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Chapter 2
Bits, Data Types,
and Operations
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How do we represent data in a computer?
At the lowest level, a computer is an electronic machine.
• works by controlling the flow of electrons
Easy to recognize two conditions:
1. presence of a voltage – we’ll call this state “1”
2. absence of a voltage – we’ll call this state “0”
Could base state on value of voltage,
but control and detection circuits more complex.
• compare turning on a light switch to
measuring or regulating voltage
We’ll see examples of these circuits in the next chapter.
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Computer is a binary digital system.
Digital system:
• finite number of symbols
Binary (base two) system:
• has two states: 0 and 1
Basic unit of information is the binary digit, or bit.
Values with more than two states require multiple bits.
• A collection of two bits has four possible states:
00, 01, 10, 11
• A collection of three bits has eight possible states:
000, 001, 010, 011, 100, 101, 110, 111
• A collection of n bits has 2n possible states.
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What kinds of data do we need to represent?
• Numbers – unsigned, signed, integers, floating point,
complex, rational, irrational, …
• Text – characters, strings, …
• Images – pixels, colors, shapes, …
• Sound
• Logical – true, false
• Instructions
• …
Data type:
• representation and operations within the computer
We’ll start with numbers…
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Unsigned Integers
Non-positional notation
• could represent a number (“5”) with a string of ones (“11111”)
• problems?
Weighted positional notation
• like decimal numbers: “329”
• “3” is worth 300, because of its position, while “9” is only worth 9
329
102 101 100
3x100 + 2x10 + 9x1 = 329
most
significant
22
101
21
least
significant
20
1x4 + 0x2 + 1x1 = 5
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Unsigned Integers (cont.)
An n-bit unsigned integer represents 2n values:
from 0 to 2n-1.
22
21
20
0
0
0
0
0
0
1
1
0
1
0
2
0
1
1
3
1
0
0
4
1
0
1
5
1
1
0
6
1
1
1
7
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Converting Binary (2’s C) to Decimal
1. If leading bit is one, take two’s
complement to get a positive number.
2. Add powers of 2 that have “1” in the
corresponding bit positions.
3. If original number was negative,
add a minus sign.
X = 01101000two
= 26+25+23 = 64+32+8
= 104ten
n 2n
0
1
2
3
4
5
6
7
8
9
10
1
2
4
8
16
32
64
128
256
512
1024
Assuming 8-bit 2’s complement numbers.
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More Examples
X = 00100111two
= 25+22+21+20 = 32+4+2+1
= 39ten
X =
-X =
=
=
X=
11100110two
00011010
24+23+21 = 16+8+2
26ten
-26ten
n 2n
0
1
2
3
4
5
6
7
8
9
10
1
2
4
8
16
32
64
128
256
512
1024
Assuming 8-bit 2’s complement numbers.
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Converting Decimal to Binary (2’s C)
First Method: Division
1. Divide by two – remainder is least significant bit.
2. Keep dividing by two until answer is zero,
writing remainders from right to left.
3. Append a zero as the MS bit;
if original number negative, take two’s complement.
X = 104ten
X = 01101000two
104/2
52/2
26/2
13/2
6/2
3/2
=
=
=
=
=
=
52 r0
26 r0
13 r0
6 r1
3 r0
1 r1
1/2 = 0 r1
bit 0
bit 1
bit 2
bit 3
bit 4
bit 5
bit 6
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Converting Decimal to Binary (2’s C)
n 2n
Second Method: Subtract Powers of Two
1. Change to positive decimal number.
2. Subtract largest power of two
less than or equal to number.
3. Put a one in the corresponding bit position.
4. Keep subtracting until result is zero.
5. Append a zero as MS bit;
if original was negative, take two’s complement.
X = 104ten
104 - 64 = 40
40 - 32 = 8
8-8 = 0
0
1
2
3
4
5
6
7
8
9
10
1
2
4
8
16
32
64
128
256
512
1024
bit 6
bit 5
bit 3
X = 01101000two
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Unsigned Binary Arithmetic
Base-2 addition – just like base-10!
• add from right to left, propagating carry
carry
10010
+ 1001
11011
10010
+ 1011
11101
1111
+
1
10000
10111
+ 111
Subtraction, multiplication, division,…
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Signed Integers
With n bits, we have 2n distinct values.
• assign about half to positive integers (1 through 2n-1)
and about half to negative (- 2n-1 through -1)
• that leaves two values: one for 0, and one extra
Positive integers
• just like unsigned – zero in most significant bit
00101 = 5
Negative integers
• sign-magnitude – set top bit to show negative,
other bits are the same as unsigned
10101 = -5
• one’s complement – flip every bit to represent negative
11010 = -5
• in either case, MS bit indicates sign: 0=positive, 1=negative
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Negative numbers (Sign magnitude)
This technique uses additional binary digit to represent
the sign, 0 to represent positive, 1 to represent negative.
Ex.:
5 = 0101
-5 = 1101
Also
5 = 00000101
-5= 10000101
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Negative numbers (1’s Complement)
1’s comp is another method used to represent negative
numbers.
In this method we invert every bit in the positive number
in order to represent its negative.
Ex.:
9 = 01001
-9= 10110
Again remember that we use additional digit to
represent the sign
We may represent 9 as 00001001 ; zeros on left have no
value
-9 in 1’s comp will be 11110110
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Negative numbers (2’s complement)
This is the method which is used in most computers to
represent integer numbers nowadays.
Remember always to represent a positive number using
any of the previous methods is the same, all what is
needed is a 0 on the left to show that the number is
positive
To represent a negative number in 2’s Comp , first we
find the 1’s Comp, then add 1 to the result
Ex:
How we represent -9 in 2’s comp
1- 9 in binary= 01001
2- invert = 10110
3 add 1 = 10111; -9 in 2’s Comp.
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Two’s Complement Representation
If number is positive or zero,
• normal binary representation, zeroes in upper bit(s)
If number is negative,
• start with positive number
• flip every bit (i.e., take the one’s complement)
• then add one
00101 (5)
11010 (1’s comp)
+
1
11011 (-5)
01001 (9)
(1’s comp)
+
1
(-9)
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Two’s Complement
Problems with sign-magnitude and 1’s complement
• two representations of zero (+0 and –0)
• arithmetic circuits are complex
How to add two sign-magnitude numbers?
– e.g., try 2 + (-3)
How to add to one’s complement numbers?
– e.g., try 4 + (-3)
Two’s complement representation developed to make
circuits easy for arithmetic.
• for each positive number (X), assign value to its negative (-X),
such that X + (-X) = 0 with “normal” addition, ignoring carry out
00101 (5)
+ 11011 (-5)
00000 (0)
01001 (9)
+
(-9)
00000 (0)
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Two’s Complement Shortcut
To take the two’s complement of a number:
• copy bits from right to left until (and including) the first “1”
• flip remaining bits to the left
011010000
100101111
+
1
100110000
011010000
(1’s comp)
(flip)
(copy)
100110000
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Two’s Complement
MS bit is sign bit – it has weight –2n-1.
Range of an n-bit number: -2n-1 through 2n-1 – 1.
• The most negative number (-2n-1) has no positive counterpart.
-23
22
21
20
-23
22
21
20
0
0
0
0
0
1
0
0
0
-8
0
0
0
1
1
1
0
0
1
-7
0
0
1
0
2
1
0
1
0
-6
0
0
1
1
3
1
0
1
1
-5
0
1
0
0
4
1
1
0
0
-4
0
1
0
1
5
1
1
0
1
-3
0
1
1
0
6
1
1
1
0
-2
0
1
1
1
7
1
1
1
1
-1
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Operations: Arithmetic and Logical
Recall:
a data type includes representation and operations.
We now have a good representation for signed integers,
so let’s look at some arithmetic operations:
• Addition
• Subtraction
• Sign Extension
We’ll also look at overflow conditions for addition.
Multiplication, division, etc., can be built from these
basic operations.
Logical operations are also useful:
• AND
• OR
• NOT
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Addition
As we’ve discussed, 2’s comp. addition is just
binary addition.
• assume all integers have the same number of bits
• ignore carry out
• for now, assume that sum fits in n-bit 2’s comp. representation
01101000 (104)
11110110 (-10)
+ 11110000 (-16) +
(-9)
01011000 (88)
(-19)
Assuming 8-bit 2’s complement numbers.
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Subtraction
Negate subtrahend (2nd no.) and add.
• assume all integers have the same number of bits
• ignore carry out
• for now, assume that difference fits in n-bit 2’s comp.
representation
01101000 (104)
11110101 (-11)
- 00010000 (16) - 11110111 (-9)
01101000 (104)
11110101 (-11)
+ 11110000 (-16) +
(9)
01011000 (88)
(-2)
Assuming 8-bit 2’s complement numbers.
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Sign Extension
To add two numbers, we must represent them
with the same number of bits.
If we just pad with zeroes on the left:
4-bit
8-bit
0100 (4)
00000100 (still 4)
1100 (-4)
00001100 (12, not -4)
Instead, replicate the MS bit -- the sign bit:
4-bit
8-bit
0100 (4)
00000100 (still 4)
1100 (-4)
11111100 (still -4)
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Overflow
If operands are too big, then sum cannot be represented
as an n-bit 2’s comp number.
01000 (8)
+ 01001 (9)
10001 (-15)
11000 (-8)
+ 10111 (-9)
01111 (+15)
We have overflow if:
• signs of both operands are the same, and
• sign of sum is different.
Another test -- easy for hardware:
• carry into MS bit does not equal carry out
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Int x=2000000000;
Int y=1000000000;
Int z=x+y;
Cout<<z;
What is the max number represented in Short type
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Logical Operations
Operations on logical TRUE or FALSE
• two states -- takes one bit to represent: TRUE=1, FALSE=0
View n-bit number as a collection of n logical values
• operation applied to each bit independently
A
0
0
1
1
B
0
1
0
1
A AND B
0
0
0
1
A
0
0
1
1
B
0
1
0
1
A OR B
0
1
1
1
A
0
1
NOT A
1
0
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Examples of Logical Operations
AND
• useful for clearing bits
AND with zero = 0
AND with one = no change
11000101
AND 00001111
00000101
OR
• useful for setting bits
OR with zero = no change
OR with one = 1
NOT
• unary operation -- one argument
• flips every bit
OR
NOT
11000101
00001111
11001111
11000101
00111010
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X=5
Y=6
Z=X&Y
Cout<< Z
Z=X|Y
Cout<< Z
Z= ~X
Cout<< Z
X=-6
Z= ~X
Cout<< Z
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Hexadecimal Notation
It is often convenient to write binary (base-2) numbers
as hexadecimal (base-16) numbers instead.
• fewer digits -- four bits per hex digit
• less error prone -- easy to corrupt long string of 1’s and 0’s
Binary
Hex
Decimal
Binary
Hex
Decimal
0000
0001
0010
0011
0100
0101
0110
0111
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
7
1000
1001
1010
1011
1100
1101
1110
1111
8
9
A
B
C
D
E
F
8
9
10
11
12
13
14
15
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Converting from Binary to Hexadecimal
Every four bits is a hex digit.
• start grouping from right-hand side
011101010001111010011010111
3
A
8
F
4
D
7
This is not a new machine representation,
just a convenient way to write the number.
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Hexadecimal
As we noticed to read a long binary number is confusing
So another numbering system was invented ( base 16)
We know base 10 ,base 2, and now base 16
Numbers in (base 16) are : 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F
There is a direct relation between base 2 and base 16,
24=16 , so the conversion from binary to hexadecimal is
quite easy.
Each 4 binary digits are converted to one hexadecimal
digit.
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Real numbers(Fixed Point)
All what we discussed before was about integers.
What about real numbers (ex.: 6.125)
To represent real numbers we take first the integer part
and convert is as we learned before, and then take the
fraction part and convert it using the following algorithm
 1- multiply fraction by 2
 2- take the integer of the result
 3- Repeat 1 and 2 until the fraction is zero, or until u reach get
enough digits.
Ex. : 6.125
6= 110
0.125x2 = 0.25
0.25x2 = 0.5
0.5x2 = 1.0
6.125= 110.001
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Another example:
Convert 9.2 to binary.
9=1001
0.2x2 = 0.4
0.4x2 = 0.8
0.8x2 = 1.6 ; take the fraction only
0.6x2 = 1.2
0.2x2 = 0.4 ; let us stop here, 5 digits after the point
The binary equivalent will be: 1001.00110
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Convert from binary
Again let us take the previous results
22 21 20 2-1 2-2 2-3
11 0.0 0 1
= 4+2+.125=6.125
Ex. 2:
23 22 21 20 2-1 2-2 2-3 2-4 2-5
1 0 0 1. 0 0 1 1 0
= 8+1+ .125+.0625 = 9.1875
Why not 9.2 !??
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Float x=0;y=0.1;
While (x!=1)
{cout<<“Hi”;
X=x+y;
}
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Float x=1,y=3;
X=x/y*y -1;
Cout<<x;
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Floating numbers
Nowadays numbers with decimal point are represended in
the computer using IEEE 754 float number format.
This format has to subtypes :
• Float: 32 bit number can represent up to 1038 .
• Double: 64 bit number can represent up to 10310 with more
number of digits after the point.
We will not cover this topic here.
It will be covered in Computer architecture course.
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Very Large and Very Small: Floating-Point
Large values: 6.023 x 1023 -- requires 79 bits
Small values: 6.626 x 10-34 -- requires >110 bits
Use equivalent of “scientific notation”: F x 2E
Need to represent F (fraction), E (exponent), and sign.
IEEE 754 Floating-Point Standard (32-bits):
1b
8b
S Exponent
23b
Fraction
N  1S  1.fraction  2exponent 127 , 1  exponent  254
N  1S  0.fraction  2 126 , exponent  0
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1000= 100.0 x21 =10.00 x22 = 1.000 x23 =0.1000 x24
10000= 1x24
101000=101x23= 1.01x25
0.125= 0.001= 1x2-3
2127= x.xxxx 10 38
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Floating Point Example
Single-precision IEEE floating point number:
10111111010000000000000000000000
sign exponent
fraction
• Sign is 1 – number is negative.
• Exponent field is 01111110 = 126 (decimal).
• Fraction is 0.100000000000… = 0.5 (decimal).
Value = -1.5 x 2(126-127) = -1.5 x 2-1 = -0.75.
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ِAnother example
-0.125 in IEEE 754 binary floating format
1) 0.125  0.001 in fixed point binary
2) 0.001 = 1.000000000 x 2 -3
after normalization
3) exponent = 127+ -3 = 124
4) 124 = 0111 1100 in binary
5) 1 0111 1100 0000 0000 0000 0000 0000 000
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Another example:
Write 9.2 in IEEE floating point format.
The binary equivalent will be:
1) 1001.0011001100110011…
2) 1.0010011001100110011…… x 23
3) Exponent = 127+3=130
4) Exponent in binary =1000 0010
5) 0 1000 0010 00100110011001100110011
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Floating-Point Operations
Will regular 2’s complement arithmetic work for
Floating Point numbers?
(Hint: In decimal, how do we compute 3.07 x 1012 + 9.11 x 108?)
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Text: ASCII Characters
ASCII: Maps 128 characters to 7-bit code.
• both printable and non-printable (ESC, DEL, …) characters
00
01
02
03
04
05
06
07
08
09
0a
0b
0c
0d
0e
0f
nul
soh
stx
etx
eot
enq
ack
bel
bs
ht
nl
vt
np
cr
so
si
10
11
12
13
14
15
16
17
18
19
1a
1b
1c
1d
1e
1f
dle
dc1
dc2
dc3
dc4
nak
syn
etb
can
em
sub
esc
fs
gs
rs
us
20
21
22
23
24
25
26
27
28
29
2a
2b
2c
2d
2e
2f
sp
!
"
#
$
%
&
'
(
)
*
+
,
.
/
30
31
32
33
34
35
36
37
38
39
3a
3b
3c
3d
3e
3f
0
1
2
3
4
5
6
7
8
9
:
;
<
=
>
?
40
41
42
43
44
45
46
47
48
49
4a
4b
4c
4d
4e
4f
@
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
50
51
52
53
54
55
56
57
58
59
5a
5b
5c
5d
5e
5f
P
Q
R
S
T
U
V
W
X
Y
Z
[
\
]
^
_
60
61
62
63
64
65
66
67
68
69
6a
6b
6c
6d
6e
6f
`
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
70
71
72
73
74
75
76
77
78
79
7a
7b
7c
7d
7e
7f
p
q
r
s
t
u
v
w
x
y
z
{
|
}
~
del
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A = $ 41 in hexa decimal = 65 in decimal
a = $ 61 in hexa decimal = 97 in decimal
Space = $20 = 32
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Interesting Properties of ASCII Code
What is relationship between a decimal digit ('0', '1', …)
and its ASCII code?
What is the difference between an upper-case letter
('A', 'B', …) and its lower-case equivalent ('a', 'b', …)?
Given two ASCII characters, how do we tell which comes
first in alphabetical order?
Are 128 characters enough?
(http://www.unicode.org/)
No new operations -- integer arithmetic and logic.
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Other Data Types
Text strings
• sequence of characters, terminated with NULL (0)
• typically, no hardware support
Image
• array of pixels
monochrome: one bit (1/0 = black/white)
color: red, green, blue (RGB) components (e.g., 8 bits each)
other properties: transparency
• hardware support:
typically none, in general-purpose processors
MMX -- multiple 8-bit operations on 32-bit word
Sound
• sequence of fixed-point numbers
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Images
•Normally Each pixel is represented as 24 bits (3 bytes)
•The file size of an image is proportional to the image dimensions and
the color quality
60
•Example
50
•Number of pixels is 60x50=3000 pixel
•If each pixel is represented using 3 bytes RGB the file size will be
3000x3= 9KByte
•if the picture is Monocrome, then the file size is 3000x1_bit= 3000 bit
= 3000/8 Byte  400 Byte
•If the number of colors used is 256 then each pixel is represented as
8 bit (1 Byte) ( remember 28 =256)  the image size is 3000 Bytes
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Digital Camera
0.3 MP= 480x640
1.3 MP = 1280x1024
2 MP =1200 x1600
3.15 MP= 1536 x 2048
5 MP =
8.1 MP=
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Image compression
•But some times we see that the size of the Image file is
not as we calculated?
•Data representing Images can be compressed
•Two types of compression:
• Lossless compression: GIF
• Loosely compression: JPG
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Cameras and Mega Pixel?
2MP = 1600x1200
=192000
3.15MP =2048x1536
=3145000
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video
Frame per second
Fps
10 fps
15 fps
30 fps
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LC-2 Data Types
Some data types are supported directly by the
instruction set architecture.
For LC-2, there is only one supported data type:
• 16-bit 2’s complement signed integer
• Operations: ADD, AND, NOT
Other data types are supported by interpreting
16-bit values as logical, text, fixed-point, etc.,
in the software that we write.
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