Download Binary - Cornell Computer Science

AnneBracy
CS3410
ComputerScience
CornellUniversity
The slides are the product of many rounds of teaching CS 3410 by
Professors Weatherspoon, Bala, Bracy, and Sirer.
inst
memory
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pc
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register file
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new pc
calculation
control
SimplifiedSingle-cycleprocessor
alu
focus
for
today
BinaryOperations
•
•
•
•
•
Numberrepresentations
One-bitandfour-bitadders
Negativenumbersandtwo’scompliment
Addition(two’scompliment)
Subtraction(two’scompliment)
3
Recall:Binary
• Twosymbols(base2):true andfalse;1 and0
• BasisofLogicCircuitsandalldigitalcomputers
So,howdowerepresentnumbersin Binary (base2)?
• WeknowrepresentnumbersinDecimal (base10).
– E.g.637
102101 100
6·102 +3·101 +7·100 =637
• Canjustaseasilyuseotherbases
– Base2— Binary
– Base8— Octal
1001111101
29 28 27 26 25 24 232221 20
0o1175
838281
– Base16— Hexadecimal
80
1·83 +1·82 +7·81 +5·80 =637
0x27d
162161160
4
Dec (base10) Bin(base2) Oct (base8) Hex (base16)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
0
1
10
11
100
101
110
111
1000
1001
1010
1011
1100
1101
1110
1111
10000
10001
10010
0
1
2
3
4
5
6
7
10
11
12
13
14
15
16
17
20
21
22
0
1
2
3
4
5
6
7
8
9
a
b
c
d
e
f
10
11
12
0b11111111=255
0b100000000=256
0o77=63
0o100=64
0xff =255
0x100=256
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Baseconversion viarepetitivedivision
• Dividebybase,writeremainder,moveleftwithquotient
637 ÷ 8 =79
79÷ 8 =9
9÷ 8 =1
1÷ 8=0
remainder5
remainder7
remainder1
remainder1
lsb (leastsignificantbit)
msb (mostsignificantbit)
637 = 0o1175
msb
lsb
6
Baseconversionviarepetitivedivision
Dividebybase,writeremainder,moveleftwithquotient
637 ÷ 2=318 remainder 1 lsb (leastsignificantbit)
318÷ 2=159 remainder 0
159÷ 2=79 remainder 1
79÷ 2=39remainder 1
39÷ 2=19remainder 1
19÷ 2=9remainder 1
9÷ 2=4
remainder 1
4÷ 2=2
remainder 0
2÷ 2=1remainder 0
1÷ 2=0remainder 1 msb (mostsignificantbit)
637 msb
= 1001111101lsb (or0b1001111101)
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Convertthenumber63710tobase16
Whatistheleastsignificantdigitofthisnumber?
a)
b)
c)
d)
e)
D
E
6
7
13
8
Binary to Octal
• Convertgroupsofthreebits frombinarytooct
• 3bits(000—111)havevalues0…7=1octaldigit
• E.g.0b1001111101
11 75
à 0o1175
Binary to Hexadecimal
• Convertnibble (groupoffourbits)frombinaryto
hex
• Nibble(0000—1111)hasvalues0…15=1hexdigit
• E.g.0b1001111101
2 7d
à 0x27d
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Thereare10typesofpeopleintheworld:
Thosewhounderstandbinary
Andthosewhodonot
And thosewhoknowthisjokewaswritteninbase3
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BinaryOperations
•
•
•
•
•
Numberrepresentations
One-bitandfour-bitadders
Negativenumbersandtwo’scompliment
Addition(two’scompliment)
Subtraction(two’scompliment)
11
Howdowedoarithmeticinbinary?
Additionworksthesameway
1
183
regardlessofbase
• Addthedigitsineachposition
+254
• Propagatethecarry
437
Carry-out
Carry-in
111
001110 Unsignedbinaryadditionisprettyeasy
•
Combinetwobitsatatime
+011100
• Alongwithacarry
1
1010 0
12
A
B
Cout
S
A
B Cout S
0
0
0
1
1
0
1
1
• Addstwo1-bitnumbers
• Computes1-bitresultand
1-bitcarry
• Nocarry-in
ClickerQuestion
WhatistheequationforCout?
a) A+B
b) AB
c) AÅ B
d) A+!B
e) !A!B
13
A
B
Cout
S
A
B Cout S
0
0
0
1
1
0
1
1
• Addstwo1-bitnumbers
• Computes1-bitresultand
1-bitcarry
• Nocarry-in
"B+AB
"
• S=A
• Cout =AB
AB
AB
Cout
S
14
A
B
• Addsthree1-bitnumbers
Cin • Computes1-bitresult,1-bitcarry
• Canbecascaded
Cout
S
A
B
Cin
0
0
0
0
0
1
0
1
0
0
1
1
1
0
0
1
0
1
1
1
0
1
1
1
Cout
S
NowYouTry:
1. FillinTruthTable
2. CreateSum-of-ProductForm
3. Minimizetheequation
• K-Maps
• AlgebraicMinimization
4. DrawtheCircuits
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A
B
• Addsthree1-bitnumbers
Cin • Computes1-bitresult,1-bitcarry
• Canbecascaded
Cout
S
A
B
Cin
0
0
0
0
0
1
0
1
0
0
1
1
1
0
0
1
0
1
1
1
0
1
1
1
Cout
S
ClickerQuestion
WhatistheequationforCout?
a) A+B+Cin
b) !A+!B+!C
c) AÅ BÅ Cin
d) AB+ACin +BCin
e) ABCin
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A[4] B[4]
Cin
Cout
S[4]
• Addstwo4-bitnumbersandcarryin
• Computes4-bitresultandcarryout
• Canbecascaded
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0011
0010
A 3 B3
A 2 B2
A 1 B1
A 0 B0
Cout 00100 Cin
S3
S2
01
S1
S0
01
• Addstwo4-bitnumbers,alongwithcarry-in
• Computes4-bitresultandcarryout
• Carry-out=resultdoesnotfitin4bits
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BinaryOperations
•
•
•
•
•
Numberrepresentations
One-bitandfour-bitadders
Negativenumbersandtwo’scompliment
Addition(two’scompliment)
Subtraction(two’scompliment)
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FirstAttempt:Sign/MagnitudeRepresentation
• 1bitforsign(0=positive,1=negative)
• N-1bitsformagnitude
Problem?
0111=7
0111=
1111=-7
1111=
• Twozero’s:+0differentthan-0
• Complicatedcircuits
• -2+1=???
0000=+0
1000=-0
IBM7090
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Positivenumbersarerepresentedasusual
• 0=0000,1=0001,3=0011,7=0111
Leading1’sfornegativenumbers
Tonegateany number:
•
•
•
•
•
•
complementall thebits(i.e.flipallthebits)
thenadd1
-1: 1Þ 0001Þ 1110Þ 1111
-3: 3Þ 0011Þ 1100 Þ 1101
-8: 8Þ 1000Þ 0111Þ 1000
-0: 0Þ 0000Þ 1111Þ 0000(thisisgood,-0=+0)21
Non-negatives
unchanged:
+0=0000
+1=0001
+2=0010
+3=0011
+4=0100
+5=0101
+6=0110
+7=0111
+8=1000
flip
Negatives
0% =1111
1% =1110
2% =1101
3% =1100
4% =1011
5% =1010
6% =1001
7% =1000
8% = 0111
thenadd1
-0=0000
-1=1111
-2=1110
-3=1101
-4=1100
-5=1011
-6=1010
-7=1001
-8=1000
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4bit
Two’s
Complement
-8…7
-1=
-2=
-3=
-4=
-5=
-6=
-7=
-8=
+7=
+6=
+5=
+4=
+3=
+2=
+1=
0=
1111
1110
1101
1100
1011
1010
1001
1000
0111
0110
0101
0100
0011
0010
0001
0000
=15
=14
=13
=12
=11
=10
=9
=8
=7
=6
=5
=4
=3
=2
=1
=0
4bit
Unsigned
Binary
0 …15
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Whatisthevalueofthe2scomplementnumber
11010
a)
b)
c)
d)
e)
26
-26
6
-6
-10
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Signedtwo’scomplement
• Negativenumbershaveleading1’s
• zeroisunique:+0=- 0
• wrapsfromlargestpositivetolargestnegative
Nbitscanbeusedtorepresent
• unsigned:range0…2N-1
– eg:8bitsÞ 0…255
• signed(two’scomplement):-(2N-1)…(2N-1 - 1)
– E.g.:8bitsÞ (10000000)…(01111111)
– -128…127
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Extending tolargersize(1st caseonslide23-24)
•
•
•
•
1111=-1
11111111=-1
0111=7
00000111=7
Truncate tosmallersize
• 00001111=15
• BUT, 0000 1111=1111=-1
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Additionasusual.Ignorethesign.Itjustworks!
-1=
1111
Examples
1+-1=
-3+-1=
-7+3=
7+(-3)=
Whatiswrongwiththefollowing?
7+1
-7+-3
-7+-1
-2=
-3=
-4=
-5=
-6=
-7=
-8=
+7=
+6=
+5=
+4=
+3=
+2=
+1=
0=
1110
1101
1100
1011
1010
1001
1000
0111
0110
0101
0100
0011
0010
0001
0000
=15
=14
=13
=12
=11
=10
=9
=8
=7
=6
=5
=4
=3
=2
=1
=0
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Whencanoverflow occur?
• addinganegativeandapositive?
– Overflowcannot occur (Why?)
• addingtwopositives?
– Overflowcan occur (Why?)
• addingtwonegatives?
– Overflowcan occur (Why?)
-1=
-2=
-3=
-4=
-5=
-6=
-7=
-8=
+7=
+6=
+5=
+4=
+3=
+2=
+1=
0=
1111
1110
1101
1100
1011
1010
1001
1000
0111
0110
0101
0100
0011
0010
0001
0000
=15
=14
=13
=12
=11
=10
=9
=8
=7
=6
=5
=4
=3
=2
=1
=0
Whencanoverflow occur?
AMSB
over
flow
BMSB
Cout_MSB
SMSB
Cin_MSB
MSB
A
B
Cin
Cout
S
0
0
0
0
0
0
0
1
0
1
0
1
0
0
1
0
1
1
1
0
1
0
0
0
1
1
0
1
1
0
1
1
0
1
0
1
1
1
1
1
Wrong
Sign
Wrong
Sign
Ruleofthumb:
• Overflowhappenediff msb’s carryin!=carryout
BinaryOperations
•
•
•
•
•
•
Numberrepresentations
One-bitandfour-bitadders
Negativenumbersandtwo’scompliment
Addition(two’scompliment)
Detectingandhandlingoverflow
Subtraction(two’scompliment)
– Whycreateanewcircuit?
– Justuseadditionusingtwo’scomplementmathHow?
Two’sComplementSubtraction
• Subtractionisaddition withanegatedoperand
– Negationisdonebyinvertingallbitsandaddingone
" +1)
A– B=A+(-B)=A+(B
B3
A3
B2
A2
B1
A1
B0
A0
Cout
S3
S2
S1
S0
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Two’sComplementSubtraction
• Subtractionisadditionwithanegatedoperand
– Negationisdonebyinvertingallbitsandaddingone
" +1)
A– B=A+(-B)=A+(B
B3
A3
B2
A2
B1
A1
B0
A0
Cout
1
S3
S2
S1
S0
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Two’sComplementAdderwithoverflowdetection
B3
mux
A3
B2
A2
mux
B1
mux
A1
B0
A0
mux
over
flow
S3
S2
S1
S0
Note:4-bitadderforillustrativepurposesandmaynotrepresenttheoptimaldesign.
0=add
1=sub
Digitalcomputersareimplementedvialogiccircuitsandthus
representall numbersinbinary(base2).
Wewritenumbersasdecimalorhexforconvenienceandneedtobe
abletoconverttobinaryandback(tounderstandwhatthecomputer
isdoing!).
Addingtwo1-bitnumbersgeneralizestoaddingtwonumbersofany
sizesince1-bitfulladderscanbecascaded.
UsingTwo’scomplementnumberrepresentationsimplifiesadder
Logiccircuitdesign(0isunique,easytonegate).Subtractionisadding,
whereoneoperandisnegated(two’scomplement;tonegate:flipthe
bitsandadd1).
OverflowifsignofoperandsAandB!=signofresultS.
CandetectoverflowbytestingCin !=Cout ofthemostsignificantbit
(msb),whichonlyoccurswhenpreviousstatementistrue.
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Wecannowimplementcombinationallogiccircuits
• Designeachblock
– Binaryencodednumbersforcompactness
• Decomposelargecircuitintomanageableblocks
– 1-bitHalfAdders,1-bitFullAdders,
n-bitAddersviacascaded1-bitFullAdders,...
• CanimplementcircuitsusingNANDorNORgates
• CanimplementgatesusingusePMOSandNMOStransistors
• Andcanaddandsubtractnumbers(intwo’s
compliment)!
• Nexttime,stateandfinitestatemachines…
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