111111111

0U00000000000000000000000000000000

214748364701111111111111111111111111111111

-214748364810000000000000000000000000000000

2147483647U01111111111111111111111111111111

-214748364810000000000000000000000000000000

-111111111111111111111111111111111

-211111111111111111111111111111110

(unsigned)-111111111111111111111111111111111

-211111111111111111111111111111110

214748364701111111111111111111111111111111

2147483648U10000000000000000000000000000000

214748364701111111111111111111111111111111

(int)2147483648U10000000000000000000000000000000


Integers


v Shiftingandarithmeticoperations– usefulforLab1av InC:Signed,UnsignedandCastingv Consequencesoffinitewidthrepresentations§ Overflow,signextension

15


ArithmeticOverflowv Whenacalculationproducesaresult

thatcan’tberepresentedinthecurrentencodingscheme§ Integerrangelimitedbyfixedwidth§ Canoccurinboththepositiveandnegative

directions

v CandJavaignoreoverflowexceptions§ Youendupwithabadvalueinyour

programandnowarning/indication…oops!

16

Bits Unsigned Signed0000 0 00001 1 10010 2 20011 3 30100 4 40101 5 50110 6 60111 7 71000 8 -81001 9 -71010 10 -61011 11 -51100 12 -41101 13 -31110 14 -21111 15 -1


Overflow:Unsigned

v Addition: dropcarrybit(−2!)

v Subtraction: borrow(+2!)

17

15+ 2171

1111+ 001010001

00000001

0011

11111110

1100

10111010

1000 01110110

0100

0010

01011001

1101

01

2

3

4

5

678

9

10

11

12

13

1415

Unsigned

1- 2-115

10001- 00101111

±2! becauseofmodulararithmetic


Overflow:Two’sComplement

v Addition: (+)+(+)=(−)result?

v Subtraction: (−)+(−)=(+)?

18

00000001

0011

11111110

1100

10111010

1000 01110110

0100

0010

01011001

1101

0+ 1

+ 2

+ 3

+ 4

+ 5

+ 6+ 7– 8

– 7

– 6

– 5

– 4

– 3

– 2– 1

Forsigned: overflowifoperandshavesamesignandresult’ssignisdifferent

Two’sComplement

6+ 39-7

0110+ 00111001

-7- 3-106

1001- 00110110


SignExtension

v Whathappensifyouconvertasigned integraldatatypetoalargerone?§ e.g. char→ short→ int→ long

v 4-bit→ 8-bitExample:§ PositiveCase

• Add0’s?

§ NegativeCase?

19

4-bit: 0010 = +28-bit: ????0010 = ?00000010 +2✓


PollingQuestion[Int II- a]

v Whichofthefollowing8-bitnumbershasthesamesigned valueasthe4-bitnumber0b1100?§ Underlineddigit=MSB§ Voteathttp://pollev.com/pbjones

A. 0b00001100B. 0b10001100C. 0b11111100D. 0b11001100E. We’relost…

20


SignExtension

v Task: Givena𝑤-bitsignedintegerX,convertitto𝑤+𝑘-bitsignedintegerX′ withthesamevalue

v Rule: Add 𝑘 copiesofsignbit§ Let𝑥" bethe𝑖-th digitofX inbinary§ X# = 𝑥$%&, … , 𝑥$%&, 𝑥$%&, 𝑥$%', … , 𝑥&, 𝑥(

21

𝑘 copiesofMSB

• • •X

Xʹ • • • • • •

• • •

𝑤

𝑘 𝑤

originalX


SignExtensionExample

v Convertfromsmallertolargerintegraldatatypesv Cautomaticallyperformssignextension

§ Javatoo

22

short int x = 12345;int ix = (int) x; short int y = -12345;int iy = (int) y;

Var Decimal Hex Binaryx 12345 30 39 00110000 00111001

ix 12345 00 00 30 39 00000000 00000000 00110000 00111001

y -12345 CF C7 11001111 11000111

iy -12345 FF FF CF C7 11111111 11111111 11001111 11000111


PracticeQuestion

v Assumeweareusing8-bitarithmetic:§ x == (unsigned char) x

§ x >= 128U

§ x != (x>>2)<<2

§ x == -x• Hint:therearetwosolutions

§ (x < 128U) && (x > 0x3F)

23

Forthefollowingexpressions,findavalueofsigned char x,ifthereexistsone,thatmakestheexpressionTRUE.Comparewithyourneighbor(s)!


Aside:UnsignedMultiplicationinC

v StandardMultiplicationFunction§ Ignoreshighorder𝑤 bits

v ImplementsModularArithmetic§ UMultw(u ,v)=u ·v mod2w

24

• • •

• • •

u

v*

• • •u · v

• • •

TrueProduct:𝟐𝒘 bits

Operands:𝒘 bits

Discard𝑤 bits:𝒘 bits

UMultw(u , v)

• • •


Aside:Multiplicationwithshiftandaddv Operationu<<k givesu*2k

§ Bothsignedandunsigned

v Examples:§ u<<3 == u * 8§ u<<5 - u<<3 == u * 24

§ Mostmachinesshiftandaddfasterthanmultiply• Compilergeneratesthiscodeautomatically

25

• • •u

2k*u · 2kTrueProduct:𝒘+ 𝒌 bits

Operands:𝒘 bits

Discard𝑘 bits:𝒘 bits UMultw(u , 2k)

0 0 1 0 0 0••• •••k

• • • 0 0 0•••

TMultw(u , 2k)0 0 0••••••


NumberRepresentationRevisited

v Whatcanwerepresentsofar?§ SignedandUnsignedIntegers§ Characters(ASCII)§ Addresses

v Howdoweencodethefollowing:§ Realnumbers(e.g. 3.14159)§ Verylargenumbers(e.g. 6.02×1023)§ Verysmallnumbers(e.g. 6.626×10-34)§ Specialnumbers(e.g.∞,NaN)

26

FloatingPoint


FloatingPointTopics

v Fractionalbinarynumbersv IEEEfloating-pointstandardv Floating-pointoperationsandroundingv Floating-pointinC

v Therearemanymoredetailsthatwewon’tcover§ It’sa58-pagestandard…

27


RepresentationofFractions

v “BinaryPoint,”likedecimalpoint,signifiesboundarybetweenintegerandfractionalparts:

Example6-bitrepresentation:

v Example: 10.10102 =1×21 +1×2-1 +1×2-3 =2.62510

28

xx.yyyy21 20 2-1 2-2 2-3 2-4


RepresentationofFractions

v “BinaryPoint,”likedecimalpoint,signifiesboundarybetweenintegerandfractionalparts:

Example6-bitrepresentation:

v Inthis6-bitrepresentation:§ Whatistheencodingandvalueof

thesmallest(mostnegative)number?

§ Whatistheencodingandvalueofthelargest(mostpositive)number?

§ Whatisthesmallestnumbergreaterthan2thatwecanrepresent?

29

xx.yyyy21 20 2-1 2-2 2-3 2-4


• • •b–1.

FractionalBinaryNumbers

v Representation§ Bitstorightof“binarypoint”representfractionalpowersof2§ Representsrationalnumber:

bi bi–1 b2 b1 b0 b–2 b–3 b–j• • •• • •124

2i–12i

• • •

1/21/41/8

2–j

30


FractionalBinaryNumbers

v Value Representation§ 5and3/4§ 2and7/8§ 47/64

v Observations§ Shiftleft=multiplybypowerof2§ Shiftright=dividebypowerof2§ Numbersoftheform0.111111…2 arejustbelow1.0

§ 1/2+1/4+1/8+…+1/2i +…➙ 1.0§ Usenotation1.0– ε

101.11210.11120.1011112

31


LimitsofRepresentation

v Limitations:§ Evengivenanarbitrarynumberofbits,canonlyexactlyrepresentnumbersoftheformx*2y (ycanbenegative)

§ Otherrationalnumbershaverepeatingbitrepresentations

Value: BinaryRepresentation:• 1/3=0.333333…10 = 0.01010101[01]…2

• 1/5= 0.001100110011[0011]…2

• 1/10= 0.0001100110011[0011]…2

32


Fixed PointRepresentation

v Impliedbinarypoint. Twoexampleschemes:#1:thebinarypointisbetweenbits2and3

b7 b6 b5b4 b3 [.] b2 b1 b0#2:thebinarypointisbetweenbits4and5

b7 b6 b5 [.] b4 b3 b2 b1 b0v Whereverweputthebinarypoint,withfixedpointrepresentationsthereisatradeoffbetweentheamountofrangeandprecisionwehave

v Fixedpoint=fixedrange andfixedprecision§ range:differencebetweenlargestandsmallestnumberspossible§ precision:smallestpossibledifferencebetweenanytwonumbers

v Hardtopickhowmuchyouneedofeach!33


Floating PointRepresentation

v Analogoustoscientificnotation§ InDecimal:

• Not12000000,but 1.2x107 InC:1.2e7• Not0.0000012,but 1.2x10-6 InC:1.2e-6

§ InBinary:• Not11000.000,but1.1x24

• Not0.000101,but1.01x2-4

v Wehavetodivvyupthebitswehave(e.g.,32)among:§ thesign(1bit)§ themantissa(significand)§ theexponent

34


ScientificNotation(Decimal)

v Normalizedform:exactlyonedigit(non-zero)toleftofdecimalpoint

v Alternativestorepresenting1/1,000,000,000§ Normalized: 1.0×10-9

§ Notnormalized: 0.1×10-8,10.0×10-10

35

6.0210 × 1023

radix(base)decimalpoint

exponentmantissa


ScientificNotation(Binary)

v Computerarithmeticthatsupportsthiscalledfloatingpoint duetothe“floating”ofthebinarypoint§ DeclaresuchvariableinCasfloat (ordouble)

36

1.012 × 2-1

radix(base)binarypoint

exponentmantissa


ScientificNotationTranslationv Convertfromscientificnotationtobinarypoint

§ Performthemultiplicationbyshiftingthedecimaluntiltheexponentdisappears• Example:1.0112×24 =101102 =2210• Example:1.0112×2-2 =0.010112 =0.3437510

v Convertfrombinarypointtonormalized scientificnotation§ Distributeoutexponentsuntilbinarypointistotherightofasingledigit

• Example:1101.0012 =1.1010012×23

v Practice:Convert11.37510 tonormalizedbinaryscientificnotation

37


Summary

v SignandunsignedvariablesinC§ Bitpatternremainsthesame,justinterpreted differently§ Strangethingscanhappenwithourarithmeticwhenweconvert/castbetweensignandunsignednumbers• Typeofvariablesaffectsbehaviorofoperators(shifting,comparison)

v Wecanonlyrepresentsomanynumbersin𝑤 bits§ Whenweexceedthelimits,arithmeticoverflow occurs§ Signextension triestopreservevaluewhenexpanding

v Floatingpointapproximatesrealnumbers§ WewilldiscussmoredetailsonMonday!

38


Someexamplesofusingshiftoperatorsincombinationwithbitmasks,whichyoumayfindhelpfulforLab1.

v Extractthe2nd mostsignificantbyteofanintv Extractthesignbitofasignedintv ConditionalsasBooleanexpressions

39


UsingShiftsandMasks

v Extractthe2nd mostsignificantbyte ofanint:§ Firstshift,thenmask:(x>>16) & 0xFF

§ Orfirstmask,thenshift:(x & 0xFF0000)>>16

40

0xFF 00000000 00000000 00000000 11111111

(x>>16) & 0xFF 00000000 00000000 00000000 00000010

x>>16 00000000 00000000 00000001 00000010

x 00000001 00000010 00000011 00000100

x & 0xFF0000 00000000 00000010 00000000 00000000

(x&0xFF0000)>>16 00000000 00000000 00000000 00000010

0xFF0000 00000000 11111111 00000000 00000000

x 00000001 00000010 00000011 00000100


UsingShiftsandMasks

v Extractthesignbitofasignedint:§ Firstshift,thenmask:(x>>31) & 0x1

• Assumingarithmeticshifthere,butthisworksineithercase• Needmasktoclear 1spossiblyshiftedin

41

x 00000001 00000010 00000011 00000100

x>>31 00000000 00000000 00000000 00000000

0x1 00000000 00000000 00000000 00000001

(x>>31) & 0x1 00000000 00000000 00000000 00000000

x 10000001 00000010 00000011 00000100

x>>31 11111111 11111111 11111111 11111111

0x1 00000000 00000000 00000000 00000001

(x>>31) & 0x1 00000000 00000000 00000000 00000001

0

0

11


UsingShiftsandMasks

v ConditionalsasBooleanexpressions§ Forint x,whatdoes(x<<31)>>31 do?

§ Canuseinplaceofconditional:• InC:if(x) {a=y;} else {a=z;} equivalenttoa=x?y:z;• a=(((x<<31)>>31)&y) | (((!x<<31)>>31)&z);

42

x=!!123 00000000 00000000 00000000 00000001

x<<31 10000000 00000000 00000000 00000000

(x<<31)>>31 11111111 11111111 11111111 11111111

!x 00000000 00000000 00000000 00000000

!x<<31 00000000 00000000 00000000 00000000

(!x<<31)>>31 00000000 00000000 00000000 00000000

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Integers II, Floating Point I - University of Washington