2’s Complement and Floating-Point - University of …...Two’s Complement •An n-bit, two’s...

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2’s Complement and Floating-Point CSE 351 Section 3

Transcript of 2’s Complement and Floating-Point - University of …...Two’s Complement •An n-bit, two’s...

Page 1: 2’s Complement and Floating-Point - University of …...Two’s Complement •An n-bit, two’s complement number can represent the range [−2!−1, 2 !−1− 1]. • Note the

2’sComplementandFloating-Point

CSE351Section3

Page 2: 2’s Complement and Floating-Point - University of …...Two’s Complement •An n-bit, two’s complement number can represent the range [−2!−1, 2 !−1− 1]. • Note the

Two’sComplement

•Ann-bit,two’scomplementnumbercanrepresenttherange[−2 𝑛−1 ,2𝑛−1 −1].

• Notetheasymmetryofthisrangeabout0– there’sonemorenegativenumberthanpositive

•Notewhathappenswhenyouoverflow

4-bittwo’scomplementrange

Page 3: 2’s Complement and Floating-Point - University of …...Two’s Complement •An n-bit, two’s complement number can represent the range [−2!−1, 2 !−1− 1]. • Note the

UnderstandingTwo’sComplement

• Aneasierwaytofindthedecimalvalueofatwo’scomplementnumber:

~x + 1 = -x

• Wecanrewritethisasx=~(-x- 1),i.e.subtract1fromthegivennumber,andflipthebitstogetthepositiveportionofthenumber.

• Example:0b11010110• Subtract1:0b11010110 - 1 = 0b11010101• Flipthebits:~0b11010101 = 0b00101010 • Converttodecimal: 0b00101010 = (32+8+2)10 = 4210• Multiplybynegativeone,Answer: -4210.

Page 4: 2’s Complement and Floating-Point - University of …...Two’s Complement •An n-bit, two’s complement number can represent the range [−2!−1, 2 !−1− 1]. • Note the

Two’sComplement:Operations

1. Convertnumberstobinary

• 0xAB = 0b10101011

• 1710 = 0b00010001

2. Computex | y 0000 ... 1010 1011

| 0000 ... 0001 0001--------------------0000 ... 1011 1011

3. Compute~(x | y) (flipthebits)~ 0000 0000 0000 0000 0000 0000 1011 1011-----------------------------------------1111 1111 1111 1111 1111 1111 0100 0100

int x = 0xAB;int y = 17;int z = 5;int result = ~(x | y) + z;

Whatisthevalueofresultindecimal?

Page 5: 2’s Complement and Floating-Point - University of …...Two’s Complement •An n-bit, two’s complement number can represent the range [−2!−1, 2 !−1− 1]. • Note the

Two’sComplement:Operations4. Compute ~(x | y) + z

• Youcaneitherconverttodecimalbeforeorafteraddingz

• 1710 = 0b00010001

• Convert111111111111111111111111010001002 to decimal(hint: -x = ~x + 1)

• ~111111111111111111111111010001002 = 0000000000000000000000000101110112

• ~0000……101110112 + 1 = 0000……101111002

• -x = 0000……101111002

• x = 1000……101111002

int x = 0xAB;int y = 17;int z = 5;int result = ~(x | y) + z;

Whatisthevalueofresultindecimal?

Page 6: 2’s Complement and Floating-Point - University of …...Two’s Complement •An n-bit, two’s complement number can represent the range [−2!−1, 2 !−1− 1]. • Note the

FloatingPoint

• value=(-1)S*M *2E

• NumericalForm• Signbits determineswhethernumberisnegativeorpositive• Significand(mantissa)M normallyafractionalvalueinrange[1.0,2.0)• ExponentE weightsvaluebya(possiblynegative)poweroftwo

• RepresentationinMemory• MSBsissignbits• exp fieldencodesE (butisnotequaltoE)– rememberthebias• Frac fieldencodesM (butisnotequaltoM)

s exp mant

Page 7: 2’s Complement and Floating-Point - University of …...Two’s Complement •An n-bit, two’s complement number can represent the range [−2!−1, 2 !−1− 1]. • Note the

FloatingPoint

• value=(-1)S*M *2E

• Value:±1× Mantissa× 2Exponent

• Bit Fields:(-1)S × 1.M × 2(E+bias)

• Bias• Readexponentasunsigned,butwithbiasof–(2w-1-1)=–127• Representableexponentsroughly½positiveand½negative• Exponent0(Exp =0)isrepresentedasE=0b01111111

• Why?• Floatingpointarithmetic=easier• Somewhatcompatiblewith2’scomplement

s exp mant

Page 8: 2’s Complement and Floating-Point - University of …...Two’s Complement •An n-bit, two’s complement number can represent the range [−2!−1, 2 !−1− 1]. • Note the

FloatingPoint

• Exponentoverflowyields+∞or-∞• Floatswithvalue+∞,-∞,andNaN canbeusedinoperations• Resultusuallystill+∞,-∞,orNaN;sometimesintuitive,sometimesnot

• Floatingpointopsdonotworklikerealmath,duetorounding!• Notassociative:

• (3.14+1e100)– 1e100!=3.14+(1e100– 1e100)• Notdistributive:

• 100*(0.1+0.2)!=100*0.1+100*0.2• Notcumulative

• Repeatedlyaddingaverysmallnumbertoalargeonemaydonothing

s exp mant

Page 9: 2’s Complement and Floating-Point - University of …...Two’s Complement •An n-bit, two’s complement number can represent the range [−2!−1, 2 !−1− 1]. • Note the

FloatingPointExamples

• Howdoyourepresent-1.5infloatingpoint?• Signbit:1• Firsttheintegralpartofthevalue:1=0b1• Nowcomputethedecimal:0.5=0b0.1• 1.510 =1.1b• Don’tneedtonormalizebecauseit’salreadyinscientificnotation:1.1x20• Exponent:0+127=12710 =011111112• Mantissa:10000000000000000000002• -1.510=10111111110000000000000000000002 =0xBFC00000

s exp mant

1 8 23

Page 10: 2’s Complement and Floating-Point - University of …...Two’s Complement •An n-bit, two’s complement number can represent the range [−2!−1, 2 !−1− 1]. • Note the

FloatingPointExamples

• Howdoyourepresent1.0infloatingpoint?• Signbit:0• Firsttheintegralpartofthevalue:1=0b1• 1.010 =>1.0b• Don’tneedtonormalizebecauseit’salreadyinscientificnotation:1.0x20• Exponent:0+127=12710 =011111112• Mantissa:000000000000000000000002• 1.010=001111111000000000000000000000002 =0x3F800000

s exp mant

1 8 23

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FloatingPointAddition

(–1)s1*M1*2E1 + (-1)s2*M2*2E2

(AssumeE1 >E2)• ExactResult:(–1)s*M*2E

• Signs,mantissaM:• M=M1+M2,resultofsignedalign&add

• ExponentE:E1

• Fixing• If M ≥ 2, shift M right, increment E• if M < 1, shift M left k positions, decrement E by k• OverflowifE outofrange• RoundM tofitfrac precision

s exp mant

1 8 23

Whatisfloatingpointresultof12.3125+1.5?12.312510 =0 10000010 100010100000000000000001.510 =0 01111111 10000000000000000000000

Inscientific notation:(1.1000101x23)+(1.1x 20 )1100.0101 x 20

+ 1.1 x 20

------------------

1101.1101 x 20

Adjusted mantissa:1.1011101=>M=1011101Ans:0 10000010 10111010000000000000000

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FloatingPointMultiplication

(–1)s1*M1*2E1 * (–1)s2*M2*2E2

• ExactResult:(–1)s*M*2E• Signs: s1 ^ s2• MantissaM: M1 * M2• ExponentE: E1 + E2

• Fixing• IfM ≥ 2,shiftM right,incrementE• IfE outofrange,overflow• RoundM tofitfrac precision

s exp mant

1 8 23

Whatisfloatingpointresultof1.1*1.0?1.110 =0 01111111 000110011001100110011011.010 =0 01111111 00000000000000000000000

Inscientific notation:(1.1x20)+(1.0x 20 )1.00011001100110011001101 x 20

* 1.0 x 20

--------------------------------------00000000000000000000000001000110011001100110011010

Result: 1.000110011001100110011010 x 20

Ans:0 01111111 00011001100110011001101

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FloatingPointWorksheet


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