UCLA CS 33 - Reinman - Spring 2014

5/21/2018 UCLA CS 33 - Reinman - Spring 2014

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SomenotesadoptedfromBryantandOHallaron

Integers

Chapter2ofB&O


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Encoding Integers

short int x = 15213;

short int y = -15213;

Cshor t 2byteslong

SignBit

For2scomplement,mostsignificantbitindicatessign

0fornonnegative

1fornegative

B2T(X) xw1 2w1 xi2

i

i0

w2

B2U(X) xi2i

i0

w1

Unsigned Twos Complement

Sign

Bit

Decimal Hex Binaryx 15213 3B 6D 00111011 01101101y -15213 C4 93 11000100 10010011


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Numeric Ranges UnsignedValues

UMin = 0

0000 UMax = 2w 1

1111

TwosComplementValues

TMin = 2w1

1000 TMax = 2w1 1

0111

OtherValues Minus1

1111

Decimal Hex BinaryUMax 65535 FF FF 11111111 11111111TMax 32767 7F FF 01111111 11111111TMi n -32768 80 00 10000000 00000000- 1 -1 FF FF 11111111 111111110 0 00 00 00000000 00000000

Values for W = 16


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Values for Different Word Sizes

Observations

|TMin| = TMax+1Asymmetricrange

UMax = 2*TMax+1

CProgramming

#i ncl ude

Declaresconstants,e.g.,

ULONG_MAX

LONG_MAX

LONG_MI N

Valuesplatformspecific

W

8 16 32 64

UMax 255 65,535 4,294,967,295 18,446,744,073,709,551,615

TMax 127 32,767 2,147,483,647 9,223,372,036,854,775,807

TMin -128 -32,768 -2,147,483,648 -9,223,372,036,854,775,808


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unsigned short int ux = (unsigned short) x;


unsigned short int uy = (unsigned short) y;

Casting Signed to Unsigned CAllowsConversionsfromSignedtoUnsigned

ResultingValue

Nochangeinbitrepresentation

Nonnegativevaluesunchanged

ux=15213

Negativevalueschangeinto(large)positivevalues

uy=50323


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T2U

T2B B2U

Twos Complement Unsigned

Maintain Same Bit Pattern

x ux

X

++ + +++

- + + +++

ux

x-

w1 0

+2w1 2w1 = 2*2w1 = 2w

ux

x x 0

x 2w x 0

Relation between Signed & Unsigned


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0

TMax

TMin

1

2

0

UMax

UMax 1

TMax

TMax + 1

2s Comp.

Range

Unsigned

Range

Illustration 2sComp.Unsigned

OrderingInversion

NegativeBigPositive


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Relation Between Signed & Unsigned

uy = y+2*32768 = y+65536

Weight -15213 50323

1 1 1 1 12 1 2 1 24 0 0 0 0

8 0 0 0 016 1 16 1 1632 0 0 0 064 0 0 0 0

128 1 128 1 128256 0 0 0 0512 0 0 0 0

1024 1 1024 1 10242048 0 0 0 04096 0 0 0 08192 0 0 0 0

16384 1 16384 1 1638432768 1 -32768 1 32768

Sum -15213 50323


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Signed vs. Unsigned in C Constants

Bydefaultareconsideredtobesignedintegers

UnsignedifhaveUassuffix

0U, 4294967259U

Casting

Explicitcastingbetweensigned&unsigned

i nt t x, t y;

unsi gned ux, uy;

t x = ( i nt ) ux;

uy = ( unsi gned) t y;

Implicitcastingalsooccursviaassignmentsandprocedurecalls

t x = ux;

uy = t y;


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0 0U == unsigned

- 1 0 < signed

- 1 0U > unsigned

2147483647 - 2147483648 > signed

2147483647U - 2147483648 < unsigned- 1 - 2 > signed

( unsi gned) - 1 - 2 > unsigned

2147483647 2147483648U < unsigned

2147483647 ( i nt ) 2147483648U > signed

Casting Surprises ExpressionEvaluation

Ifmixunsignedandsignedinsingleexpression,signedvaluesimplicitlycast

tounsigned

Includingcomparisonoperations,==,= ExamplesforW=32

Constant1 Constant2 Relation Evaluation

0 0U

- 1 0

- 1 0U

2147483647 - 2147483648

2147483647U - 2147483648- 1 - 2

( unsi gned) - 1 - 2

2147483647 2147483648U

2147483647 ( i nt ) 2147483648U


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Sign Extension Task:

Givenwbitsignedintegerx

Convertittow+kbitintegerwithsamevalue

Rule:

Makekcopiesofsignbit:

X= xw1

,,x

w1,x

w1,x

w2,,x

0

k copies of MSB

X

X

w

wk


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Sign Extension Example

Convertingfromsmallertolargerintegerdatatype

Cautomaticallyperformssignextension


int ix = (int) x;


int iy = (int) y;

Decimal Hex Binary

x 15213 3B 6D 00111011 01101101

i x 15213 00 00 3B 6D 00000000 00000000 00111011 01101101y -15213 C4 93 11000100 10010011i y -15213 FF FF C4 93 11111111 11111111 11000100 10010011


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Justification For Sign Extension ProveCorrectnessbyInductiononk

InductionStep:extendingbysinglebitmaintainsvalue

Keyobservation: 2w1 =2w+2w1

Lookatweightofupperbits:

X 2w1 xw1

X 2

w

xw1+

2

w1

xw1 = 2

w1

xw1

- X

X - +

w+1

w


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Beware When Using Unsigned DontUseJustBecauseNumberNonzero

Easytomakemistakes

f or ( i = cnt - 2; i >= 0; i - - )a[ i ] += a[ i +1] ;


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Negating with Complement & Increment Claim:FollowingHoldsfor2sComplement

~x + 1 == - x

Complement

Observation:~x + x == 1111112 == - 1

Increment

~x + x + ( - x + 1) ==- 1 + ( - x + 1) ~x + 1 == - x

1 0 0 1 0 11 1x

0 1 1 0 1 00 0~x+

1 1 1 1 1 11 1-1


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Comp. & Incr. Examples

Decimal Hex Binary

x 15213 3B 6D 00111011 01101101~x -15214 C4 92 11000100 10010010

~x+1 -15213 C4 93 11000100 10010011y -15213 C4 93 11000100 10010011

x = 15213

Decimal Hex Binary

0 0 00 00 00000000 00000000~0 -1 FF FF 11111111 11111111~0+1 0 00 00 00000000 00000000

0


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Unsigned Addition

StandardAdditionFunction

Ignorescarryoutput

ImplementsModularArithmetic

s = UAddw(u,v) =u+v mod2w

UAddw(u,v) u v u v 2w

u v 2w u v 2w

u

v+

u + v

True Sum: w+1 bits

Operands: wbits

Discard Carry: wbits UAddw(u , v)


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0

2 4 68

1012

14

0

2

4

6

8

10

12

14

0

4

8

12

16

20

24

28

32

Integer Addition

Visualizing Integer AdditionIntegerAddition

4bitintegersu,v

ComputetruesumAdd4(u,v)

Valuesincrease

linearlywithuandv

Formsplanar

surface

Add4(u , v)

u

v


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02

46

810

1214

0

2

4

6

8

10

12

14

0

2

4

6

8

10

12

14

16

Visualizing Unsigned Addition WrapsAround

Iftruesum 2w

Atmostonce

0

2w

2w+1

UAdd4(u , v)

u

v

True Sum

Modular Sum

Overflow

Overflow


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Mathematical Properties ModularAddition

Closedunderaddition

0 UAddw(u,

v) 2w

1 Commutative

UAddw(u,v) = UAddw(v,u)

AssociativeUAddw(t,UAddw(u,v)) = UAddw(UAddw(t,u),v)

0isadditiveidentity

UAddw(u,0) = u

Everyelementhasadditiveinverse

Let UCompw(u) = 2wuUAddw(u,UCompw(u)) = 0


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Twos Complement Addition

TAdd

and

UAdd

have

Identical

Bit

Level

Behavior Signedvs.unsignedadditioninC:

i nt s, t , u, v;

s = ( i nt ) ( ( unsi gned) u + ( unsi gned) v) ;

t = u + v

Willgive s == t

u

v+

u + v

True Sum:w

+1 bits

Operands: wbits

Discard Carry: wbits TAddw(u , v)


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Characterizing TAdd Functionality

Truesumrequiresw+1

bits

DropoffMSB

Treatremainingbitsas

2scomp.integer

2w1

2w

0

2w1

2w1

True Sum

TAdd Result

10000

11000

00000

01000

01111

1000

0000

0111

PosOver

NegOver

TAddw (u,v)

u v 2w1 u v TMinwu v TMinw u v TMaxwu v 2w1 TMaxw u v

(NegOver)

(PosOver)u

v

< 0 > 0

< 0

> 0

NegOver

PosOver

TAdd(u , v)


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-8-6

-4-2

02

46

-8

-6

-4-2

0

2

4

6

-8

-6

-4

-2

0

2

4

6

8

Visualizing 2s Comp. Addition Values

4bittwoscomp.

Rangefrom 8to+7

WrapsAround

Ifsum2w1 Becomesnegative

Atmostonce

Ifsum


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Detecting 2s Comp. Overflow Task

Givens = TAddw(u,v)

Determineifs = Addw(u,v) Example

i nt s, u, v;

s = u + v;

Claim

Overflowiffeither:

u,v


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Multiplication ComputingExactProductofwbitnumbersx,y

Eithersignedorunsigned

Ranges

Unsigned:0x*y (2w 1)2 = 22w 2w+1 +1

Upto2wbits

Twoscomplementmin:x*y (2w1)*(2w11) =22w2 +2w1

Upto2w1bits

Twoscomplementmax:x*y (2w1) 2 = 22w2

Upto2wbits,butonlyfor(TMinw)2

Maintaining

Exact

Results Wouldneedtokeepexpandingwordsizewitheachproductcomputed

Doneinsoftwarebyarbitraryprecisionarithmeticpackages

(GMP,BigNum,etc.)


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Unsigned Multiplication in C

StandardMultiplicationFunction

Ignoreshighorderwbits

ImplementsModularArithmeticUMultw(u,v) = u v mod2

w

u

v*

u

v

True Product: 2*

w bits

Operands: wbits

Discard wbits: wbits UMultw(u , v)


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Unsigned vs. Signed Multiplication UnsignedMultiplication

unsi gned ux = ( unsi gned) x;

unsi gned uy = ( unsi gned) y;unsi gned up = ux * uy

Truncatesproducttowbitnumberup =UMultw(ux,uy)

Modulararithmetic:up=uxuy mod2w

TwosComplementMultiplication

i nt x, y;

i nt p = x * y;

Computeexactproductoftwowbitnumbersx,y

Truncateresulttowbitnumberp =TMultw(x,y)


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Unsigned vs. Signed Multiplication UnsignedMultiplication

unsi gned ux = ( unsi gned) x;

unsi gned uy = ( unsi gned) y;unsi gned up = ux * uy

TwosComplementMultiplication

i nt x, y;

i nt p = x * y;

Relation Signedmultiplicationgivessamebitlevelresultas

unsigned

up == ( unsi gned) p


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Power-of-2 Multiply with Shift Operation

u


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Unsigned Power-of-2 Divide with Shift QuotientofUnsignedbyPowerof2

u >> k gives u / 2k Useslogicalshift

Division Computed Hex Binary

x 15213 15213 3B 6D 00111011 01101101x >> 1 7606.5 7606 1D B6 00011101 10110110

x >> 4 950.8125 950 03 B6 00000011 10110110

x >> 8 59.4257813 59 00 3B 00000000 00111011

0 0 1 0 0 0

u

2k/

u / 2kDivision:

Operands:

k

0

u / 2k Result:

.

Binary Point

0

S d P f d h Sh f


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Signed Power-of-2 Divide with Shift QuotientofSignedbyPowerof2

x >> k gives x / 2k Usesarithmeticshift Roundswrongdirectionwhenu < 0

0 0 1 0 0 0

x

2k/

x / 2kDivision:

Operands:

k

0

RoundDown(x / 2k) Result:

.

Binary Point

0

Division Computed Hex Binary

y -15213 -15213 C4 93 11000100 10010011y >> 1 -7606.5 -7607 E2 49 11100010 01001001

y >> 4 -950.8125 -951 FC 49 11111100 01001001

y >> 8 -59.4257813 -60 FF C4 11111111 11000100

C P f D d


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Correct Power-of-2 Divide QuotientofNegativeNumberbyPowerof2

Want x / 2k (RoundToward0) Compute

as

( x+2k

- 1) / 2k InC:( x + ( 1 k

Biasesdividendtoward0

Case

1:

No

rounding

Divisor:

Dividend:

0 0 1 0 0 0

u

2k/

u / 2k

k

1 0 0 0

1 0 1 1 .

Binary Point

1

0 0 0 1 1 1+2k +1

1 1 1

1 1 1 1

Biasing has no effect

C P f D d (C )


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Correct Power-of-2 Divide (Cont.)

Divisor:

Dividend:

Case 2: Rounding

0 0 1 0 0 0

x

2k/

x / 2k

k

1

1 0 1 1 .

Binary Point

1

0 0 0 1 1 1+2k +1

1

Biasing adds 1 to final result

Incremented by 1

Incremented by 1

C P l


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C Puzzles Assumemachinewith32bitwordsize,twoscomplement

integers

ForeachofthefollowingCexpressions,either:

Arguethatitistrueforallargumentvalues

Giveexamplewherenottrue x < 0 ((x*2) < 0)

ux >= 0

x & 7 == 7 (x y -x < -y

x * x >= 0

x > 0 && y > 0 x + y > 0

x >= 0 -x


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C Puzzle Answers Assumemachinewith32bitwordsize,twoscomp.

integers

TMinmakesagoodcounterexampleinmanycases

x < 0 ((x*2) < 0) False: TMin

ux >= 0 True: 0 = UMin

x & 7 == 7

(x y -x < -y False: - 1, TMin

x * x >= 0 False: 30426

x > 0 && y > 0 x + y > 0 False: TMax, TMax

x >= 0 -x = 0

x & 7 == 7

(x y -x < -y

x * x >= 0

x > 0 && y > 0 x + y > 0

x >= 0 -x


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ezThreeFourths/*

*ezThreeFourthsmultipliesby3/4roundingtoward0,

* ShouldexactlyduplicateeffectofCexpression(x*3/4),

* includingoverflowbehavior.

* Examples:ezThreeFourths(11)=8

* ezThreeFourths(9)= 6

* ezThreeFourths(1073741824)= 268435456(overflow)

* Legalops:!~&^|+

* Maxops:12* Rating:3

*/

int ezThreeFourths(int x){

}

Th F h


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ezThreeFourths/*

*ezThreeFourthsmultipliesby3/4roundingtoward0,

* ShouldexactlyduplicateeffectofCexpression(x*3/4),

* includingoverflowbehavior.

* Examples:ezThreeFourths(11)=8

* ezThreeFourths(9)= 6

* ezThreeFourths(1073741824)= 268435456(overflow)

* Legalops:!~&^|+

* Maxops:12* Rating:3

*/

int ezThreeFourths(int x){

int temp=x+x+x;

int sign=temp>>31;

int temp2=sign&((12;

}

UCLA CS 33 - Reinman - Spring 2014

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