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Faculty of Computer Science
Representing Information
Numbers, Numbers, and
Numbers
CMPUT 229
© 2006
Department of Computing Science
Slide’s source
Yale N. Patt and Sanjay J. Patel, Introduction to Computing Systems:
From bits & gates to C & Beyond, McGrawHill Press, 2001, Chapter2.
CMPUT 229
© 2006
Department of Computing Science
Positional Number System
329
3
2
9
923
9
2
3
9
9
2
2
3
3
923
329
CMPUT 229
© 2006
Department of Computing Science
Positional Number System
329
923
3102 + 2101 + 9100
9102 + 2101 + 3100
3100 + 210 + 91
300 + 20 + 9
9100 + 210 + 31
900 + 20 + 3
329
923
CMPUT 229
© 2006
Department of Computing Science
Positional Number System
The same positional system works with different basis:
32913
92313
3132 + 2131 + 9130
9132 + 2131 + 3130
316910 + 21310 + 9110
50710 + 2610 + 910
916910 + 21310 + 3110
152110 + 2610 + 310
54210
155010
CMPUT 229
© 2006
Department of Computing Science
Binary System
In computers we are mostly interested on bases 2, 8, and 16.
1102
92316
122 + 121 + 020
1410 + 1210 + 0110
410 + 210 + 010
610
9162 + 2161 + 3160
925610 + 21610 + 3110
230410 + 3210 + 310
233910
CMPUT 229
© 2006
Department of Computing Science
Signed Integers
– Problem: given 2k distinct patterns of bits, each pattern
with k bits, assign integers to the patterns in such a way
that:
• The numbers are spread in an interval around zero without
gaps.
• Roughly half of the patterns represent positive numbers,
and half represent negative numbers.
• When using standard binary addition, given an integer n,
the following property should hold:
pattern(n+1) = pattern(n) + pattern(1)
CMPUT 229
© 2006
Department of Computing Science
Sign-Magnitude Representation
Pattern Value Represented
Sign Magnitude
0000
0
0001
1
0010
2
0011
3
0100
4
0101
5
0110
6
0111
7
1000
1001
1010
1011
1100
1101
1110
1111
In a sign-magnitude representation we
use the first bit of the pattern to indicate
if it is a positive or a negative number.
CMPUT 229
© 2006
Department of Computing Science
Sign-Magnitude Represetation
Pattern Value Represented
Sign Magnitude
0000
0
0001
1
0010
2
0011
3
0100
4
0101
5
0110
6
0111
7
1000
1001
-1
1010
-2
1011
-3
1100
-4
1101
-5
1110
-6
1111
-7
What do we do with the pattern 1000?
CMPUT 229
© 2006
Department of Computing Science
Sign-Magnitude Representation
Pattern Value Represented
Sign Magnitude
0000
0
0001
1
0010
2
0011
3
0100
4
0101
5
0110
6
0111
7
1000
-0
1001
-1
1010
-2
1011
-3
1100
-4
1101
-5
1110
-6
1111
-7
Having two patterns to represent
0 is wasteful.
The sign-magnitude representation
has the advantage that it is easy to
read the value from the pattern.
But does it have the binary
arithmetic property?
For instance, what is the result
of pattern(-1) + pattern(1)?
1001
+ 0001
??
CMPUT 229
pattern(-1)
pattern(1)
PattPatel, pp. 20
© 2006
Department of Computing Science
Sign-Magnitude Representation
Pattern Value Represented
Sign Magnitude
0000
0
0001
1
0010
2
0011
3
0100
4
0101
5
0110
6
0111
7
1000
-0
1001
-1
1010
-2
1011
-3
1100
-4
1101
-5
1110
-6
1111
-7
Having two patterns to represent
0 is wasteful.
The sign-magnitude representation
has the advantage that it is easy to
read the value from the pattern.
But does it have the arithmetic
property?
For instance, what is the result
of pattern(-1) + pattern(1)?
1001 pattern(-1)
+ 0001 pattern(1)
1010 = ??
CMPUT 229
PattPatel, pp. 20
© 2006
Department of Computing Science
Sign-Magnitude Representation
Pattern Value Represented
Sign Magnitude
0000
0
0001
1
0010
2
0011
3
0100
4
0101
5
0110
6
0111
7
1000
-0
1001
-1
1010
-2
1011
-3
1100
-4
1101
-5
1110
-6
1111
-7
Having two patterns to represent
0 is wasteful.
The sign-magnitude representation
has the advantage that it is easy to
read the value from the pattern.
But does it have the arithmetic
property?
For instance, what is the result
of pattern(-1) + pattern(1)?
1001 pattern(-1)
+ 0001 pattern(1)
1010 = pattern(-2)
CMPUT 229
PattPatel, pp. 20
© 2006
Department of Computing Science
1’s-Complement Representation
Pattern
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
Value Represented
Sign Magnitude 1’s complement
0
0
1
1
2
2
3
3
4
4
5
5
6
6
7
7
-0
-7
-1
-6
-2
-5
-3
-4
-4
-3
-5
-2
-6
-1
-7
-0
A negative number is represented
by “flipping” all the bits of a positive
number.
We still have two patterns for 0.
It is still easy to read a value from
a given pattern.
How about the arithmetic property?
Suggestion: try the following
-1 + 1 = ??
-0 + 1 = ??
0 + 1 = ??
CMPUT 229
PattPatel, pp. 20
© 2006
Department of Computing Science
2’s-Complement Representation
Pattern
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
Value Represented
Sign Magnitude 1’s complement 2’s complement
0
0
0
1
1
1
2
2
2
3
3
3
4
4
4
5
5
5
6
6
6
7
7
7
-0
-7
-8
-1
-6
-7
-2
-5
-6
-3
-4
-5
-4
-3
-4
-5
-2
-3
-6
-1
-2
-7
-0
-1
A single pattern
for 0.
1111 pattern(-1)
+ 0001 pattern(1)
0000 = pattern(0)
It holds the arithmetic property.
But the reading of a negative pattern is not trivial.
CMPUT 229
PattPatel, pp. 20
© 2006
Department of Computing Science
Binary to Decimal Conversion
Problem: Given an 8-bit 2’s complement
binary number:
a7 a6 a5 a4 a3 a2 a1 a0
find its corresponding decimal value.
Because the binary representation has
8 bits, the decimal value must be in the
[-27; +(27-1)] =[-128;+127] interval.
CMPUT 229
PattPatel, pp. 23
© 2006
Department of Computing Science
Binary to Decimal Conversion
a7 a6 a5 a4 a3 a2 a1 a0
Solution:
negative  false
if (a7 = 1)
then negative  true
flip all bits;
compute magnitude using:
V  a6  26  a5  25  a4  2 4  a3  23  a2  2 2  a1  21  a0  20
if (negative = true)
then V  V  1
V  V
CMPUT 229
PattPatel, pp. 24
© 2006
Department of Computing Science
Binary to Decimal Conversion (Examples)
Convert the 2’s complement integer 11000111 to its
decimal integer value.
1. a7 is 1, thus we make a note that this is a negative
number and invert all the bits, obtaining:
00111000
2. We compute the magnitude:
V  0  26  1 25  1 2 4  1 23  0  2 2  0  21  0  20
 0  64  1 32  116  1 8  0  4  0  2  0 1
 0  32  16  8  0  0  0  56
3. Now we remember that it was a negative number, thus:
V  56  1  57;
V  57
CMPUT 229
PattPatel, pp. 24
© 2006
Department of Computing Science
Decimal to Binary Convertion
We will start with an example. What is the binary
representation of 10510?
105  a6  26  a5  25  a4  2 4  a3  23  a2  2 2  a1  21  a0  20
Our problem is to find the values of each ai
Because 105 is odd, we know that a0 = 1
Thus we can subtract 1 from both sides to obtain:
104  a6  26  a5  25  a4  2 4  a3  23  a2  2 2  a1  21
CMPUT 229
PattPatel, pp. 24
© 2006
Department of Computing Science
Decimal to Binary Convertion (cont.)
104  a6  26  a5  25  a4  2 4  a3  23  a2  2 2  a1  21
Now we can divide both sides by 2
52  a6  25  a5  2 4  a4  23  a3  2 2  a2  21  a1  20
Because 52 is even, we know that a1 = 0
52  a6  25  a5  2 4  a4  23  a3  2 2  a2  21
26  a6  2 4  a5  23  a4  2 2  a3  21  a2  20
a2 = 0
26  a6  2 4  a5  23  a4  2 2  a3  21
13  a6  23  a5  2 2  a4  21  a3  20
a3 = 1
12  a6  23  a5  2 2  a4  21
CMPUT 229
PattPatel, pp. 24
© 2006
Department of Computing Science
Decimal to Binary Convertion (cont.)
12  a6  23  a5  2 2  a4  21
6  a6  2 2  a5  21  a4  20
a4 = 0
6  a6  2 2  a5  21
3  a6  21  a5  20
a5 = 1
2  a6  21
1  a6  2 0
Thus we got:
a6 = 1
a5 = 1
a6 = 1
a4 = 0
a3 = 1
a2 = 0
a1 = 0
a0 = 1
10510 = 011010012
CMPUT 229
PattPatel, pp. 25
© 2006
Department of Computing Science
Decimal to Binary Conversion (Another Method)
We can also use repeated long division:
105/2 = 52 remainder 1
52/2 = 26 remainder 0
26/2 = 13 remainder 0
13/2 = 6 remainder 1
6/2 = 3 remainder 0
3/2 = 1 remainder 1
1/2 = 0 remainder 1
CMPUT 229
© 2006
Department of Computing Science
Decimal to Binary Conversion (Another Method)
We can also use repeated long division:
105/2 = 52 remainder 1
rightmost digit
52/2 = 26 remainder 0
26/2 = 13 remainder 0
13/2 = 6 remainder 1
10510 = 011010012
6/2 = 3 remainder 0
3/2 = 1 remainder 1
1/2 = 0 remainder 1
CMPUT 229
© 2006
Department of Computing Science
Decimal to Binary Conversion (Negative Numbers)
What is the binary representation of -10510 in 8 bits?
We know from the previous slide that:
+10510 =011010012
To obtain the binary representation of a negative
number we must flip all the bits of the positive
representation and add 1:
10010110
+ 00000001
10010111
CMPUT 229
Thus:
-10510 =100101112
PattPatel, pp. 22-23
© 2006
Department of Computing Science
Hexadecimal Numbers (base 16)
Binary Decimal Hexadecimal
0000
0
0
0001
1
1
0010
2
2
0011
3
3
0100
4
4
0101
5
5
0110
6
6
0111
7
7
1000
8
8
1001
9
9
1010
10
A
1011
11
B
1100
12
C
1101
13
D
1110
14
E
1111
15
F
In the 68K assembler, by convention,
the character $
is printed in front of an hexadecimal
number to indicate base 16.
If the number $FACE represents
a 2’s complement binary number,
what is its decimal value?
First we need to look up the binary
representation of F, which is 1111.
Therefore $FACE is a negative number,
and we have to flip all the bits.
CMPUT 229
PattPatel, pp. 26-27
© 2006
Department of Computing Science
Hexadecimal Numbers (base 16)
It is best to write down the binary
representation of the number first:
Binary Decimal Hexadecimal
0000
0
0
0001
1
1
0010
2
2
0011
3
3
0100
4
4
0101
5
5
0110
6
6
0111
7
7
1000
8
8
1001
9
9
1010
10
A
1011
11
B
1100
12
C
1101
13
D
1110
14
E
1111
15
F
$FACE = 1111 1010 1100 1110
Now we flip all the bits and add 1:
0000 0101 0011 0001
+ 0000 0000 0000 0001
0000 0101 0011 0010 = $0532
Then we convert 0x0532 from base 16 to base 10:
$0532 = 0163 + 5162 + 3161 + 2160
= 0 + 5256 + 316 + 21
= 1280 + 48 + 2
= 133010
$FACE = -133010
CMPUT 229
PattPatel, pp. 26-27
© 2006
Department of Computing Science
Binary Arithmetic
Decimal
Binary
19
+ 3
22
010011
+ 000011
010110
Binary
Decimal
14
- 9
5
910 = 0010012
-910 = 1101112
CMPUT 229
001110
+ 110111
000101
PattPatel, pp. 25
© 2006
Department of Computing Science
Overflow
What happens if we try to add +9 with +11 in a 5-bit 2-complement
representation?
Decimal
Binary
9
+ 11
20
01001
+ 01011
10100 = -12 ?
The result is too large to represent in 5 digits,
i.e. it is larger than 01111 = +1510.
When the result is too large for the representation
we say that the result has OVERFLOWed the
capacity of the representation.
CMPUT 229
PattPatel, pp. 27
© 2006
Department of Computing Science
Overflow Detection
What happens if we try to add +9 with +11 in a 5-bit 2-complement
representation?
Decimal
Binary
9
+ 11
20
01001
+ 01011
10100 = -12 ?
We can easily detect the overflow by detecting that the
addition of two positive numbers resulted in a negative result.
CMPUT 229
PattPatel, pp. 28
© 2006
Department of Computing Science
Overflow (another example)
Could overflow happen when we add two negative numbers?
Decimal
Binary
- 12
+ -6
-18
10100
+ 11010
01110 = +14 ?
Again we can detect overflow by detecting that we
added two negative numbers and got a positive result.
Could we get overflow when adding a positive and
a negative number?
CMPUT 229
PattPatel, pp. 28
© 2006
Department of Computing Science
Sign-extension
What is the 8-bit representation of +510?
0000 0101
What is the 16-bit representation of +510?
0000 0000 0000 0101
What is the 8-bit representation of -510?
1111 1011
What is the 8-bit representation of -510?
1111 1111 1111 1011
CMPUT 229
PattPatel, pp. 27
© 2006
Department of Computing Science
Sign-extension
What is the 8-bit representation of +510?
0000 0101
What is the 16-bit representation of +510?
0000 0000 0000 0101
What is the 8-bit representation of -510?
1111 1011
To sign-extend a
number to a larger
representation, all
we have to do is to
replicate the sign bit
until we obtain the
new length.
What is the 8-bit representation of -510?
1111 1111 1111 1011
CMPUT 229
PattPatel, pp. 27
© 2006
Department of Computing Science
Some Useful Numbers to Remember
20 = 110 = $0001
210 = 102410 = 0100 0000 0000 =
$0400 = 1K
21 = 210 = $0002
22 = 410 = $0004
23 = 810 = $0008
220 = 210 × 210 = 102410 × 102410 =
0001 0000 0000 0000 0000 0000 =
$0010 0000 = 1M
24 = 1610 = $0010
25 = 3210 = $0020
26 = 6410 = $0040
230 = 210 × 210 × 210 = 1 G
27 = 12810 = $0080
28
240 = 210 × 210 × 210 × 210 = 1 T
= 25610 = $0100
29 = 51210 = $0200
CMPUT 229
© 2006
Department of Computing Science
Data and Addresses
“She is 104.”
In each case, how do we know
what 104 is?
“I am staying at 104.”
From the context in which it
is used! The same is true for
data and addresses. The same
number is data on one instance
and address on another.
“It costs 104.”
“There is 104.”
“My office is 104.”
“Write 100 at $0780.”
“Stop in front of 104.”
“Write 100 at 100.”
“My house is 104.”
CMPUT 229
© 2006
Department of Computing Science
Representing Fractions
In base 10 we know that:
+0.10510 = 10510/100010 = 10510/103
Fractions are represented in a similar fashion in base 2:
+0.101012 = 101012/1000002 = 101012/25
101012 = 2110 and 25 = 32, thus
0.101012 = 2110/3210 = 0.6562510
CMPUT 229
Clements, pp. 152
© 2006
Department of Computing Science
Fractions: Binary to Decimal
Another way to convert a binary fraction into
a decimal value is as follows:
0.101012 = 12-1 + 02-2 + 12-3 + 02-4 + 12-5
= 10.5 + 00.25 + 10.125 + 00.0625 + 10.03125
= 0.5 + 0.125 + 0.03125
= 0.65625
CMPUT 229
© 2006
Department of Computing Science
Fractions: Decimal to Binary
The following method works to convert decimal factions
to binary:
0.687510 = ??2
0.6875  2 = 1.3750
0.3750  2 = 0.7500
0.7500  2 = 1.5000
0.5000  2 = 1.0000
0.0000  2 Done!
0.687510 = 0.10112
CMPUT 229
© 2006
Department of Computing Science
CMPUT 229
© 2006