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William Stallings Computer Organization and Architecture

Chapter 9Computer Arithmetic

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The von Neumann computer

MainMemory

Arithmetic and Logic Unit

Program Control Unit

InputOutputEquipment

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CPU Components: Top Level View

Control Unit

ALU

Registri

CP

U I

nte

rna

l Bu

s

Control Signals

PC

MAR

IR

MBRAC

Dat

a Li

nes

Add

ress

Lin

es

Con

trol

Lin

es

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Arithmetic & Logic Unit

• Does the calculations• Almost everything else in the computer is

there to service this unit• Handles integers• May handle floating point (real) numbers

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ALU Inputs and Outputs

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Reminder

• Let A=an-1an-2…a1a0 be a (pure) binary number

• Its (decimal) value is:

1

0

2n

ii

i aA

Example: 10010 = 24 + 21 = 16 + 2 = 18

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Integer Representation

• Only have 0 & 1 to represent everything• Positive numbers stored in binary

e.g. 41=00101001 Binary digit = Bit

• No minus sign• Negative integer representations:

Sign-Magnitude Two’s complement

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Sign-Magnitude

• Left most bit is sign bit• 0 means positive• 1 means negative

+18 = 00010010 -18 = 10010010

• Problems Need to consider both sign and magnitude in

arithmetic Two representations of zero (+0 and -0)

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A better representation:Two’s Complement

• Let A=an-1an-2…a1a0 be a n-length binary string

• It’s value is in Two’s Complement representation is:

2

01

1 22n

ii

in

n aaA

if an-1=0 then A is a non-negative number

if an-1=1 then A is a negative number

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Two’s Complement: examples

-128 64 32 16 8 4 2 1

0 1 0 0 0 1 1 0

-128 64 32 16 8 4 2 1

1 0 0 0 0 0 1 1

-128 64 32 16 8 4 2 1

1 0 0 0 1 0 0 0

+64 +4 +2 = 70

-128 +2 +1 = -125

-128 +8 = -120

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Benefits

• Only one representation of zero+0 = 0000 -1 = 1111+1 = 0001 -2 = 1110+2 = 0010 -3 = 1101+3 = 0011 -4 = 1100+4 = 0100 -5 = 1011+5 = 0101 -6 = 1010+6 = 0110 -7 = 1001+7 = 0111 -8 = 1000

• Arithmetic works easily (see later)

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Geometric Depiction of Two’s Complement Integers

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Range of Numbers

• 8 bit 2’s complement+127 = 01111111 = 27 -1 -128 = 10000000 = -27

• 16 bit 2’s complement+32767 = 01111111 11111111 = 215 - 1 -32768 = 10000000 00000000 = -215

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…when you cannot sleep…

…don’t count sheep in two’s complement…

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2’s Complement representation for negative numbers

• To represent a negative number using the “two’s complement” technique:1. First decide how many bits are used for

representation2. Then write the modulo of the negative

number (in pure binary)3. Then, change each 0 in 1, each 1 in 0

(Boolean Complement or “one’s complement”)

4. Finally, add 1 (as the result of Step 3 was a pure binary number)

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Examples• E.g.: how to represent -3 with 4 bits:

Start from +3 = 0011 Boolean complement gives 1100 Add 1 to LSB gives -3 1101

• Represent -20 with 8 bits: Start from +20 = 00010100 Bolean complement gives 11101011 Add 1 11101100

• Negation works in the same way, e.g. negation of -3 is obtained by the “two’s complement” of -3: Representation of -3 1101 Boolean complement gives 0010 Add 1 to LSB gives -(-3)=+3 0011

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Negation Special Case 1

• 0 = 0000• Bitwise NOT 1111 (Boolean

complement)• Add 1 to LSB +1• Result 1 0000• Carry is ignored, so:• - 0 = 0 OK !

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Negation Special Case 2

• -8 = 1000• Bitwise NOT 0111 (Boolean

complement)• Add 1 to LSB +1• Result 1000• So:• -(-8) = -8 WRONG !• Monitor MSB (sign bit)• If it does not change during negation there

is a mistake (but for 0!)

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Conversion Between Lengths

• Positive number: pack with leading zeroes+18 = 00010010+18 = 00000000 00010010

• Negative number: pack with leading ones-18 = 11101110-18 = 11111111 11101110

• i.e. pack with MSB (sign bit)

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Addition and Subtraction

• Addition: standard• Overflow: when the result needs more bits to be

represented• Monitor MSB: if it changes there may be an

overflow When Pos + Pos or Neg + Neg the sign bit should not

change: if it does there is an overflow

• Subtraction: take two’s complement of subtrahend and add to minuend i.e.: a - b = a + (-b)

• So we only need addition and complement circuits

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Addition: examples

1001 + -70101 = 51110 -2

0011 + 30100 = 40111 7

1100 + -40100 = 40000 01

1100 + -41111 = -11011 -51

1001 + -71010 = -60011 overflow1

0101 + 50100 = 41001 overflow

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Hardware for Addition and Subtraction

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Real Numbers

• …very informally, numbers with the fractional point

• Actually, only finite precision real numbers

• we need to code the binary point• we need to code the binary point

position• Solution: floating point numbers

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Reminder

• Let A=an-1an-2…a1a0 ,a-1,a-2,…,a-m be a binary number

• Its value is:

11

0

22mj

jj

n

ii

i aaA

Example:1001,1011 = 23 + 20 +2-1 + 2-3 + 2-4 =9,6875

2-1 2-2 2-3 2-4 2-5

0,5 0,250 0,125 0,0625 0,03125

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Real Numbers

• Where is the binary point? Fixed?

• Very limited

Moving?• How do you show where it is?

• Example:

976.000.000.000.000

0,0000000000000976

= 9,76 * 1014

= 9,76 * 10-14

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Floating Point

• Represents the value+/- .<significand> * <base><exponent>

• “Floating” refers to the fact that the true position of the point “floats” according to the value of the exponent

Sig

n bi

t

BiasedExponent

Significand or Mantissa

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Normalization

• Floating Point numbers are usually normalized exponent is adjusted so that there is a single bit equal

to 1 before the binary point

• Similar to scientific notation for decimal numbers where numbers are normalized to give a single digit before the decimal point

3123 = 3.123 x 103

(10,0101)2= 10,0101* 20 = 10010,1 * 2-3

= 1001010 * 2-5

= 0,00100101 * 24

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Normalization

• A normalized number ( 0) has the following form:

• The base (or radix) for the exponent is suppose to be 2 (so it is not represented)

• Since MSB of the mantissa is always 1 there is no need to represent it

+/- 1,bbb…b * 2 +/- E

where b {0,1}

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Typical Representation of Floating Point with 32 bits

• Mantissa uses 23 bits to store a 24 bits pure binary number in the interval [1,2)

• Sign is stored in the first bit• Exponent value is represented in excess or biased notation with 8 bits

• Excess (bias) 127 means 8 bit exponent field Nominal exponent value has range 0-255 Subtract 127 (= 28-1-1) to get correct exponent value Real range of exponent value is -127 to +128

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excess (bias) notation

• Two parameters are specified: the number of bits

n the bias value K

(usually, K=2n -1 -1)

• the string consisting of the number i in pure binary represents the value i–K

n = 4 K=7

0000 = -70001 = -60010 = -50011 = -40100 = -30101 = -20110 = -10111 = 0

1000 = 11001 = 21010 = 31011 = 41100 = 51101 = 61110 = 71111 = 8

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Floating Point Examples

147 = 127+20

107 = 127-20

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Expressible Numbers

2-127-2-127 (2-2-23)*2128-(2-2-23)*2128

Notice: we are representing232 different numbers in both cases!!

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density• Precision decreases with the increase of modulo

(representation is denser for smaller numbers)• How many numbers can we represent in the

interval [2,4)?• And in the range [4,8)?• they are the same: 223

Note: Finite precision: (10.0/3.0)*3.0 may be different from 10.0

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IEEE 754

• Standard for floating point storage• 32 and 64 bit standards• 8 and 11 bit exponent respectively• Extended formats (both mantissa and

exponent) for intermediate results

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IEEE 754 Formats

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Special cases (single format)• exponent [-126,127]: normalized numbers• Some strings are interpreted as special strings• Two representations for zero (posit. and negat.)

+/- 0 = 0/1 00000000 00000000000000000000000• Able to represent infinity (posit. and negat.)

+/- = 0/1 11111111 00000000000000000000000• Non-normalized numbers:

0/1 00000000 (string different to all 0s) exponent=-126 form of significand: 0.bbbb..

• NAN = Not A Number Result of an operation which has no solution Propagated as NAN through subsequent operations representation: 0/1 11111111 (string different to all 0s)

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FP Arithmetic: addition and subtraction

• Check for zero• Align significands (adjusting exponents)• Add or subtract significands• Normalize result

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FP Arithmetic: multiplication and division

• Check for zero• Add/subtract exponents • Multiply/divide significands (watch sign)• Normalize• Round• All intermediate results should be in

double length storage