Computer Arithmetic
4-3
Some of the the slides in this chapter have been compiled using the material provided by the publisher and
the lecture notes of the text book’s author Prof. Dave Patterson (CS, UC Berkeley) and Prof. Randy Brynt
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(ECE, CMU).
1
Divide: Paper & Pencil
1001
Divisor 1000
1001010
–1000
10
101
1010
–1000
10
Quotient
Dividend
Remainder (or Modulo result)
See how big a number can be subtracted, creating
quotient bit on each step
Binary => 1 * divisor or 0 * divisor
Dividend = Quotient x Divisor + Remainder
=> | Dividend | = | Quotient | + | Divisor |
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D I V I D E H A R D W A R E – Use what we have
u 32-bit
Divisor r e g, 32 - bit ALU, 64- bit Remainder r e g
Divisor
32 bits
32-bit ALU
“HI”
“LO”
Shift Left
Remainder (Quotient)
64 bits
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Control
Write
3
Divide Algorithm
Remainder
0000 0111
Divisor
Start: Place Dividend in Remainder
0010
1. Shift the Remainder register left 1 bit.
2. Subtract the Divisor register from the
left half of the Remainder register, & place the
result in the left half of the Remainder register.
Remainder >= 0
3a. Shift the
Remainder register
to the left setting
the new rightmost
bit to 1.
Test
Remainder
Remainder < 0
3b. Restore the original value by adding the Divisor
register to the left half of the Remainder register,
& place the sum in the left half of the Remainder
register. Also shift the Remainder register to the
left, setting the new least significant bit to 0.
nth
repetition?
No: < n repetitions
Yes: n repetitions (n = 4 here)
Done. Shift left half of Remainder right 1 bit.
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Observations on Divide
u
Same Hardware as Multiply: just need ALU to add or subtract, and
6 3 -bit register to shift left or shift right
u
H i a n d L o r e g i s t e r s i n M I P S c o m b i n e t o a c t a s 6 4 - bit register for
multiply and divide
u
S i g n e d D i v i d e s : S i m p l e s t i s t o r e m e m b e r s i g n s , m a k e p o s i t i v e , a nd
complement quotient and remainder if necessary
– Note: Dividend and Remainder must have same sign
– Note: Quotient negated if Divisor sign & Dividend sign disagree
e.g., –7 ÷ 2 = –3, remainder = –1
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5
Floating- Point Review
u
What can be represented in N bits?
u
Unsigned
u
2s Complement
- 2N
u
1s Complement
-2 N - 1 +1 to
u
fractions
.fffff …
u
But, what about?
0
-1
to
2N
to
2 N-1 - 1
2 N - 1 -1
to
n n n n n n.f
– v e r y l a r g e n u m b e r s ? 9,349,398,989,787,762,244,859,087,678
– very small number?
0.0000000000000000000000045691
– numbers with fractions, e.g., 3.1416
– very small numbers, e.g., .000000001
– very large numbers, e.g., 3.15576
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× 10 9
6
Recall Scientific Notation
exponent
Sign, magnitude
decimal point
23
6.02 x 10
Mantissa
-24
1.673 x 10
radix (base)
Sign, magnitude
u
Issues:
– Arithmetic (+, -, *, / )
– Representation, Normal form
– Range and Precision
– Rounding
– Exceptions (e.g., divide by zero, overflow, underflow)
– Errors
– Properties ( negation, inversion )
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Floating Point Review
u
u
Representation:
( – 1 )s i g n
×
–
sign, exponent, mantissa:
mantissa
–
more bits for mantissa gives more precision
–
more bits for exponent increases range
×
2exponent
How many different numbers can be represented
using N bits?
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–
No more than 2 N
–
2 * 2 *
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bits_in_mantissa
* 2 *
bits_in_exponent
8
Typical 32 bit Floating Number Representation
s
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8-bit exponent
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23-bit mantissa
9
IEEE 754 floating-point standard
u
IEEE 754 floating point standard:
single precision: 8 bit exponent, 23 bit mantissa
–
double precision: 11 bit exponent, 52 bit mantissa
CPEG
mantissa
Leading “1” bit of mantissa is implicit
Exponent is “biased” to make sorting easier
–
all 0s is smallest exponent all 1s is largest
–
bias of 127 for single precision and 1023 for double precision
s u m m a r y : ( – 1) s i g n × (1+ mantissa) × 2 e x p o n e n t – b i a s
–
u
exponent
–
–
u
s
Example:
–
–
d e c i m a l : -.75 = -3/4 = -3/22
b i n a r y : -.11 = -1.1 x 2 -1
–
floating point: exponent = 126 = 01111110
–
IEEE single precision:
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10111111010000000000000000000000
10
Floating- Point Arithmetic
Representation of floating point numbers in IEEE 754 standard:
single precision
sign
actual exponent is
e = E - 127
1
8
23
S
E
M
exponent:
excess 127
binary integer
mantissa:
sign + magnitude, normalized
binary mantissa w/ hidden
integer bit: 1.M
0 < E < 255
S E-127
N = (-1) 2
(1.M)
0 = 0 00000000 0 . . . 0
-1.5 = 1 01111111 10 . . . 0
Magnitude of numbers that can be represented is in the range:
-126
2
(1.0)
to
2
127
(2 - 2
23
)
which is approximately:
1.8 x 10
-38
to
3.40 x 10
38
(integer comparison valid on IEEE Fl.Pt. numbers of same sign!)
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Interpretation of IEEE 754 Floating Point Numbers
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12
FP Addition Algorithm
For addition (or subtraction) this translates into the following steps:
(1) compute Ye - Xe (needed to align binary point)
(2) right shift Xm that many positions to form Xm 2
(3) compute Xm 2
Xe-Ye
Xe-Ye
+ Ym
if representation demands normalization, then a normalization step
follows:
(4) left shift result, decrement result exponent (e.g., 0.001xx…)
right shift result, increment result exponent (e.g., 101.1xx…)
continue until MSB of data is 1 (NOTE: Hidden bit in IEEE Standard)
(5) doubly biased exponent must be corrected:
Xe = 7
Ye = -3
Excess 8
Xe = 1111
= 15
Ye = 0101
= 5
10100
20
extra subtraction step of the bias amount
=7+8
= -3 + 8
4+8+8
(6) if result is 0 mantissa, may need to set the exponent to zero by
special step
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FP Adder Block Diagram
Sign
Exponent
Significand
Sign
Exponent
Significand
Compare
exponents
Small ALU
Exponent
difference
0
1
0
Control
1
0
Shift smaller
number right
Shift right
Big ALU
0
1
0
Increment or
decrement
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Exponent
Add
1
Shift left or right
Rounding hardware
Sign
1
Normalize
Round
Significand
14
FP Multiply
Start
1. Add the biased exponents of the two
numbers, subtracting the bias from the sum
to get the new biased exponent
2. Multiply the significands
3. Normalize the product if necessary, shifting
it right and incrementing the exponent
Overflow or
underflow?
Yes
No
Exception
4. Round the significand to the appropriate
number of bits
No
Still normalized?
Yes
5. Set the sign of the product to positive if the
signs of the original operands are the same;
if they differ make the sign negative
Done
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