Advanced ALU
Lecture 5
CS365
RoadMap

Implementation of MIPS ALU
 Signed
and unsigned numbers (Section 3.2)
 Addition and subtraction (Section 3.3)
 Constructing an arithmetic logic unit (Appendix
B.5, B.6)
 Multiplication (Section 3.4)
This lecture
 Division (Section 3.5)
 Floating point (Section 3.6)
Advanced ALU
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Unsigned Multiply

Paper and pencil example (unsigned):
Multiplicand
Multiplier
Product


1000
1001
1000
0000
0000
1000
01001000
m bits x n bits = m+n bit product (at most)
Binary makes it easy:
0 => place 0
 1 => place a copy


( 0 x multiplicand)
( 1 x multiplicand)
4 versions of multiply hardware & algorithm:

Successive refinement
Advanced ALU
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Unsigned Combinational Multiplier
0
A3
0
A2
0
A1
0
A0
B0
A3
A3
A2
A2
A1
A1
A0
B1
A0
B2
A3
P7

P6
A2
A1
P5
A0
P4
B3
P3
P2
P1
P0
Stage i accumulates A * 2i if Bi == 1
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How Does It Work?
0
0
0
0
A3
A3
A3
A3
P7



P6
A2
P5
A2
A1
P4
A2
A1
0
A2
A1
0
A1
0
A0
A0
B1
A0
B2
A0
P3
B0
B3
P2
P1
P0
At each stage shift A left ( x 2)
Use next bit of B to determine whether to add in
shifted multiplicand
Accumulate 2n bit partial product at each stage
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Unsigned Multiplier (Version 1)

Shift-add multiplier
 64-bit
Multiplicand reg, 64-bit ALU, 64-bit
Product reg, 32-bit Multiplier reg
Shift Left
Multiplicand
64 bits
Multiplier
64-bit ALU
Product
Shift Right
32 bits
Write
Control
64 bits
Figure 3.5
Advanced ALU
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Multiply Algorithm Version 1
Start
Figure 3.6
Multiplier0 = 1
1. Test
Multiplier0
Multiplier0 = 0
1a. Add multiplicand to product &
place the result in Product register
0010 x 0011
Product
0000 0000
0000 0010
0000 0110
0000 0110
0000 0110
Multiplier Multiplicand
0011
0000 0010
0001
0000 0100
0000
0000 1000
0000
0001 0000
2. Shift the Multiplicand register left 1 bit.
3. Shift the Multiplier register right 1 bit.
32nd
repetition?
Yes: 32 repetitions
Done
Advanced ALU
No: < 32 repetitions
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Observations on Multiply Version 1

One clock cycle per step => ~100 clocks per
multiply

Half bits in multiplicand always 0 => 64-bit adder
is wasted
0 is inserted when shift the multiplicand left
=> least significant bits of product never changed
once formed
Instead of shifting multiplicand to left, shift
product to right?


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How Does It Work?
0
A3
0
A2
0
A1
0
A0
B0
A3
A2
A1
A0
B1
A3
A2
A1
A0
B2
A3
P7

A2
A1
P6
A0
P5
B3
P4
P3
P2
P1
P0
Multiplicand stays still and product moves right
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Unsigned Multiplier Version 2

Refined from the first version: 32-bit
Multiplicand reg, 32 -bit ALU, 64-bit
Product reg, 32-bit Multiplier reg
Multiplican
d
32 bits
Multiplier
32-bit ALU
Shift Right
32 bits
Shift Right
Product
64 bits
Advanced ALU
Control
Write
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Multiply Algorithm Version 2
Start
Multiplier0 = 1
1:
2:
3:
1:
2:
3:
1:
2:
3:
1:
2:
3:
1. Test
Multiplier0
0010 x 0011
1a. Add multiplicand to the left half of product &
place the result in the left half of Product register
Product
Multiplier
0000 0000 0011
0010 0000 0011
0001 0000 0011
0001 0000 0001
0011 0000 0001
0001 1000 0001
0001 1000 0000
0001 1000 0000
0000 1100 0000
0000 1100 0000
0000 1100 0000
0000 0110 0000
0000 0110 0000
Multiplicand
0010
0010
0010
0010
0010
0010
0010
0010
0010
0010
0010
0010
0010
Advanced ALU
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Multiplier0 = 0
2. Shift the Product register right 1 bit.
3. Shift the Multiplier register right 1 bit.
32nd
repetition?
No: < 32 repetitions
Yes: 32 repetitions
Done
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Observations on Multiply Version 2
Useful portion of multiplier shrinks while
valid portion of product grows in the
process of multiplication
 Product register wastes space that exactly
matches size of multiplier
 =>Combine Multiplier register and Product
register?

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MULTIPLY Hardware Version 3

32-bit Multiplicand reg, 32 -bit ALU, 64-bit
Product reg, (0-bit Multiplier reg)
Multiplican
d
32 bits
32-bit ALU
Shift Right
Product (Multiplier)
Write
64 bits
Advanced ALU
Control
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Multiply Algorithm Version 3
Start
Product0 = 1
1. Test
Product0
Product0 = 0
0010 x 0011
1a. Add multiplicand to the left half of product &
place the result in the left half of Product register
Multiplicand
0010
0010
0010
0010
0010
Advanced ALU
Product
0000 0011
0010 0011
0001 0001
0011 0001
0001 1000
0000 1100
0000 0110
2. Shift the Product register right 1 bit.
32nd
repetition?
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No: < 32 repetitions
Yes: 32 repetitions
14
Done
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Multiply Version 3
Two steps per bit because Multiplier &
Product combined
 MIPS multiplication (unsigned)

multu $t1, $t2
# t1 * t2
 No destination register: product could be ~2 64;
need two special registers to hold it
 Registers Hi and Lo are left and right half of
Product
 To use the result:
mfhi $t3
mflo $t4
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Signed Multiplication

What about signed multiplication?
 Easiest
solution is to make both positive &
remember whether to complement product
when done
 Apply definition of 2’s complement

Need to sign-extend partial products
 Booth’s
algorithm is an elegant way to multiply
signed and unsigned numbers using the same
hardware and save cycles

Can handle multiple bits at a time
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Motivation for Booth’s Algorithm

Example 2 x 6 = 0010 x 0110:
Can get the same
result in more than
one way:
6= – 2 + 8
0110 = – 0010 + 1000

0010
x
0110
+
0000
+
0010
+ 0010
+ 0000
00001100
shift (0 in multiplier)
add (1 in multiplier)
add (1 in multiplier)
shift (0 in multiplier)
Basic idea: replace a string of 1s with an initial subtract
on seeing a one and add after last one
0010 x (-2)
0010 x 8
Advanced ALU
0010
x
0110
0000
–
0010
0000
+ 0010
00001100
shift (0 in multiplier)
sub (first 1 in multiplier)
shift (mid string of 1s)
add (prior step had last 1)
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Booth’s Algorithm
end of run
Current
bit
1
1
0
0

Bit to
right
0
1
1
0
middle of run
beginning of run
0 1 1 1 1 0
Explanation
Begins run of 1s
Middle of run of 1s
End of run of 1s
Middle of run of 0s
Example
Op
0001111000
0001111000
0001111000
0001111000
sub
none
add
none
Originally for speed when shift was faster than
add
Big advantage: it handles signed number
–1
Why it works?
+ 10000
01111
Advanced ALU
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
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Booth’s Algorithm
1. Depending on the current and previous
bits, do one of the following:
00:Middle of a string of 0s, no arithmetic op
01:End of a string of 1s, so add multiplicand to
the left half of the product
10:Beginning of a string of 1s, so subtract
multiplicand from the left half of the product
11: Middle of a string of 1s, so no arithmetic op
2. As in the previous algorithm, shift the
Product register right (arithmetically) 1 bit
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Booths Example (2 x 7)
step
Multiplicand Product
0. initial value 0010
1. P = P - m
2.
3.
4.
1110
0010
0010
0010
0010
0010
Advanced ALU
0000 0111 0
+1110
1110 0111 0
1111 0011 1
1111 1001 1
1111 1100 1
+0010
0001 1100 1
0000 1110 0
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operation
10 -> sub
shift P(sign ext)
11 -> nop, shift
11 -> nop, shift
01 -> add
shift
done
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Booths Example (2 x -3)
step
Multiplicand Product
0. initial value 0010
1. P = P - m
2.
1110
0010
0010
3.
0010
1110
0010
0010
4.
0010
Advanced ALU
0000 1101 0
+1110
1110 1101 0
1111 0110 1
+ 0010
0001 0110 1
0000 1011 0
+1110
1110 1011 0
1111 0101 1
1111 0101 1
1111 1010 1
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operation
10 -> sub
shift P(sign ext)
01 -> add
shift P
10 -> sub
shift
11 -> nop
shift
done
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Outline

Implementation of MIPS ALU
 Signed
and unsigned numbers (Section 3.2)
 Addition and subtraction (Section 3.3)
 Constructing an arithmetic logic unit (Appendix
B.5, B.6)
 Multiplication (Section 3.4)

Booth’s algorithm: CD: 3.23 In More Depth
 Division
(Section 3.5)
 Floating point (Section 3.6)
Advanced ALU
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DIVIDE: Paper & Pencil
1001
Divisor 1000 1001010
–1000
10
101
1010
–1000
10


Remainder (or Modulo result)
See how big a number can be subtracted,
creating quotient bit on each step


Quotient
Dividend
Binary => 1 * divisor or 0 * divisor
Dividend = Quotient x Divisor + Remainder
3 versions of divide, successive refinement
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DIVIDE Hardware Version 1

64-bit Divisor reg, 64-bit ALU, 64-bit
Remainder reg, 32-bit Quotient reg
Shift Right
Divisor
64 bits
Quotient
64-bit ALU
Remainder
Shift Left
32 bits
Write
Control
64 bits
Advanced ALU
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Divide Algorithm Version 1

Start: Place Dividend in Remainder
Takes n+1 iterations for nbit Quotient & Remainder
Quot. Divisor
Rem.
0000 00100000 00000111
11100111
00000111
0000 00010000 00000111
11110111
00000111
0000 00001000 00000111
11111111
00000111
0000 00000100 00000111
00000011
0001
00000011
0001 00000010 00000011
00000001
0011
00000001
0011 00000001 00000001
Advanced ALU
1. Subtract the Divisor register from the
Remainder register, and place the result
in the Remainder register.
Remainder  0
2a. Shift the
Quotient register
to the left, setting
the new rightmost
bit to 1.
Test
Remainder
Remainder < 0
2b. Restore the original value by adding the
Divisor register to the Remainder register, &
place the sum in the Remainder register. Also
shift the Quotient register to the left, setting
the new least significant bit to 0.
3. Shift the Divisor register right 1 bit.
n+1
repetition?
No: < n+1 repetitions
Yes: n+1 repetitions (n = 4 here)
Done
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Observations on Divide Version 1



1st step cannot produce a 1 in quotient bit
(otherwise quotient is too big for the register)
=> switch order to shift first and then subtract
=> save 1 iteration
1/2 bits in divisor always 0
=> 1/2 of 64-bit adder is wasted
=> 1/2 of divisor is wasted
Instead of shifting divisor to right, shift remainder
to left?
Advanced ALU
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DIVIDE Hardware Version 2

32-bit Divisor reg, 32-bit ALU, 64-bit
Remainder reg, 32-bit Quotient reg
Divisor
32 bits
Quotient
32-bit ALU
Shift Left
32 bits
Shift Left
Remainder
64 bits
Advanced ALU
Control
Write
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Divide Algorithm Version 2
Start: Place Dividend in Remainder
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
Quotient 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 Quotient register to the left,
setting the new least significant bit to 0.
nth
repetition?
No: < n repetitions
Yes: n repetitions
Done
Advanced ALU
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Observations on Divide Version 2
Remainder shrinks while Quotient grows
 Eliminate Quotient register by combining
with Remainder register

 Start
by shifting the Remainder left as before
 Only shifting once for both Remainder(left half)
and Quotient(right half): 2 steps each iteration
 Another consequence is that the remainder
will be shifted left one more time

Thus the final correction step must shift back only
the remainder in the left half of the register
Advanced ALU
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DIVIDE HARDWARE Version 3

32-bit Divisor reg, 32 -bit ALU, 64-bit
Remainder reg, (0-bit Quotient reg)
Divisor
32 bits
32-bit ALU
“HI”
“LO”
Shift Left
Remainder (Quotient)
Write
64 bits
Advanced ALU
Control
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Divide Algorithm Version 3
Step
0
1.1
1.2
1.3b
2.2
2.3b
3.2
3.3a
Remainder Div.
0000 0111 0010
0000 1110
1110 1110
0001 1100
1111 1100
0011 1000
0001 1000
0011 0001
3a. Shift the
Remainder register
to the left setting
the new rightmost
bit to 1.
4.2 0001 0001
4.3a 0010 0011
0001 0011
Advanced ALU
Start: Place Dividend in Remainder
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
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
No: < n repetitions
repetition?
Yes: n repetitions (n = 4 here)
Done. Shift left half of Remainder right 1 bit
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More Division Issues

Signed divides: simplest is to remember signs,
make positive, and complement quotient and
remainder if necessary
Satisfy Dividend =Quotient x Divisor + Remainder
 Note: Dividend and Remainder must have same sign
 Note: Quotient negated if Divisor sign & Dividend sign
disagree, e.g., –7 ÷ 2 = –3, remainder = –1



Quotient = 4 and remainder =1 is not correct
Possible for quotient to be too large
If divide 64-bit integer by 1, quotient is 64 bits (“called
saturation”)
 MIPS ignores overflow, SW detection and processing

Advanced ALU
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Recap

Same hardware for Divide as Multiply: just need
ALU to add or subtract, and 64-bit register to shift
left or shift right
Hi and Lo registers in MIPS combine to act as 64-bit
register for multiply and divide
 div $s2, $s3
#Lo=$S2/$s3, Hi=$s2 mod $s3


Successive refinement to see final design


Booth’s algorithm to handle signed multiplies
Overflow
Both MIPS multiply and divide instructions ignore
overflow
 Software must check for overflow scenarios

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Outline

Implementation of MIPS ALU
 Signed
and unsigned numbers (Section 3.2)
 Addition and subtraction (Section 3.3)
 Constructing an arithmetic logic unit (Appendix
B.5, B.6)
 Multiplication (Section 3.4, CD: 3.23 In More
Depth)
 Division (Section 3.5)
 Floating point (Section 3.6)
Advanced ALU
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Floating-Point: Motivation

What can be represented in N bits?
Unsigned
 2’s Complement
 1’s Complement
 Binary Coded Decimal


0
to
- 2N-1 to
-2N-1+1to
0
to
2N-1
2N-1 - 1
2N-1-1
10N/4 –1
But, what about?
Very large numbers?
9,349,398,989,787,762,244,859,087,678
 Very small number?
0.0000000000000000000000045691
 Rational numbers
2/3
 Irrational numbers
pi

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Recall Scientific Notation: Binary
binary point
exponent
1.01 x 223
Significand(Mantissa)


Normalized form: no leading 0s, exactly one digit to left of
decimal/binary point
Alternatives to represent 1/1,000,000,000



radix (base)
Normalized:
1.0 x 10-9
Not normalized: 0.1 x 10-8, 10.0 x 10-10
Computer arithmetic that supports scientific notation
numbers is called floating point, because the binary point
is not fixed, as it is for integers
Advanced ALU
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FP Representation
Normalized format: 1.xxxxxxxxxxtwo  2yyyytwo
 Want to put it into multiple words: 32 bits for
single-precision and 64 bits for double-precision
 A simple single-precision representation:
31 30
23 22
0
S Exponent
Fraction

1 bit 8 bits
23 bits
S represents sign
Exponent represents y’s
Fraction represents x’s
 Represent numbers as small as 2.0 x 10-38 to
as large as 2.0 x 1038
Advanced ALU
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Double Precision Representation

Next multiple of word size (64 bits)
31 30
20 19
S
Exponent
1 bit
Fraction
11 bits
0
20 bits
Fraction (cont’d)
32 bits

Double precision (vs. single precision)
Represent numbers almost as small as
2.0 x 10-308 to almost as large as 2.0 x 10308
 But primary advantage is greater accuracy
due to larger fraction

Advanced ALU
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IEEE 754 Standard (1/4)



Regarding single precision, DP similar
Sign bit:
1 means negative
0 means positive
Significand:
To pack more bits, leading 1 implicit for normalized
numbers
 Significand = 1 + fraction: 1 + 23 bits single, 1 + 52
bits double
 Always true: 0 < fraction < 1
(for normalized numbers)


Note: 0 has no leading 1, so reserve exponent
value 0 just for number 0
Advanced ALU
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IEEE 754 Standard (2/4)

Exponent:
Need to represent positive and negative exponents
 Also want to compare FP numbers as if they were
integers, to help in value comparisons
 If use 2’s complement to represent?
e.g., 1.0 x 2-1 versus 1.0 x2+1 (1/2 versus 2)

1/2 0 1111 1111 000 0000 0000 0000 0000 0000
2 0 0000 0001 000 0000 0000 0000 0000 0000
If we use integer comparison for these two
words, we will conclude that 1/2 > 2!!!
Advanced ALU
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Biased (Excess) Notation


Remapping: shift all values by subtracting a bias
from the unsigned representation
Biased 7
Bit-pattern biased-7
value
0000
0001
0010
0011
0100
0101
0110
0111
Advanced ALU
-7
-6
-5
-4
-3
-2
-1
0
Bit-pattern biased-7
value
1000
1
•All-zero
pattern is the
1001
2
smallest; all-one
1010
3
pattern is the
1011
4
largest
1100
5
•Now integer
1101
6
comparison can
be used directly
1110
7
to compare
1111
8
exponents
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IEEE 754 Standard (3/4)

Use biased notation for exponents, where bias is
the number subtracted to get the real number
IEEE 754 uses bias of 127 for single precision:
subtract 127 from Exponent field to get actual value
for exponent
 1023 is the bias for double precision

1/2 0 0111 1110 000 0000 0000 0000 0000 0000
2 0 1000 0000 000 0000 0000 0000 0000 0000

All-zero and all-one exponent patterns are
reserved for special numbers (Fig 3.15)
Advanced ALU
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IEEE 754 Standard (4/4)

Summary (single precision):
31 30
23 22
S Exponent
1 bit
Fraction
8 bits
0
23 bits
(-1)S x (1.Fraction) x 2(Exponent-127)

Double precision identical, except longer
fraction and exponent segments with
exponent bias of 1023
Advanced ALU
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Example: FP to Decimal
0 0110 1000 101 0101 0100 0011 0100 0010


Sign: 0 => positive
Exponent:
0110 1000two = 104ten
 Bias adjustment: 104 - 127 = -23


Fraction:


1+2-1+2-3 +2-5 +2-7 +2-9 +2-14 +2-15 +2-17 +2-22
= 1.0 + 0.666115
Represents: 1.666115ten2-23  1.986  10-7
Advanced ALU
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Example 1: Decimal to FP



Number = - 0.75
= - 0.11two  20 (scientific notation)
= - 1.1two  2-1 (normalized scientific
notation)
Sign: negative => 1
Exponent:
 Bias adjustment: -1 +127 = 126
 126ten = 0111 1110two
1 0111 1110 100 0000 0000 0000 0000 0000
Advanced ALU
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Example 2: Decimal to FP

A more difficult case: representing 1/3?
= 0.33333…10 = 0.0101010101… 2  20
= 1.0101010101… 2  2-2
 Sign: 0
 Exponent = -2 + 127 = 12510=011111012
 Fraction = 0101010101…
0 0111 1101 0101 0101 0101 0101 0101 010
Advanced ALU
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Special Numbers

What have we defined so far? (single precision)
Exponent
0
0
1-254
255
255

Fraction
0
nonzero
anything
0
nonzero
Object
0
???
+/- floating-point
???
???
Range:
1.0  2-126  1.8  10-38
What if result too small? (>0, < 1.8x10-38 =>
Underflow!)
(2 – 2-23)  2127  3.4  1038
What if result too large? (> 3.4x1038 => Overflow!)
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Denormalized Numbers

For single-precision f. p. (page 217)
0 is 00000000000000000000000000000000
 Smallest positive normalized number is
1.0000 0000 0000 0000 0000 000 x 2-126
 This is actually a pretty big gap...
 Gradual underflow: allow a number to degrade in
significance until it becomes 0
 By getting rid of implicit leading 1 in front of the
fraction... we can represent a smaller number
 We denote a denormalized number by 0 exponent and
a non-zero fraction



The smallest denormalized number is
0.0000 0000 0000 0000 0000 001 x 2-126 or 1.0 x 2-149
For double-precision f. p.

Smallest denormalized number: 1.0 x 2-1074
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Special Numbers

What have we defined so far? (single
precision)
Exponent
0
0
1-254
255
255
Advanced ALU
Fraction
0
nonzero
anything
0
nonzero
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Object
0
denorm
+/- floating-point
???
???
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Representation for +/- Infinity
In FP, divide by zero should produce +/infinity, not overflow
 Why?

 OK
to do further computations with infinity, e.g.,
X/0 > Y may be a valid comparison

IEEE 754 represents +/- infinity by all-one
exponent and all-zero fraction
S 1111 1111 0000 0000 0000 0000 0000 000
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Representation for Not a Number

What do I get if I calculate sqrt(-4.0) or 0/0?
If infinity is not an error, these should not be either
 They are called Not a Number (NaN)
 IEEE 754: all-one Exponent (255), nonzero Fraction


Why is this useful?
A symbol for invalid operations: allow programmers to
postpone some tests and decisions to later time
 They contaminate: op(NaN,X) = NaN

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Special Numbers

What have we defined so far? (single-precision)
Exponent
0
0
1-254
255
255
Fraction
0
nonzero
anything
0
nonzero
Object
0
denom
+/- fl. pt. #
+/- infinity
NaN
Fig. 3.15
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FP Operations Complexities


Operations are somewhat more complicated
In addition to overflow we can have “underflow”


Absolute value too small
Accuracy can be a big problem
IEEE 754 keeps two extra bits, guard and round
 Four rounding modes
 Special cases:





Positive divided by zero yields “infinity”
Zero divide by zero yields “not a number”
Implementing the standard can be tricky
Not using the standard can be even worse

See text for description of 80x86 and Pentium bug!
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Basic Addition Algorithm
(1) We'll need to convert from fraction to significand
(2) We'll need to shift the smaller number so that we have
the same exponent:
num of shifts= Larger Number's exponent - Smaller Number's exponent
-- right shift smaller number that many positions
(3) Add or subtract the two numbers
(4) Normalize:
a. left shift result, decrement result exponent (e.g. 0.0001)
b. right shift result, increment result exponent (e.g. 101.1)
continue until MSB of data is 1 (will not be stored)
(5) Check for overflow or underflow
(6) If the result is a 0 significand, set the exponent to zero
by special step
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Floating Point Addition Example
Example: Add 9.999  101 and 1.610  10-1 assuming 4 decimal digits
 Align decimal point of number with smaller exponent



1.610  10-1 = 0.161  100 = 0.0161  101
Shift smaller number to right
Add significands
9.999
+ 0.016
10.015
 SUM = 10.015  101
 NOTE: One digit of precision lost during shifting


Shift sum to put it in normalized form 1.0015  102
Since significand only has 4 digits, we need to round the
sum


SUM = 1.002  102
NOTE: normalization maybe needed again after
rounding, e.g, rounding 9.9999 you get 10.000
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Start
FP Addition

Algorithm: Fig 3.16
1. Compare the exponents of the two numbers.
Shift the smaller number to the right until its
exponent would match the larger exponent
2. Add the significands
3. Normalize the sum, either shifting right and
incrementing the exponent or shifting left
and decrementing the exponent
Overflow or
underflow?
Yes
No
Exception
4. Round the significand to the appropriate
number of bits
No
Still normalized?
Yes
Done
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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
Advanced ALU
Exponent
Add
1
Shift left or right
Rounding hardware
Sign
1
Significand
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Normalize
Round
Fig 3.17
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Accurate Arithmetic



Round accurately requires HW to include extra
bits in the calculation
IEEE 754 standard specifies the use of 2 extra
bits on the right during intermediate calculations
– Guard bit and Round bit
Example: Add 2.56  100 and 2.34  102,
assuming 3 significant digits

Shift the smaller number


Without guard and round bits


2.56  100 = 0.0256  102(guard holds 5 and round holds 6)
2.34+ 0.02 = 2.36  102
With guard and round bits


2.34+ 0.0256 = 2.3656  102
ROUND  2.37102
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Floating-Point Multiplication

Algorithm: Z = X * Y
 Product exponent = X's exponent + Y's
exponent


Doubly-biased exponent must be corrected
Product exponent = sum of stored exponents – bias
Xe = 7
Ye = -3
Bias 8
Xe = 1111
Ye = 0101
10100
= 15
= 5
20
= 7+8
= -3 + 8
4+8+8
 Multiply
the significands: Zm = Xm x Ym
 Normalization, rounding and exception
checking (overflow and underflow)
 Determine the sign bit
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MIPS Floating-Point Instructions

Floating-point instructions
Computation: add.s, add.d, sub.s, sub.d, mul.s, mul.d,
div.s, div.d
 Comparison: c.x.s, c.x.d (x=eq, neq, lt, le, gt, ge)
 FP conditional branch: bc1t, bc1f
 Floating-point data transfer: lwc1, swc1, ldc1, sdc1


32 floating-point registers: $f0, …, $f31


Double precision: by convention, even/odd pair
contain one DP FP number: $f0/$f1, $f2/$f3
See page 206 for details
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Summary

Multiplication
 Refined
hardware and algorithms
 Booth’s algorithm

Division
 Similar

hardware as multiplication
Floating point
 Representing
numbers
 Addition and multiplication
 Precision is a big issue
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Next Lecture

Topic:
 Assessing

and understanding performance
Reading
 Ch4
Patterson and Hennessy
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