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AnneBracy
CS3410
ComputerScience
CornellUniversity
The slides are the product of many rounds of teaching CS 3410 by
Professors Weatherspoon, Bala, Bracy, and Sirer.
See:P&HChapter2.4,3.2,B.2,B.5,B.6
inst
memory
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32
pc
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)
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Recall:Binary
• Twosymbols(base2):true andfalse;1 and0
• BasisofLogicCircuitsandalldigitalcomputers
So,howdowerepresentnumbersin Binary (base2)?
• WeknowrepresentnumbersinDecimal (base10).
– E.g.637
6·102 +3·101 +7·100
=637
102101 100
• Canjustaseasilyuseotherbases
– Base2— Binary 1001111101
9 8
7 6 5 4
3 2 1 0
2 2
– Base8— Octal
2 2 2 2
0o1175
2 2 2 2
838281 80
– Base16— Hexadecimal
0x27d
162161160
1·83 +1·82 +7·81 +5·80
=637
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Howdowecountindifferentbases?
• 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
.
.
99
100
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=
0b100000000=
0o77=
0o100=
0xff =
0x100=
.
.
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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 (least significant bit)
msb (mostsignificant bit)
637 = 0o1175
msb
lsb
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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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Baseconversion viarepetitivedivision
Dividebybase,writeremainder,moveleftwithquotient
637 ÷ 16=39 remainder13 lsb
39÷ 16=2 remainder7
2÷ 16=0remainder 2 msb dec =hex =bin
637 =0x2713=0x27d
?
Thus, 637 = 0x27d
10=0xa
11=0xb
12=0xc
13=0xd
14=0xe
15=0xf
=1010
=1011
=1100
=1101
=1110
=1111
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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)
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Howdowedoarithmeticinbinary?
1
183
+254
437
Carry-in
Additionworksthesameway
regardlessofbase
• Addthedigitsineachposition
• Propagatethecarry
Carry-out
111
001110 Unsignedbinaryadditionisprettyeasy
+011100 • Combinetwobitsatatime
• Alongwithacarry
101010
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A
B
Cout
S
A
B Cout S
0
0
0
1
1
0
1
1
HalfAdder
• Addstwo1-bitnumbers
• Computes1-bitresultand
1-bitcarry
• Nocarry-in
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A
B
FullAdder
• 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[4] B[4]
4-BitFullAdder
• Addstwo4-bitnumbersandcarryin
Cin • Computes4-bitresultandcarryout
• Canbecascaded
Cout
S[4]
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0011
0010
A3 B3
A2 B2
A1 B1
A0 B0
Cout 00100 Cin
S3
S2
01
S1
S0
01
• Addstwo4-bitnumbers,alongwithcarry-in
• Computes4-bitresultandcarryout
• Carry-out=resultdoesnotfitin4bits
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SubfrequencyCode1à A
SubfrequencyCode2à B
Easytoremember:AnneBracy
Areyou:
A. CSMinor
B. CSMajorinArts&Sciences
C. CSMajorinEngineering
D. MEngstudent
E. Other
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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
-1:1⇒ 0001⇒ 1110⇒ 1111
-3:3⇒ 0011⇒ 1100 ⇒ 1101
-7:7⇒ 0111⇒ 1000⇒ 1001
-8:8⇒ 1000⇒ 0111⇒ 1000
-0:0⇒ 0000⇒ 1111⇒ 0000(thisisgood,-0=+0)23
Non-negatives Negatives
(asusual):
+0=0000
+1=0001
+2=0010
+3=0011
+4=0100
+5=0101
+6=0110
+7=0111
+8=1000
(two’scomplement:flipthenadd1):
0" =1111
1" =1110
2" =1101
3" =1100
4" =1011
5" =1010
6" =1001
7" =1000
8" = 0111
-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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Signedtwo’scomplement
• Negativenumbershaveleading1’s
• zeroisunique:+0=- 0
• wrapsfromlargestpositivetolargestnegative
Nbitscanbeusedtorepresent
• unsigned:range0…2N-1
– eg:8bits⇒ 0…255
• signed(two’scomplement):-(2N-1)…(2N-1 - 1)
– E.g.:8bits⇒ (10000000)…(01111111)
– -128…127
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Extending tolargersize
•
•
•
•
1111=-1
11111111=-1
0111=7
00000111=7
Truncate tosmallersize
• 00001111=15
• BUT, 0000 1111=1111=-1
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=Additionasusual,ignorethesign(itjustworks)
Examples
1+-1=
-3+-1=
-7+3=
7+(-3)=
Whatiswrongwiththefollowingadditions?
7+1-7+-3-7+-1
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Whycreateanewcircuit?
Justuseadditionusingtwo’scomplementmath
• How?
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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
1
Cout
S3
S2
S1
S0
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BinaryOperations
•
•
•
•
•
•
Numberrepresentations
One-bitandfour-bitadders
Negativenumbersandtwo’scompliment
Addition(two’scompliment)
Subtraction(two’scompliment)
Detectingandhandlingoverflow
Whencanoverflow occur?
• addinganegativeandapositive?
• addingtwopositives?
• addingtwonegatives?
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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
Two’sComplementAdderwithoverflowdetection
B3
mux
A3
B2
A2
mux
B1
mux
A1
B0
A0
mux
over
flow
S3
S2
S1
S0
Note:4-bitadderforillustrative purposesandmaynotrepresenttheoptimaldesign.
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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