Csci1802 Computer Systems
Computer Number Systems
CSCI 1802
Computer Systems and Networks
Part 1: Computer Number Systems
References and further reading:
Blundell B: Computer Systems and Networks
Patt & Patel: Introduction to Computing Systems pp 1 – 50
Ceri, Mandroli, Sbattella: The Art & Craft of Computing
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Csci1802 Computer Systems
Computer Number Systems
© De Montfort University, Dowson, Zarzycki, Birkenhead
csci1802 Computer Systems
Computer Number Systems
Computer Number Systems
1
Motivation
At the lowest level a computer has an electronic component that stores either a high voltage, or a
low voltage. We take a high voltage to represent the digit 1, and a low voltage to represent the
digit 0. So now we have a very simple and very limited counting device.
With two such components we have four states: 00,01,10 and 11.
These four states can represent the numbers 0, 1, 2, 3.
Similarly with 3 components we can represent eight states:
000,001,010,011,100,101,110 and 111.
These eight states can represent the numbers 0, 1, 2, 3, 4, 5, 6 and 7.
1.1
Question
How many states are there with:
four components?
five components?
six components?
seven components?
eight components?
What is the pattern?
When computer scientists and engineers work at the level of the computer, they represent
numbers using binary digits 0, 1. As shorthand they use octal digits (0,1,2,3,4,5,6,7) to represent
a group of 3 binary digits; or hexadecimal digits (0,1,2,3,4,5,6,7,8,9,A, B, C, D, E, F) to
represent a group of 4 binary digits.
The first section explains how to work with these number systems.
First, we must be able to understand raising a number to a power.
Can you do the following (without a calculator!)?
(Note that any number to the power 0 equals 1 i.e. n0 = 1)
22 =
23 =
32 =
33 =
20 =
50 =
10-2 =
2-2 =
10-5 =
3-3 =
21/2 =
20.5 =
21/3 =
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104 =
1
Csci1802 Computer Systems
2
Computer Number Systems
Radix (or Base)
Numbers are written using a place system, e.g. Hundreds, Tens and Units. For example, 123 is
the number where 1 represents one 100, 2 represents 2 tens, and 3 represents 3 units (or ones).
Altogether the number is 100+20+3. Each place in the number system is 10 times larger than the
previous place. This idea can be extended to other number bases.

Decimal is base 10 (or radix 10) and uses the symbols 0 1 2 3 4 5 6 7 8 9
The number 271 is 2×102 + 7×101 + 1×100

Binary is base 2 (or radix 2) and uses the symbols 0 1
The number 011012 is 0×24 + 1×23 + 1×22 + 0×21 + 1×20

Octal is base 8 (or radix 8) and uses the symbols 0 1 2 3 4 5 6 7
The number 51728 is 5×83 + 1×82 + 7×81 + 2×80

Hexadecimal is base 16 (or radix 16) and uses the symbols 0 1 2 3 4 5 6 7 8 9 A B C D E F
where A = 10, B = 11, C = 12, D = 13, E = 14, F = 15
The number 5A1B16 is
5×163 + 10×162 + 1×161 + 11×160
= 20480 + 2560 + 16
+
11
= 2306710
Ref: Blundell pp10-13, pp36-39
Exercises
1. List some of the different number bases you use in everyday life. Hint: think about time …
2
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csci1802 Computer Systems
Computer Number Systems
3
To Translate from one base to another
3.1
Binary to Decimal
As illustrated above, starting at the right hand side, the digits represent increasing powers of 2.
111010012
7
6
Decimal value
of each bit
Example
binary
number
Value of each
bit in the
example
Sum of the
values
4
3
2
1
0
e.g. to convert 111010012
The easiest way is to use a table.
Powers of 2
5
= 1×2 + 1×2 + 1×2 + 0×2 + 1×2 + 0×2 + 0×2 + 1×2
= 128 + 64 + 32 + 0 + 8
+ 0
+ 0 + 1
= 23310
27
26
25
24
23
22
21
20
128
64
32
16
8
4
2
1
1
1
1
0
1
0
0
1
Here’s a more complicated example:
01011011111010012
= 0+16384+0+4096+2048+0+512+256+128+64+32+0+8+0+0+1
= 2352910
We can write this as a table as shown below.
Powers of
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
2
2
2
2
2
2
2
2
2
2 2 2 2 2 2 2
2
Decimal
32768 16384 8192 4096 2048 1024 512 256 128 64 32 16 8
4
2
1
value of
each bit
Example
0
1
0
1
1
0
1
1
1
1
1
0
1
0
0
1
binary
number
Value of
each bit in
0
16384
0
4096 2048
0
512 256 128 64 32 0
8
0
0
1
the
example
Sum of the
= 16384 + 4096 + 2048 + 512 + 256 + 128 + 64 + 32 + 8 + 1 = 23529
values
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3.2
Computer Number Systems
Binary to Octal
Starting from the right hand side, divide the number into groups of 3 digits, and then convert
each group into an octal value. This works because 3 binary digits represent the range of values
from 0 to 7, i.e. all the octal digits.
01011011111010012
3.3
= 0 101 101 111 101 0012
= 0 5
5
7
5
1
= 0557518
Binary to Hexadecimal
Starting from the right hand side, divide the number into groups of 4 digits, and then convert
each group into a hexadecimal value. This works because 4 binary digits represent the range of
values from 0 to 15, i.e. all the hexadecimal digits.
01011011111010012
3.4
=
=
=
0101 1011 1110 10012
5
B
E
9
5BE916
Decimal to Binary
3.4.1 Either...
Subtract from the original number the highest power of 2 that does not exceed the value of the
number, put a 1 in the corresponding bit position. Then continue down through the powers of 2
inserting 1's or 0's in the corresponding positions. (A table of powers of 2 helps here.)
Example: To convert 75 to binary:
Powers of 2 27
Decimal
value of each 128
bit
26
25
24
23
22
21
20
64
32
16
8
4
2
1
64 is the highest power of 2 not greater than 75
6
64 = 2
01000000
75 - 64 = 11
Using 64
1
8 is the highest power of 2 not greater than 11
3
8 = 2
01001000
11 - 8 = 3
Using 8
1
1
then, following on in the same way
1
2 = 2
01001010
3 - 2 = 1
Using 2
1
1
1
Using 1
1
1
1
1
1
1
1 = 2
1 - 1 = 0
0
01001011
Put 0 in gaps
4
0
1
0
0
1
0
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csci1802 Computer Systems
Computer Number Systems
3.4.2 ...Or
Successively divide the number by 2 and use the remainders as the binary digits (from right to
left)
Thus
75/2
37/2
18/2
9/2
4/2
2/2
1/2
gives 37 remainder 1
gives 18 remainder 1
gives 9 remainder 0
gives 4 remainder 1
gives 2 remainder 0
gives 1 remainder 0
gives 0 remainder 1
1
11
011
1011
01011
001011
1001011
Pad out with 0's (on the left hand side) to make the binary number the width you require.
7510
=
01001011
Note: If you use this method, it is a good idea to always check your answer by converting your
result back to decimal.
3.5
Decimal to Octal
Convert the number to binary then convert the binary number to octal.
3.6
Decimal to Hexadecimal
Convert the number to binary then convert the binary number to hexadecimal.
3.7
Exercise
What are the smallest and largest positive whole numbers represented by:
4 bits? i.e. 0000 to 1111 ?
8 bits?
16 bits?
32 bits?
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4
Computer Number Systems
Representation of Negative numbers
So far when looking at binary numbers we have been looking at unsigned integers. We have to
be able to handle negative numbers. In decimal notation we use the minus symbol (-) to denote
a negative number. But a computer can only use binary symbols i.e. 0's and 1's. There is no
extra symbol to represent a minus sign. So we must use one of the bits to represent or indicate a
negative value.
There are four common ways of representing negative binary numbers.
 Signed magnitude
 One's complement
 Two's complement
 Biased value
4.1
Signed magnitude (or Sign/modulus).
The leftmost bit is the sign bit. It is 0 for positive, or 1 for negative.
+29 is represented by 000111012
sign
(+)0
64
0
32
0
16
1
8
1
4
1
2
0
1
1
(+) 0×64 + 0×32 + 1×16 +1×8+ 1×4 + 0×2 + 1×1
-29 is represented
sign
(-)1
64
0
32
0
16
1
8
1
4
1
2
0
1
1
by 100111012
(-) 0×64 + 0×32 + 1×16 +1×8+ 1×4 + 0×2 + 1×1
4.2
One's complement.
The leftmost bit represents a negative value. But instead of representing the minus symbol, in an
8-bit one’s complement binary number, a 1 in the leftmost place represents the value, -127.
-29 is represented by 111000102
-127
1
i.e.
64
1
32
1
16
0
8
0
4
0
2
1
1
0
1×-127 + 1×64 + 1×32 + 0×16 + 0×8+ 0×4 + 1×2 + 0×1
= -127 + 98 = -29
This system is popular because it is easy to change a binary number to a negative equivalent.
To negate a binary number in one’s complement,
- just work out the positive representation
- then flip the bits. i.e. change all 0's to 1's and all 1's to 0's.
6
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Computer Number Systems
A computer can do this very quickly.
Example: -84 in 1’s complement:
+84 is represented by = 010101002
-84 is represented by = 101010112
i.e. -84 = -127 + 43
Of course this will only work for numbers in the appropriate range you need to know how many
bits are being used.
4.3
Two's complement.
This is very similar to one's complement. The only difference is that the leftmost bit is -2n rather
than -2n +1, e.g. for an 8-bit number it is -128 rather than -127. This is a little more consistent
with the place value system, so that the leftmost place value represents a negated power of two.
e.g. -29 is represented by -127 + 98 = 111000102
= -128 + 99 = 111000112
in one's complement.
in two's complement.
Another example: +84 and -84
1. Using the place values method
In One’s complement (and in Two’s complement):
+84 is represented by 64+16+4 = 010101002
-127
0
64
1
32
0
16
1
8
0
4
1
2
0
1
0
.
In One’s complement:
-84 is represented by -127+43 = -127+32+8+2+1 = 101010112
-127
1
64
0
32
1
16
0
8
1
4
0
2
1
1
1
In Two’s complement:
-84 is represented by -128+44 = -128+32+8+4
-128
1
64
0
32
1
16
0
8
1
4
1
2
0
= 101011002
1
0
2. An easier way to get a negative value in two’s complement is to use the algorithm:
- create the positive binary number using one's complement
- convert it to negative by flipping the bits
- then add one to the result to get the two’s complement value
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Computer Number Systems
Find +84 in 1’s Comp
1’s Comp
-127
64
32
16
+84
0
1
0
1
Flip the bits to get –ve value in 1’s complement
1’s Comp
-127
64
32
16
-84
1
0
1
0
Add one to get –ve value in 2’s complement
2’s Comp
-128
64
32
16
-84
1
0
1
0
4.4
8
0
4
1
2
0
1
0
8
1
4
0
2
1
1
1
8
1
4
1
2
0
1
0
Biased value (Excess 127)
Excess 127 is usually used for 8 bit representations, other values can be used for different
numbers of bits. (e.g. Excess 1023 for 11 bits.)
In Excess 127, we represent every number by a value that is 127 larger. i.e. (the number + 127)
For example:
-127 is represented by -127 + 127 = 0
-50 is represented by -50 + 127 = 77,
+20 is represented by 20 + 127 = 147
This means that numbers from -127 upwards will be represented by a positive value.
So, in Excess 127 the binary representation is always 127 larger than it really is. We need to
remember this when converting back from binary to decimal !!
-29 is represented by -29 + 127 = 9810
-84 is represented by -84 + 127 = 4310
+29 is represented by +29 + 127 = 15610
4.5
= 011000102
= 001010112
= 100111002
Exercise
1. Give five different interpretations of the binary number 101010102 .





2. Complete the table below to show the smallest and largest values of 8-bit binary numbers
using each of the four representations stated.
8
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csci1802 Computer Systems
8 bits
Computer Number Systems
Smallest value
Binary
Decimal
1. Signed magnitude
2. One's complement
3. Two's complement
4. Excess 127
10000000
=
Largest value
Binary
Decimal
01111111 =
127
-127
3. Repeat this for 16-bit numbers.
[Hints: 215 = 32768 ; Would we use Excess 127 or something else?]
5
Addition and Subtraction in Binary
5.1
Simple addition of positive integers
In decimal addition we 'carry' when the result is greater than 9. 'Carrying' works because when
we get ten values in the 10n place then we have 10×10n which is the same as 1×10n+1 .
e.g. 10×102 =1×103 .
In binary addition we 'carry' when the result is greater than 1,
e.g.
1 +
1 +
1 +
0
1
1
1
1 0
1
1 1
Thus,
27
0
0
= 1
26
1
1
1
25
0
1
0
24
1
1
0
23
1
0
1
22
0
1
1
21
0
0
1
20
1 +
1
0
102 101 100
8
9 +
1
1
7
2
0
6
This works well for unsigned positive numbers.
5.2
Arithmetic using positive and negative integers
First, try the following simple decimal additions and subtractions.
Note we are using the convention here that a raised minus sign indicates a negative number
e.g. –3 means negative 3, whereas -3 means subtract 3
1+3=
3–1=
1–3=
–
1 – –3 =
3 + –1 =
1 + –3 =
–
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1+3=
1 – –3 =
9
Csci1802 Computer Systems
Computer Number Systems
1 + – (–3) =
Question: Is subtracting X the same as adding –X?
Yes. Extending this to binary arithmetic we can simplify subtraction of binary numbers by
adding a negative representation of the number (e.g. two's complement).
Using two’s complement representation we can add and subtract both positive and negative
numbers by always using the add operation.
REMEMBER: In 2’s complement:
 a positive value is unchanged
e.g. +40
= 32 + 8
= 00101000
 a negative value has the positive value changed to the negative value by the process we’ve
just learnt
e.g. -40
= - (32 + 8)
= - 00101000
= 11010111+1
= 11011000
5.3
Addition and subtraction examples
Example 1: adding 2 positive values
e.g. 5 + 12
decimal
2’s complement
+5
00000101
+12
00001100 +
+17
00010001
Check the answer is correct
Example 2: subtracting second number from first number
e.g. 5 – 12 = 5 + –12
decimal
2’s complement
+5
00000101
–
12
11110100 +
–
7
11111001
Check the answer is correct
Example 3:
Here is an example where there is a
‘carry’ on the left hand side of the result.
Because the two operands have different
signs, the carried 1 is simply ignored.
decimal
–
5
+12
+7
2’s complement
11111011
00001100 +
[1]00000111 sum = +7
Example 4:
However, in this example, where the
two operands both have the same sign,
the carried 1 on the left hand side of the
result and the fact that result is a different
decimal
–
80
–
64
–
144
2’s complement
10110000
11000000 +
[1]01110000 sum= +112
10
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csci1802 Computer Systems
Computer Number Systems
sign to both operands indicates that there
has been an overflow and the result is wrong.
5.4
Exercise
By considering the column values, (i.e. –128, 64, 32 etc) explain what has occurred in the last
two examples.
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Computer Number Systems
6 Some effects when using finite-precision numbers (i.e.
using computers)
6.1
Overflow, Underflow and Rounding
Say a computer uses 8 bits to represent integers then the range of values that can be stored
(assuming we use 2's complement) is -128 to +127.
Thus a computer using 8 bits to store integers cannot do every arithmetic calculation you might
want to give it.
Examples:
2+2
=
4
(correct)
70 + 70
=
140
(too big to be represented: integer overflow)
5×5
=
25
(correct)
5 × 50
=
250
(too big: integer overflow)
50 – 95
=
-
50 – 95
6.2
=
-
-
45
(correct)
145
(too big negative: negative overflow)
4/2
=
2
(correct)
5/2
=
2.5
(not an integer, so cannot be stored as an integer except by
rounding or truncating it)
Definitions
Integer overflow:
Negative overflow:
Result too big (+ve)
Result too big (–ve)
Rounding:
Result cannot be stored exactly, so nearest valid value is stored.
Rounding error: Difference between correct value and calculated value.
Relative error: Rounding error (i.e. Rounding error divided by Correct value)
Correct value
12
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csci1802 Computer Systems
Computer Number Systems
6.3 Effect of order of calculations when using finite-precision numbers (i.e.
using computers)
In these examples, assume 8-bit two's complement representation of integers. (Therefore, the
range of values is -128 to +127)
6.3.1 Example 1:
Calculate a + b – c
Method 1) calculate b – c then add a
Method 2) calculate a + b then subtract c
[ Associative law states: a + ( b – c ) = ( a + b ) – c]
Using a = 80, b = 50, c = 30
Method 1)
Method 2)
80 + (50 – 30)
= 80 + 20
= 100
Method 1 gives correct answer
(80 + 50) – 30
= (overflow) – 30
=?
Method 2 gives wrong answer!
6.3.2 Example 2:
Calculate a * ( b – c )
Method 1) calculate b – c then multiply by a
Method 2) calculate a * b then subtract a * c
[ Distributive law says: a * ( b – c ) = ( a * b ) – ( a * c ) ]
Using a = 8, b = 40, c = 30
Method 1)
Method 2)
a *( b – c )
= 8 * (40 – 30 )
= 8 * 10
= 80
(a * b )– (a * c)
= ( 8 * 40 ) – ( 8 * 30 )
= 320
–
240
(overflow) – (overflow)
= (garbage) ?
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Method 1 gives correct answer.
Method 2 gives wrong answer!
13
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6.4
Computer Number Systems
Arithmetic – computers vs humans
The conclusions from section 6 of this booklet are:
1. Arithmetic using computers is not the same as arithmetic done by people.
2. It is important to understand
 how computers work
 their limitations
14
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Computer Number Systems
7
Binary Representation of Characters
7.1
ASCII Character codes
Characters like 'a', 'b', 'c', '2', '!' are represented by number codes. ASCII (American Standard
Code for Information Interchange) is a code used to encode characters. Standard ASCII uses 7
bits and allows 128 characters to be represented. Extended ASCII uses 8 bits and allows 256
characters to be represented. An additional 8th or 9th bit might be used as a parity bit, which is
used to detect errors.
 odd parity - the parity bit is set to 0 or 1 to give an odd number of 1's in total.
 even parity – the parity bit is set to 0 or 1 to give an even number of 1's in total.
Standard characters that can be encoded are:
 Command characters
Non-printing transmission and printer control codes 00..1F Hex
 Alphanumeric characters {0 .. 9} codes 30.. 39 Hex
{A .. Z} codes 41..5A Hex
{a .. z} codes 61..6A Hex
 Symbols
punctuation and arithmetic characters
e.g. the space character is 20 in Hex
ASCII codes for printable characters
(Note that the codes are given here in Hexadecimal - why is that?)
Hex
2
3
4
5
6
7
0
0
@
P
`
p
1
!
1
A
Q
a
q
2
"
2
B
R
b
r
3
#
3
C
S
c
s
4
$
4
D
T
d
t
5
%
5
E
U
e
u
6
&
6
F
V
f
v
7
'
7
G
W
g
w
8
(
8
H
X
h
x
9
)
9
I
Y
i
y
A
*
:
J
Z
j
z
B
+
;
K
[
k
{
C
,
<
L
\
l
|
D
=
M
]
m
}
E
.
>
N
^
n
~
F
/
?
O
_
o
•
Example: To get the ASCII binary code for the letter Q
1. Get the hex code for Q
 use the value down the left = 5
 use the value along the top = 1
 this gives the value 51 in hex
2. Convert the hex value 51 into binary
 5 is 0101
 1 is 0001
 put them together to get 01010001
3. Hence the ASCII code for the letter Q is 01010001 in binary.
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7.2
Computer Number Systems
Example
Assuming 8-bit extended ASCII encoding, give the binary pattern that represents the sequence of
characters "Hello Joe".
Answer:
Char
Hex
Binary
H
48
e
65
l
6C
l
6C
o
6F
01001000
01100101
01101100
01101100
01101111
20
J
4A
o
6F
e
65
00100000
01001010
01101111
01100101
Doing the same example using 'odd parity' we would get:
Char
Hex
Binary
Parity bit
7.3
H
48
e
65
l
6C
l
6C
o
6F
20
J
4A
o
6F
e
65
01001000
1
01100101
1
01101100
1
01101100
1
01101111
1
00100000
0
01001010
0
01101111
1
01100101
1
o
6F
e
65
01101111
01100101
Exercise
Fill in the gaps to do the same example using 'even parity'
Char
Hex
Binary
Parity bit
7.4
H
48
e
65
l
6C
l
6C
o
6F
20
J
4A
01001000
01100101
01101100
01101100
01101111
00100000
01001010
Discussion question
How do you know whether a number in a computer's memory cell represents a number or a
character code?
16
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7.5
Computer Number Systems
Unicode
Here is a brief timeline showing character representation developments
ASCII
 dominant since 1970’s
 7 bits -> 128 characters
ISO 646 (in 1972)
 national variants e.g $, £, é
 but £ in UK .. # in US
8-bit ASCII
 use all 8 bits for character representation
o data transmission more reliable
o better ways of transmission error checks
o so don’t need parity bit
 but manufacturers developed their own extensions to the standard
 standard ASCII characters
+ extra characters – non-compatible!
Multi-part ISO 8859 (1980’s)
 ISO 8859-1: (ISO Latin-1): most W Europe
 ISO 8859-2: E Europe ) Czech, Slovak
 ISO 8859-7: Greek, etc
ISO 10646
 ISO standard
 32 bit character set => 232 characters
o hypercube
o groups, planes, rows, columns
o character -> (g,p,r,c)
Unicode
 industry consortium
 16 bit character set => 65,536 characters
 CJK Consolidation (Chinese, Japanese, Korean)
 contemporary major languages
 symbols, maths, dingbats …
 mechanism to build composite characters
 reserved codes for expansion & private symbols
In 1991:
 Unicode & ISO 10646 made compatible
 ISO agreed BMP (Basic Multilingual Plane) (0,0,*,*) identical to Unicode
Current:
 Unicode 5.1 provided codes for 100,713 characters.
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
Computer Number Systems
Now matches ISO 10646 standard 32 bit representation
The 8 bit character representation on most computers is now Latin-1. However this misses out
large chunks of the world. For this reason other coding formats based on the Unicode/ISO 10646
standard are used. In order to save storage space a 32 bit coding is not in standard use. Instead
people tend to use multi-byte encoding schemes such as UTF-8. In some areas of computing
(HTML, XML, Java … ) direct reference to Unicode is made via expressions like &#xABCD,
&#140 or \uXXXX.
7.6 UTF-8
Rules for UTF-8 are designed to reduce the size of files in order to save both storage and
transmission time over networks. The way it does this is to store the ASCII range of characters
as themselves and store other characters as 2, 3 or 4 bytes as needed. The formats look like this:
0XXXXXXX
This represents the ascii char 0XXXXXXX.
110XXXXX10YYYYYY represents the Unicode character 00000XXXXXYYYYYY
1110XXXX10YYYYYY10ZZZZZZ represents the Unicode character
XXXXYYYYYYZZZZZZ.
An important property of Unicode encodings like UTF-8 is that there are no overlaps – when
searching for a character if you encounter a matching byte-sequence you know that it is the
character you are searching for. In some coding schemes you might have found part of another
character.
Here is how the coding is described in the Unicode UTF-8 specification.
Notice how the first byte gives the information on how many bytes are used to code the
character. The saving in file size is obviously dependant on the language used (and for some
languages it will have a negative effect). Other encodings are also available (UTF-16 and UTF32).
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8
Storage of data in computer memory
8.1
Bits, Bytes and Words
The examples of rounding errors and overflow described in Section 6 arise because of the way
that data is stored in computer memory.
We are already familiar with binary digits (called bits). One bit is one binary digit which can
take a value of 0 or 1. Another unit of data is the byte which is a sequence of 8 bits. Thus one
byte can produce 28 (= 256) different values (0 to 255). These values can be used as codes.
Different representations are used to code different kinds of data.
A word is the smallest addressable unit of computer memory, and depending on the machine
and software might be 16 or 32 or 48 or 60 or 64 bits.
1 Kilobyte (1Kb) = 1024 bytes = 210 bytes
1 Megabyte = 1,048,576 bytes = a kilobyte of kilobytes = 1024 x 1024 bytes = 220 bytes
1 Kilobyte is approximately 1000 bytes,
1 Megabyte is approximately 1 000 000 bytes
We now frequently use Gigabytes and Terabytes as measures of size. How big are they?
8.2
Representation of primitive C types
The C programming language has several primitive data types. The way they are represented is
machine- and implementation-dependent. However, typical descriptions (ref: Kernighan &
Ritchie: The C Programming Language) are given below.
Datatype
char
int
Description
Character, not necessarily
printable
Integer value
short
long
A short int but …
A long int
float
Real (or floating point) value
double
Double precision floating
point value
long double
Extended precision floating
point
© De Montfort University, Dowson , Zarzycki, Birkenhead
Description/Typical values
1 byte ASCII code
-32767 to +32767
i.e. -215-1..215-1
often the same as int
-2147483647 to +2147483647
i.e. -231-1..231-1
6 decimal digits of precision,
min value 1E-37, max value 1E+37,
smallest value x such that 1+x  1 is 1E-5
10 decimal digits of precision,
min value 1E-37, max value 1E+37,
smallest value x such that 1+x  1 is 1E-9
machine/implementation dependent
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8.3
Computer Number Systems
Exercise
How many bytes are required for each of these data types? Use the final column to help you fill
in the blanks.
Datatype
No of bytes required
Description/Typical values
char
1 byte ASCII code
int
-32767 to +32767
i.e. -215...215-1
short
often the same as int
long
float
to be discussed in the next
section
double
to be discussed in the next
section
long double
machine/implementation
dependent
8.4
-2147483647 to +2147483647
i.e. -231...231-1
6 decimal digits of precision,
min value 1E-37, max value 1E+37,
smallest value x such that 1+x  1 is 1E-5
10 decimal digits of precision,
min value 1E-37, max value 1E+37,
smallest value x such that 1+x  1 is 1E-9
machine/implementation dependent
Further reading
Blundell, B: pp 67-82
Ceri, Mandroli, Sbattella: The Art & Craft of Computing, Sections 13.1-13.4.1, pp 312-320
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9
Computer Number Systems
Representation of Fractional Numbers
Fractional numbers are numbers between 0 and 1. Their representation in decimal or binary
simply uses negative powers of the base. Recall that a-n = 1/an
For example:
10-1 = 1/101
10-2 = 1/102
10-3 = 1/103
10-4 = 1/104
10-5 = 1/105
=
=
=
=
=
1/10
1/100
1/1000
1/10000
1/100000
=
=
=
=
=
0.1
0.01
0.001
0.0001
0.00001
2-1
2-2
2-3
2-4
2-5
=
=
=
=
=
1/21
1/22
1/23
1/24
1/25
=
=
=
=
=
1/2 =
1/4 =
1/8 =
1/16 =
1/32 =
0.5
0.25
0.125
0.0625
0.03125
etc.
The number 0.42610 is 4×10-1 + 2×10-2 + 6×10-3 + 1×10-4
The number 0.011012 is 0×2-1 + 1×2-2 + 1×2-3 + 0×2-4 + 1×2-5
9.1
To translate fractional binary to fractional decimal
Simply work out the sum using the decimal or fractional values of the bits.
e.g.
0.011012
= 0×2-1 + 1×2-2 + 1×2-3 + 0×2-4 + 1×2-5
= 0
+ 0.25 + 0.125 +
0 + 0.03125
= 0.40625
(or
0.011012
= 1/4 + 1/8 + 1/32 = 13/32 = 0.40625)
A table can be used, similar to the one used to work out integer values.
-3
-4
-5
-6
-7
-8
. 2-1 2-2
Powers of 2
2
2
2
2
2
2
Fractional value
1/2 1/4 1/8 1/16
1/32
1/64
1/128
1/256
of each bit
Decimal value of
0.5 0.25 0.125 0.0625 0.03125 0.015625 0.0078125 0.00390625
each bit
Example binary
.
0
1
1
0
1
0
0
0
number
0.25 0.125
0.03125
Value of each bit
0
or
or
0
or
0
0
0
in the example
1/4
1/8
1/32
Sum of the
0.25 + 0.125 + 0.03125 = 0.40625 or 1/4 + 1/8 + 1/32 = 13/32 ( = 0.40625 )
values
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9.2
Computer Number Systems
To translate fractional decimal to fractional binary
9.2.1 Either...
Successively subtract the highest possible value of negative power of 2
0.42510 - 0.25
= 0.175
(2-2 = 0.012)
0.175 - 0.125
= 0.05
(2-3 = 0.0012)
0.05
- 0.03125 = 0.01875
(2-5 = 0.000012)
etc
which gives 0.01101 to an accuracy of 5 binary digits after the point.
This can be done using a table as before:
e.g.
To convert
0.425 to binary
Powers of 2 . 2-1
2
2
Decimal value
of each bit
0.25
0.125
0.425 - 0.25
= 0.175
Using 0.25
0.175 - 0.125
= 0.05
Using 0.125
0.05 -0.03125
= 0.00875
Using 0.03125
0.00875 - ?
0.5
-2
-3
-4
2
0.0625
-5
2
0.03125
-6
2
0.015625
-7
2
0.0078125
1
1
1
etc
Put 0 in gaps
.
0
1
1
0
1
9.2.2 ...Or
Successively multiply the fractional part of the decimal value by 2 and move the integer part of
the result onto the end of the binary number (from left to right after the point)
e.g.
0.425
0.85
0.7
0.4
0.8
etc
×
×
×
×
×
2
2
2
2
2
=
=
=
=
=
0.85
1.7
1.4
0.8
1.6
gives
gives
gives
gives
gives
0.0
0.01
0.011
0.0110
0.01101
Note: If you use this method, it is a good idea to always check your answer by converting your
result back to decimal.
(Exercise: explain why this method works!).
9.2.3 Rounding Error
This example also demonstrates that it is not always possible to translate a decimal value exactly
into a specific number of binary digits. (Continue with the conversion until you convince
yourself that this is the case.) The result of the translation here, if left with 5 binary digits, will
have a rounding error of 0.0187510.
 We tried to convert 0.42510 and the result was 0.011012 (to 5 binary digits)
 We can calculate that 0.011012 = 0.4062510
 Therefore, the rounding error is: 0.425 - 0.40625 = 0.0187510
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10 Representation of Real Numbers
Real numbers (or rational numbers) contain both an integer part and a fractional part. For
example, 23.714 has an integer part of 23 and a fractional part of 0.714 .
A binary fractional number may look like 01101.10012. How can this 'real' binary number be
stored in a computer which does not have a decimal point symbol?
10.1 Fixed point representation
One solution is to assume that the decimal point is in a fixed position. This is called a fixed
point representation with a fixed number of bits each for the integer and fractional part.
For example, we could assume that in an eight bit number the decimal point is always such that
there are 5 bits for the integer part and the remaining 3 bits for the fractional part. So if we put
the point in, then 10011101 represents 10011.101,
Integer part 10011 gives 19,
Fractional part 0.101 gives 0.625
Thus 10011101 would represent the real number 19.625
10.2 Exercise
1. What number would 10011101 represent if the (integer, fractional) split is
(6, 2) ?
(4, 4) ?
2. What are the smallest and largest values that can be represented with each possible split:
(0,8), (1,7), (2,6), etc.
10.3 Normalised floating point representation
The more usual encoding of real numbers in binary is using a normalised floating point
representation. We will explain the term ‘floating point’ using both decimal and binary notation
and then go on to explain ‘normalized floating point representation’ of binary real numbers.
10.3.1 Floating point representation
In decimal we can write values as a number multiplied by 10 to some power.
e.g. –241.65 =
=
=
=
=
=
=
=
–0.0024165 × 105
–0.024165 × 104
–0.24165 × 103
–2.4165 × 102
–24.165 × 101
–241.65 × 100
–2416.5 × 10-1
–24165.0 × 10-2
etc
Thus the position of the decimal point can ‘float around’, giving the term ‘floating point’.
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In binary, we are now working on powers of 2 instead of powers of 10. Thus,
110.11
=
=
=
=
=
=
1.1011 × 22
0.11011 × 23
0.011011 × 24
0.0011011 × 25
11011.0 × 2-2
1101100 × 2-3
etc
We are interested in the value that starts with a single 1 followed by the decimal point (or more
correctly the binary point). This is what we call the normalized binary value.
10.3.2 Mantissa and Exponent
Floating point representation uses 2 values to represent a number and these are given the names:
the mantissa and the exponent.
The mantissa (sometimes called the fraction) is the fractional part of the number.
The exponent (sometimes called the characteristic) is the power of 2 the mantissa must be
multiplied by to get the original value.
10.3.3 Binary Normalised Floating Point
First, we normalize our binary numbers so that they always start with 1.
Because they always start with 1. , for efficiency of storage, the 1 and the point don’t have to be
stored (but the system has to know that they have to be replaced when extracted from storage).
Binary value Normalised value Mantissa
11.010
1.1010 × 2+1
1010
0.00011
1.1
× 2-4
1000
-2
0.010101
1.0101 × 2
0101
1.101
1.101 × 20
1010
Exponent
+1
-4
-2
0
Thus, each of the values in the table above can be stored using the values of the mantissa and the
exponent.
This is the basis of how real numbers are stored in computers using the IEEE 754 standard.
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10.4 IEEE 754 Standard Representation of Real Numbers
This is now the standard way of representing real (binary) numbers in computers. For single
precision, there are a total of 32 bits used as shown here.
No of bits
1
8
23
Component
S(sign) E (exponent)
F (fraction or mantissa)
Bit Number(s)
31
30 -------- 23 22 -------------------------- 0
S - the sign of the number, is 0 for positive or 1 for negative
E - the exponent, uses Excess 127 representation (i.e. Excess 27-1) and can take values between
1 and 254, equivalent to a range of -126 to 127. (The values 0 and 255 are reserved for special
cases.)
F - the mantissa, uses a normalised representation as described above. The leading one and the
decimal point are omitted (because they can be assumed to be present). F should more correctly
be called the significand.
Zero is a special number and is represented by values of E=0 and F=0 (with S taking the value of
either 0 or 1).
For more detailed understanding of the IEEE 754 Standard Representation of Real
Numbers you are recommended to read the reprint of a web article which is given after the
following exercises.
This article will also help you with the tutorial exercises.
10.5 Exercises:
A reprint of a web article is included next.
a) Read this to find out what is meant by NaN and how NaN is represented in the IEEE
standard.
b) Use the example given to help you to represent the decimal value 132.2 as a single
precision IEEE 754 number.
c) Find out if there is a C library that allows you to use infinity and NaN.
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Representing Real Numbers
1. Background
Real number representation in a fixed-size word represents a challenge because they can have a varying
number of decimals and an enormous range. As early as 1945, John von Neumann realised that,
because of round-off errors, it would be impossible to represent real numbers exactly, as was done for
integers, and he even suggested that computers should not deal with such numbers! On the hardware
side, designers were reluctant to incorporate real number support in their processors because that would
have a severe impact on performance and occupy valuable real estate on the processor chip. The
evolution of real number representation went through a number of controversies and backward steps
before a standard (IEEE 754 / 854) finally emerged in 1985. And while so many people contributed to this
development over the years, it is widely believed that the work of William Kahan, which started at the
University of Toronto in 1953, was instrumental, and for that, he was awarded the Turing Award in 1989.
2. Objectives
The goals of the IEEE standard are twofold:
1. Strike a compromise between range and accuracy
2. Minimize floating-point hardware by exploiting the integer unit h/w.
3. Single-Precision IEEE 754
The number is represented in 32 bits (4 bytes) using the following format:

S is the sign of the number (1=minus, 0=plus), contributing (-1)S.

E is the biased exponent, contributing 2(E-127). It is represented as an 8-bit unsigned integer, and
thus, have the range 0 through 255.

F is the binary fraction (mantissa or significand), contributing 1.F; i.e. an implicit 1 and a binary
point are not represented but assumed present. Together with the implicit 1, 24 bits are thus
significant. Since a decimal digit can be represented in between 3 and 4 bits, we can see that this
single-precision representation produces about 7 significant digits.
This standard achieves the 1st goal by using powers of 2 instead of 10. It also realises the 2 nd goal
because of the way it orders the field and represents the exponent: Since S is in the most significant bit, it
is easy to perform a test of less than, greater than, or equal to 0 using integer unit hardware. Placing the
exponent E before the mantissa also allows us to sort floating-point numbers using integer comparison
instructions, as long as all exponents are non-negative. In fact, this is why the standard uses biased (or
excess 127) notation instead of 2's complement.
The exponent E ranges between 0 and 255 but the end points of this interval are reserved for special
values (see below). Hence, the allowed range of E is 1 though 254 which gives a magnitude range (in
absolute value) of about:
2(1-127) = 2-126 = 10-38 to 2(254-127) = 2(127) = 10+38
Special Values:
26

The fact that 1. is implicit means that the number zero cannot be represented and, hence, a
special value must be set aside for zero.
This (zero) value is: E = F = 0 (regardless of S).

Furthermore, instead of issuing an exception when a divide by 0 is attempted, the standard
allows software to set special values to indicate plus or minus infinity, thus providing for
topological closure in real arithmetic.
These (infinity) values are: E = 255, F = 0, S = 0 / 1.
Reprinted from: http://www.cs.yorku.ca/~roumani/g98/ieee.htm
csci1802 Computer Systems
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
The results of invalid operations (such as 0/0 or infinity minus infinity) are indicated by a special
value in the standard known as NaN (Not a Number), thus allowing program to proceed with a
number of operations and postpone some tests and decisions to a later time in the program.
This (NaN) value is: E = 255, F not = 0 (regardless of S).

The standard realises that there is a gap between 0 and the smallest positive (and largest
negative) number that can be represented, and allows some numbers in this gap to be
represented. Supporting such a gradual underflow means that we abandon our implicit 1.
premise and replace it by 0. Such numbers are denormalized and can be as small (in absolute
value) as about 2-23 x 2-126 = 2-149 = 10-45. Due to their departure from the premise, most
processor makers do not support this optional feature of the standard.
These (denormalized) numbers have: E = 0, F not = 0, S = 0 / 1.
4. Double-Precision IEEE 754
This 64-bit (8 bytes) representation provides added accuracy and increased range but uses the same
idea as single-precision. Here are the differences:

11 bits for E (with a bias of 1023) and 52 bits for F.

Aside from special values, E ranges between 1 and 2046. The range is thus: 2-1022 (about 10-308)
to 2-1023 (about 10+308).

Based on 52+1 significant bits, the accuracy is about 15 (decimal) digits.

Special Values:
E = 0, F = 0 (zero)
E = 0, F not 0 (denormalized)
E = 2047, F = 0 (infinity)
E = 2047, F not 0 (NaN)
5. Example
Represent 19.2 as a single-precision IEEE 754 number:
19 / 2 = 9 R 1
9 / 2 = 4 R 1
4 / 2 = 2 R 0
2 / 2 = 1 R 0
1 / 2 = 0 R 1
Hence 19 = 10011
0.2 * 2 = 0.4
0.4 * 2 = 0.8
0.8 * 2 = 1.6
0.6 * 2 = 1.2
0.2 * 2 = 0.4
0.4 * 2 = 0.8
Hence 0.2 = 001100110011..
19.2 = + 10011 . 001100110011...
= + 1. 0011 0011 0011 0011 0011 001 X 24
S = 0, E = 4+127 = 131 = 10000011 and F = 001100110011001100110
Reprinted from: http://www.cs.yorku.ca/~roumani/g98/ieee.htm
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11 Discreteness of real number representation
11.1 Range and accuracy
Not all possible values can be represented by floating point representation of numbers. Only a
discrete set of values can be represented. This has implications on the accuracy of computer
arithmetic. e.g. try adding together a large number and a very small number.
However, the big advantage of using floating point representation is that a vast range of numbers
can be represented with both very small and very large values being possible. This is because
the exponent can take both positive values (to represent large values) and negative values (to
represent small values).
A sign bit is used to represent the sign of the number, so both positive and negative values are
possible.
The accuracy of the number represented depends on the number of bits used to represent the
mantissa and real numbers will usually contain some rounding error (the difference between
the correct value and its nearest expressible value).
11.2 Errors
Overflow will occur when the number is too big to be represented. This is when the exponent
does not contain enough bits to hold the power of 2 that is required.
e.g, in decimal if the exponent could hold 2 digits, then there would be overflow for any value
greater than 1099.
Underflow will occur when the number is too small to be represented. This is when the
exponent does not contain enough bits to hold the negative power of 2 that is required.
e.g, in decimal if the exponent could hold 2 digits, then there would be underflow for any value
smaller than 10-99.
The spacing between adjacent expressible numbers is not constant
e.g (using decimal notation) 0.999 × 10+99 –0.998 × 10+99 ( = 0.001 × 10+99)
is much greater than
0.999 × 10+01 – 0.998 × 10+01 (= 0.001 × 10+01)
but the relative error (caused by rounding) is approximately constant throughout the range of
expressible numbers.
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11.3 Further Reading
Floating-Point Fallacies by Dan Zuras: http://www.zuras.net/FPFallacies.html
This article starts with the following:
The puzzling behavior of some programs that use floating-point arithmetic to solve problems can
often elicit some interesting questions. "Why is it that 0.2 x 5.0 results in 0.9999999 not 1.0?"
"Why is sin(3.14159265) equal to -8.742276e-8 not 0.0 ?" "How can the total system usage be
100.1% ?"
In the next paragraph it continues with:
Those who use and write programs that use floating-point to solve problems have a common
problem. Floating point numbers and operations are expected to behave like real numbers and
operations. This quite reasonable expectation is, unfortunately, never true.
Download and read the whole article!
© De Montfort University, Dowson & Zarzycki
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Tutorial Work Sheet 1
Binary, Octal, Hex
These exercises should be attempted during the staffed tutorial session. You can ask for help at
any time during the tutorial.
If you do not finish the exercises during the tutorial, you should complete them in your own time
before the next tutorial. If you have any questions, ask at the start of the next tutorial.
Tutorial Exercises
1. Convert the binary numbers below to decimal.
(a) 1010
(b) 1111
(c) 00010000
(d) 00010101
(e) 00000101
(f) 11000110
2. Convert each decimal number below to binary.
(a) 64
(b) 15
(c) 234
(d) 1
(e) 100
(f) 456
3. Perform the following binary additions. (Hint: write each pair of numbers one below the
other and, if necessary, extend both to 8 bits by adding extra 0’s at the left hand side.)
(a) 1+1
(b) 00110100 + 00001011
(c) 001010 + 001111
(d) 00110111 + 11000
4. Perform the following binary subtractions.
Difficult isn’t it! Don’t worry if you can’t do this question, you’ll learn a better way of doing
this soon.
(a) 0111 - 0101 (b) 1101 - 1010
(c) 1111 - 0110
(d) 01110110 - 01001101
5. What are the 16 digits of the hexadecimal system? Now convert the hex numbers below to
binary.
(a) FFFF
(b) AF12
(c) 3457
6. Convert the following decimal numbers to binary and then to hexadecimal.
(a) 59
(b) 29
(c) 842
(d) 3595
(e) 32
/ continues on next page
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7. Convert the binary numbers below to their octal and hex equivalents. (Hint: you may need to
put extra 0’s on the left hand side to get the right number of bits.)
(a) 1110
(b) 11011
(c) 110110101
(d) 1010111101110010
8. Some computers represent numeric data using Binary Coded Decimal (BCD) where 4 bits are
used to represent each decimal digit. For example, 0010 1001 is the BCD form of the decimal
number 29. What decimal numbers do each of the following BCDs represent?
(a) 0101 0001
(b) 1000 1000 1000
(c) 0110 0011
(d) 0111 0001 0010
Questions from Blundell:
Activities 1.4 to 1.8
Discussion Questions
1. Given a binary number, is it true that the 1's complement of its 1's complement is the original
number? Can you give a reason? Is the same true for 2's complement?
2. Assume that X, Y and Z are digits. Discuss whether it is always true that XYZ8 <= XYZ10.
Explain your reasoning.
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Tutorial Work Sheet 2
Negative Numbers and Binary Arithmetic
These exercises should be attempted during the tutorial. You can ask for help at any time during
the tutorial. If you do not finish the exercises during the tutorial, you should complete them in
your own time before the next tutorial. If you have any questions, ask at the start of the next
tutorial.
Exercises
1. Represent -105 (negative 105) as an 8-bit binary number using:
(a) Signed magnitude (b) One's complement (c) Two's complement (d) Excess 127
2. Perform the following subtractions using 8-bit two's complement binary arithmetic. Show
you have checked your answers.
(a) 75 - 30
(b) 22 - 54
(c) -85 - 10
(d) -100 -50
3. Assume that a machine uses 4 digits to store integers in binary form. Complete the following
table.
Decimal value Signed Magnitude Two's Complement
-8
-7
-6
-5
-4
-3
-2
1001
-1
0
0000
0000
+1
+2
+3
+4
+5
+6
+7
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Questions from Blundell
Activity 4.2
Discussion Questions
1. Given a binary number, is it true that the 1's complement of its 1's complement is the
original number? Can you give a reason? Is the same true for 2's complement?
2.
Discuss the meaning of the following terms:
(a) Data
(b) Information
(c) Interpretation
(d) Meaning
(e) Representation
3. What sort of items would you expect to find in the memory of a computer?
Are all of them data?
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Tutorial Work Sheet 3
ASCII, Parity, Real Numbers, Floating Point
Exercises
ASCII & Parity
1. What is the internal bit pattern representation (i.e. the binary representation) of
(a) the number 13
(b) the characters 13
2. Produce the binary representation of the message:
My name’s Ada!
3. For the binary representation of the message in Q2, show what the parity bit for each
character would be if using (a) odd parity and (b) even parity.
Unicode
Code the following short strings using UTF-8 and indicate the number of bytes used to encode
each string (you will need to look some symbols up before the tutorial).
1. fred
2. ¼+½
3. ૧૨૩૪ (These symbols appear in the range 0A80 to 0AFF of the Unicode code-charts)
Real Numbers and Floating Point Representation
1. Convert these fractional binary numbers to decimal.
(a) 0.101
(b) 0.1111
(c) 101.11 (d) 110.1011
2. Convert each decimal number below to binary.
(a) 0.5
(b) 0.875
(c) 0.2
(d) 2.25
(e) 100.01
3. Calculate the relative error between 0.310 and 0.010012
4. What is the decimal value of binary 0.101000 x 24 ?
5. Use the binary value 0.101000 x 24 to explain what are the mantissa and exponent of a
floating point number.
6. Represent the following values as single precision IEEE 754 numbers.
(a) decimal value 132.2
(b) binary value 0111010101.10011
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(c) decimal value -16.25
Questions from Blundell
Activities 4.3 to 4.6
Discussion Questions and Further Reading
1. Explain underflow and overflow in floating point numbers.
2. Read Section 11 in this booklet on ‘Discreteness of real number representation’ and then
discuss the concept of relative errors in the representation of floating point numbers.
3. Earlier in these notes there is a reprint of a very interesting article written by Dan Zuras (who
was a researcher with Hewlett Packard and who did a lot of excellent work on computation
and the representation of numbers). Use the article to help you explain what is meant by
NaN and how NaN is represented in the IEEE standard.
4. Using decimal notation, and assuming we can store




the sign of the mantissa,
3 digits for the mantissa,
the sign of the exponent,
2 digits for the exponent
(a) What is the range of numbers we can represent?
(b) Mark on the diagram below the ranges of (for numbers represented as described at the
start of question 4) :
 expressible positive numbers
 expressible negative numbers
 positive overflow
 negative overflow
 +ve underflow
 -ve underflow.
-
-10100
-10-100
0
10-100
10100
+
Computer Exercises
1. Using a spreadsheet, experiment to find out the maximum and minimum values of integer
and real numbers that can be displayed.
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2. Using your spreadsheet, find the smallest value x such that (1.0+x) is not equal to 1.0.
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Computer Number Systems – Sample Test
Note: This is an example of the format and types of questions that might be included in a test on
Computer Number Systems. In an actual test or examination there could be questions on any
part of the group of topics covered in that section of the syllabus.
No calculators are allowed.
1.
Number conversions.
(a) Convert binary 00011011 to decimal.
(b) Convert decimal 278 to binary.
(c) Convert decimal 0.625 to binary.
(d) Convert 19B116 to binary.
(e) Convert 00110111110111112 to hexadecimal.
(f) Convert decimal 8.58 to binary
(g) Convert binary 0.00110011 to decimal
2.
Negative numbers. Express the negative number: –19, in each of the following binary
representations of negative numbers (using 8 bits). Explain how you have done the
conversions.
(a) Signed magnitude (sometimes called ‘sign and modulus’).
(b) Two’s complement.
3.
Binary arithmetic using 2’s complement. Do the following sums, showing your working.
Also show you have checked your answers.
(a) 01001010 + 00001111
(b) 00010001 + 10001010
(c) 00001010 - 00110011
4.
Character codes. The ASCII code for the letter ‘A’ is 4116, the letter ‘a’ is 6116, the space
character is 2016. Code the binary representation of the message ‘Hi Ada’ using 9 bits for
each character including an odd parity bit.
5.
Encode Ainto Utf-8 given that  has code point 00DB.
6.
Floating point numbers. Select the correct answers:
(a) Floating Point Overflow occurs when
The value in the mantissa is too big and positive
The value in the exponent is too big and positive
(b) Floating Point Underflow occurs when
The value in the mantissa is too small and positive
The value in the mantissa is too large and negative
The value in the exponent is too small and positive
The value in the exponent is too large and negative
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





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7.
Errors.
(a) Convert the decimal number 0.56 to binary using 4 binary digits after the point.
(b) Using your answer to part (a) as an example, explain the concept of rounding errors.
8.
IEEE Standard Representation of Real Numbers
(a) Convert 12.5 to a standard binary fractional number
(b) Normalise the result (so that it starts with 1. and has a multiplier of 2 to some power)
(c) Assuming the IEEE standard uses
- excess 127 representation of the exponent,
- the fraction without the leading 1 and binary point
give the values of S (the sign bit), E (the exponent) and F (the fraction or mantissa)
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CSCI1802 - Lab Worksheet 1
If you need help with any of these exercises, ask your lab tutor or the person sitting next
to you.
1. Log on and change password
Log on to a computer in your laboratory. If you haven’t already done so, change your password to
something you can remember, but which is not obvious to anyone else. (Use cntl/alt/del to get to change
password.)
2. myDMU – Managed Learning Environment
Knowing about the MLE is useful because this is where you will get your end of year results from ….. and
many other things too.
(a) Go to the MLE (Managed Learning Environment) at:
http://my.dmu.ac.uk
(b) Save the site as one of your ‘favourites’ or ‘bookmarks’.
(c) Log in using your ‘P number’ as your ID and your date of birth (dd/mm/yy) as your password.
(d) Change your myDMU password (to something you can remember!).
(e) Log out of myDMU and then log in again using your new password.
(f) Explore the site and see what it offers.
3. Email
It is vital that you can use your DMU email and that you check it regularly. We use email as a means of
communicating with you.
Your university email is available via the web from DMU or at home.
Check that you can send and read emails by sending an email to the person sitting next to you, asking
them to reply to you. Use your university email account not hotmail or any other account.
4. Blackboard
Knowing how to use Blackboard is very useful because this is the main source of information about this
module. It will be used as the only ‘Notice Board’ for the module. Other notes and handouts will also be
published here. You should check this site regularly for announcements and other information about the
modules you are studying.
(a) Go to the Blackboard site via the link from myDMU.
(b) Save the site as one of your ‘favourites’ or ‘bookmarks’.
(c) Log in the same way as for the MLE.
(d) Find this laboratory worksheet on the site.
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(e) You will be required to take tests using Blackboard during this module. So, to give you some
experience of the way tests are done in this system, there is a short ‘diagnostic’ test using the different
types of questions you may encounter during the course.
 Find and take the test. Check your result and the feedback.
(f) Explore the Blackboard site and see what else it offers.
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CSCI1802 - Lab Worksheet 2
If you haven’t use Microsoft Excel before, you can get a tutorial booklet called 'Excel 2007 Quick Guide'
from the CSE Student Advice Centre on the ground floor of Gateway House. Or your tutor may have a
copy for you. The document is also available on the web, from inside DMU only, at www.training.dmu.ac.uk
then follow the link to office 2007documents.
If you are new to Excel, It’s a good idea to work through this booklet before trying the exercises below.
1.
(a) Using a spreadsheet program (such as Microsoft Excel) enter the following headings and the data in
the left hand column.
x
0
1
2
2x
10x
Note: 2x is 2 to the power of x and 10x is 10 to the power of x.
The x's are raised up using the superscript option:
Format/Font/superscript
(b) Now insert formulae to calculate 2x and 10x in the first row. (You can either use the symbol ^ for
raising to a power, or the function POWER). To find out more use the ‘help’ function. (And then ask your
tutor if you need more help.) Check the results you get in your spreadsheet are correct.
(c) Copy the formulae down into the rows below. Again check the results carefully.
(d) Extend the table to include the values of x up to 10. Check it !
(e) Extend the table to include negative values of x. Check it !!
2. Use the Excel Chart Wizard (following the instructions below) to produce a graph of the function y = 2 x
from your table of values. The finished graph should look like this:
a) Select all the data, including the headings, in the
x
y=2
first 2 columns.
1200
b) Click on the Chart Wizard
c) Select XY (Scatter) and the middle left-hand chart
1000
sub-type, which is 'scatter with data points connected
by smoothed lines'.
800
d) Click on the Next button until you get to Step 3 of 4
y 600
– Chart Options
e) To make the layout look like the example here, you
400
will need to use the tabs:
Titles: to enter the chart title and x and y axis labels
200
Gridlines: to draw the horizontal and vertical gridlines
0
Legend: to remove the legend which we don't want,
0
2
4
6
8
10
since we've given the chart a title
x
f) Click on Finish and the graph will be drawn in your
spreadsheet.
g) You'll notice that the y axis label is not as in the illustration. To rotate the label, right click on it, then
select 'format axis title', click on alignment and then adjust as required. You can also change the font and
size of the title in a similar way.
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CSCI1802 - Lab Worksheet 3
1. DEC2BIN is an Excel function that converts values from decimal to binary. Use the help button to see
exactly how it works.
2. Using a similar step by step development technique to that used in Lab Worksheet 2, produce a table to
convert between the different bases. Start by entering the following headings, values and the function
DEC2BIN in your spreadsheet:
1
2
3
4
5
A
Decimal value
0
1
2
B
Binary value
= DEC2BIN(A2,8)
C
Cell B2 should then display the value 00000000
3. Extend the decimal value column to value 32 and copy the formula from B2 all the way down. You will
now have a lookup table for converting between decimal and binary for decimal values between 0 and 32.
You can extend this still further.
4. Now work out how to convert decimal or binary to hexadecimal (you can use help if necessary) and
extend your spreadsheet to calculate the hexadecimal values. Your spreadsheet might now look like:
1
2
3
4
5
A
Decimal value
0
1
2
3
B
Binary value
00000000
00000001
00000010
C
Hexadecimal value
00
01
02
You can now use your spreadsheet to check your answers to most of the questions in the first tutorial.
5. Experiment to find out what happens if you enter negative decimal values. How do the answers differ
from your tutorial answers? (Hint: count the number of bits)
6. Look up the functions CHAR and CODE in 'help'. Might one of these functions help you check the
answers to the ASCII tutorial worksheet? Would using the DEC2HEX and HEX2BIN functions help?
Produce a spreadsheet that uses these functions to give you a lookup table to convert between characters
and their ASCII codes
e.g.
A
B
C
D
1 Character Ascii value (decimal) Ascii value (hex)
Ascii value (binary)
2
A
65
41
01000001
3
B
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These last two exercises illustrate some of the limitations of computers in being able to represent numbers.
7.
Using a spreadsheet, experiment to find out the maximum and minimum values of integer and real
numbers that can be displayed.
9.
Using your spreadsheet, find the smallest value x such that (1.0+x) is not equal to 1.0.
10.
Open the file UnicodeExample.html in a browser and look at what displays. Open the file in notepad
and work out the relationship between the two ways of displaying the character. Can you work out
where the characters appear in the Unicode charts and look them up? Use the program excel to help
you to write a web page to display the whole page table of Gujerati characters with the code points
identified in decimal. Look at other code pages (e.g. the mathematical symbols at code points 2200
to 22FF) and see if they will display.
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