### Lecture 5 - Integers and Arithmetic4

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CS 325: CS Hardware and Software
Organization and Architecture
Integers and Arithmetic
Part 4
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Outline
 Binary

Multiplication
Booth’s Algorithm
 Number
Representations
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2’s Complement Binary
Multiplication – Booth’s Algorithm

Multiplication by bit shifting and addition.


Removes the need for multiply circuit
Requires:
 A way to compute 2’s Complement
 Available as fast hardware instructions
 X86 assembly instruction: NEG
 A way to compare two values for equality
A
B
A
 How to do this quickly?
0
0
Exclusive Not OR (NXOR) Gate
Compare all sequential bits of bit string A
0
1
and bit string B. Values are equal if the
1
0
comparison process produces all 1s.
 A way to shift bit strings.
1
1
 Arithmetic bit shift, which preserves the sign bit when shifting to the
right.
10110110
arithmetic shift right
11011011
 x86 assembly instruction: SAR
NXOR B
1
0
0
1
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2’s Complement Binary
Multiplication – Booth’s Algorithm

Example: 5 x -3

First, convert to 2s comp bin:
 5 = 0101
 -3 = 1101
 If
we add 0 to the right of both values, there are
4 0-1 or 1-0 switches in 0101, and 3 in 1101. Pick
1101 as X value, and 0101 as Y value
 Next,

2s Comp of Y: 1011
for bin subtraction.
 Next,
set 2 registers, U and V, to 0. Make a table
using U, V, and 2 additional registers X, and X-
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2’s Complement Binary
Multiplication – Booth’s Algorithm


Register X is set to the predetermined value of x, and X-1 is set to 0
U
V
X
X-1
0000
0000
1101
0
Rules: Look at the LSB of X and the number in the X-1 register.

If the LSB of X is 1, and X-1 is 0, we subtract Y from U.

If LSB of X is 0, and X-1 is 1, then we add Y to U.

If both LSB of X and X-1 are equal, do nothing and skip to shifting stage.
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2’s Complement Binary
Multiplication – Booth’s Algorithm

In our case, the LSB of X is one, and X-1 is zero, so we subtract Y from U.
U
V
X
X-1
0000
0000
1101
0
1000
1110
1
+1011
1011
1101

Next, we do an arithmetic right shift on U and V

1011  1101, 0000  1000

Copy the LSB of X into X-1

And then perform a circular right shift on X


1101  1110
Repeat the process three more times.
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2’s Complement Binary
Multiplication – Booth’s Algorithm

The LSB of X is zero, and X-1 is one, so we add Y to U.
U
V
X
X-1
1101
1000
1110
1
0100
0111
0
+0101
0010
0001

Next, we do an arithmetic right shift on U and V

0010 0001, 1000  0100

Copy the LSB of X into X-1

And then perform a circular right shift on X


1110  0111
Repeat the process two more times.
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2’s Complement Binary
Multiplication – Booth’s Algorithm

The LSB of X is one, and X-1 is zero, so we subtract Y from U.
U
V
X
X-1
0001
0100
0111
0
0010
1011
1
+1011
1100
1110

Next, we do an arithmetic right shift on U and V

1100 1110, 0100  0010

Copy the LSB of X into X-1

And then perform a circular right shift on X


0111  1011
Repeat the process one more time.
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2’s Complement Binary
Multiplication – Booth’s Algorithm


The LSB of X is one, and X-1 is one, begin shifts.
U
V
X
X-1
1110
0010
1011
1
1111
0001
1101
1
Next, we do an arithmetic right shift on U and V

1110 1111, 0010  0001

Copy the LSB of X into X-1

And then perform a circular right shift on X

1011  1101
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2’s Complement Binary
Multiplication – Booth’s Algorithm

The result is stored in U followed by V.
U
V
X
X-1
1111
0001
1101
1
11110001

This result is stored in 2’s complement notation.

Convert to decimal:
11110001  00001111  -1510

This gives the correct result of 3 x -5
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2’s Complement Binary
Multiplication – Booth’s Algorithm

Another Example: 7 x -4

First, convert to 2s comp bin:

7  0111, add zero to right gives 01110, 2 switches

-4  1100, add zero to right gives 11000, 1 switch

X = 1100

Y = 0111

-Y = 1001, for easy bin subtract
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2’s Complement Binary
Multiplication – Booth’s Algorithm
U
V
X
X-1
0: 0000
0000
1100
0
1: 0000
0000
0110
0
2: 0000
0000
0011
0
+1001
1001
3: 1100
1000
1001
1
4: 1110
0100
1100
1

Result of 7 x -4: UV
11100100  00011100  -2810
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2’s Complement Binary
Multiplication – Booth’s Algorithm
Try:
-9 x 7
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Memory
A
is a place to store bits
word is a fixed number of bits
 Ex: 32 bits, or 4 bytes
 An address is also a fixed
number of bits
 Represented as
unsigned numbers
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Numbering Bits and Bytes
 Need
to choose order for:
 Storage
in physical memory system
 Transmission over serial/parallel medium
(data network)
 Bit
order
 Handled
by hardware
 Usually hidden from programmer
 Byte
order
 Affects
multi-byte data items such as
integers
 Visible and important to programmers
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Possible Byte Orders
Least
significant byte of integer in
lowest memory location

Little endian
Most
Significant byte of integer in
lowest memory location.

Big endian
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Byte Order Illustration

Note: Difference is especially important when transferring
data between computers for which the byte ordering
differs.
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Sign Extension
Convert
2’s comp number using N bits
to more than N bits (int to long int):
 Replicate
the MSB (sign bit) of the smaller
number to fill new bits.
 2’s comp positive number has infinite 0s
 2’s comp negative number has infinite 1s
 Ex: 16bit
-410 to 32-bit:
1111 1111 1111 1100
1111 1111 1111 1111 1111 1111 1111 1100
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Conclusion
 We
represent “things” in computers as
particular bit patterns: N bits  2N
 Decimal
for human calculations, binary
for computers, hex for convenient way to
write binary
 2’s
comp universal in computing: so
make sure to learn!
 Number
are infinite, computers are not,
so errors can occur (overflow,
underflow)
 Know
the powers of 2.
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