CMPUT 229 - Computer Organization and Architecture I
1
2. Number Systems
Decimal Binary Addition HeadecimalTwo’s complement ASCII characters
CMPUT 229 - Computer Organization and Architecture I
2
Positional Number System
329 923
329
923
9 2 3
9 23
329 923
CMPUT 229 - Computer Organization and Architecture I
3
Decimal numbers
329 923
3102 + 2101 + 9100 9102 + 2101 + 3100
3100 + 210 + 91 9100 + 210 + 31
300 + 20 + 9 900 + 20 + 3
923329
CMPUT 229 - Computer Organization and Architecture I
4
Positional Number System
32913 92313
3132 + 2131 + 9130 9132 + 2131 + 3130
3169 + 213 + 91 9169 + 213 + 3150710 + 2610 + 910 152110 + 2610 + 310
The same positional system works with different basis:
15501054210
CMPUT 229 - Computer Organization and Architecture I
5
Binary System
1102 92316
122 + 121 + 020 9162 + 2161 + 3160
14 + 12 + 01 9256 + 216 + 31410 + 210 + 010 230410 + 3210 + 310
In computers we are mostly interested on bases 2, 8, and 16.
233910610
CMPUT 229 - Computer Organization and Architecture I
6
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)
Signed MagnitudeValue Represented Pattern
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 signed magnitude representation weuse the first bit of the pattern to indicateif it is a positive or a negative number.
Signed MagnitudeValue Represented Pattern
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?
Signed MagnitudeValue Represented Pattern
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 represent0 is wasteful.
The signed magnitude representationhas the advantage that it is easy toread the value from the pattern.
But does it have the binary arithmetic property?
For instance, what is the resultof pattern(-1) + pattern(1)?
1001 pattern(-1)+ 0001 pattern(1) ??
PattPatel pp. 20
Signed MagnitudeHaving two patterns to represent0 is wasteful.
The signed magnitude representationhas the advantage that it is easy toread the value from the pattern.
But does it have the arithmetic property?
For instance, what is the resultof pattern(-1) + pattern(1)?
1001 pattern(-1)+ 0001 pattern(1) 1010 = ??
Value Represented Pattern 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
PattPatel pp. 20
Signed MagnitudeValue Represented Pattern
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 represent0 is wasteful.
The signed magnitude representationhas the advantage that it is easy toread the value from the pattern.
But does it have the arithmetic property?
For instance, what is the resultof pattern(-1) + pattern(1)?
1001 pattern(-1)+ 0001 pattern(1) 1010 = pattern(-2)
PattPatel pp. 20
1’s ComplementValue Represented Pattern
Sign Magnitude 1’s complement 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 -0 -7 1001 -1 -6 1010 -2 -5 1011 -3 -4 1100 -4 -3 1101 -5 -2 1110 -6 -1 1111 -7 -0
A negative number is representedby “flipping” all the bits of a positivenumber.
We still have two patterns for 0.
It is still easy to read a value froma given pattern.
How about the arithmetic property?
Suggestion: try the folllowing-1 + 1 = ??-0 + 1 = ?? 0 + 1 = ??
PattPatel pp. 20
CMPUT 229 - Computer Organization and Architecture I
13
2’s Complement
a representation for negative numbers The leftmost bit is used to indicate +/-
Positive number starts with 0, negative 1
A negative number is obtained by Convert the corresponding positive
decimal number to a binary toggle all bits ( all 10 and all 0 1) Add 1 to the the binary obtained in the
previous step
2’s ComplementValue Represented Pattern
Sign Magnitude 1’s complement 2’s complement 0000 0 0 0 0001 1 1 1 0010 2 2 2 0011 3 3 3 0100 4 4 4 0101 5 5 5 0110 6 6 6 0111 7 7 7 1000 -0 -7 -8 1001 -1 -6 -7 1010 -2 -5 -6 1011 -3 -4 -5 1100 -4 -3 -4 1101 -5 -2 -3 1110 -6 -1 -2 1111 -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.
PattPatel pp. 20
CMPUT 229 - Computer Organization and Architecture I
15
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.
PattPatel pp. 23
Because the binary representation has8 bits, the decimal value must be in the [-27; +(27-1)] =[-128;+127] interval.
CMPUT 229 - Computer Organization and Architecture I
16
if (negative = true)then
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:
00
11
22
33
44
55
66 2222222 aaaaaaaV
VV
VV
1
PattPatel pp. 24
CMPUT 229 - Computer Organization and Architecture I
17
Binary to Decimal Conversion (Examples)
Convert the 2’s complement integer 11000111 to itsdecimal 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:
56000816320
10204081161321640
20202021212120 0123456
V
3. Now we remember that it was a negative number, thus:57 ;57156 VV
PattPatel pp. 24
CMPUT 229 - Computer Organization and Architecture I
18
Decimal to Binary Convertion
We will start with an example. What is the binaryrepresentation of 10510?
00
11
22
33
44
55
66 2222222105 aaaaaaa
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:
11
22
33
44
55
66 222222104 aaaaaa
PattPatel pp. 24
CMPUT 229 - Computer Organization and Architecture I
19
Decimal to Binary Convertion (cont.)
11
22
33
44
55
66 222222104 aaaaaa
Now we can divide both sides by 2
01
12
23
34
45
56 22222252 aaaaaa
Because 52 is even, we know that a1 = 0
12
23
34
45
56 2222252 aaaaa
02
13
24
35
46 2222226 aaaaa a2 = 0
13
24
35
46 222226 aaaa
03
14
25
36 222213 aaaa a3 = 1
14
25
36 22212 aaa
PattPatel pp. 24
CMPUT 229 - Computer Organization and Architecture I
20
Decimal to Binary Convertion (cont.)
14
25
36 22212 aaa
04
15
26 2226 aaa a4 = 0
15
26 226 aa
05
16 223 aa a5 = 1
16 22 a
06 21 a a6 = 1
Thus we got:a1 = 0a4 = 0a5 = 1a6 = 1 a2 = 0a3 = 1 a0 = 1
10510 = 011010012
PattPatel pp. 25
CMPUT 229 - Computer Organization and Architecture I
21
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 - Computer Organization and Architecture I
22
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
rightmost digit
10510 = 011010012
CMPUT 229 - Computer Organization and Architecture I
23
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 negativenumber we must flip all the bits of the positive
representation and add 1:
10010110+ 00000001
10010111
Thus:-10510 =100101112
PattPatel pp. 22-23
CMPUT 229 - Computer Organization and Architecture I
24
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
If the number 0xFACE representsa 2’s complement binary number,
what is its decimal value?
By convention, the characters 0xare printed in front of an hexadecimal
number to indicate base 16.
First we need to look up the binaryrepresentation of F, which is 1111.
Therefore 0xFACE is a negative number, and we have to flip all the bits.
PattPatel pp. 26-27
CMPUT 229 - Computer Organization and Architecture I
25
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
It is best to write down the binaryrepresentation of the number first:
0xFACE = 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 = 0x0532
Then we convert 0x0532 from base 16 to base 10:
0x0532 = 0163 + 5162 + 3161 + 2160
= 0 + 5256 + 316 + 21
= 1280 + 48 + 2
= 133010
0xFACE = -133010
PattPatel pp. 26-27
CMPUT 229 - Computer Organization and Architecture I
26
Binary Arithmetic
Decimal
19+ 3
22
Binary
010011+ 000011
010110
Decimal
14- 9
5
Binary
001110+ 110111
000101
910 = 0010012
-910 = 1101112
PattPatel pp. 25
CMPUT 229 - Computer Organization and Architecture I
27
Overflow
What happens if we try to add +9 with +11 in a 5-bit 2-complementrepresentation?
Decimal
9+ 11
20
Binary
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 representationwe say that the result has OVERFLOWed the
capacity of the representation?
PattPatel pp. 27
CMPUT 229 - Computer Organization and Architecture I
28
Overflow Detection
What happens if we try to add +9 with +11 in a 5-bit 2-complementrepresentation?
Decimal
9+ 11
20
Binary
01001+ 01011
10100 = -12 ?
We can easily detect the overflow by detecting that theaddition of two positive numbers resulted in a negative result.
PattPatel pp. 28
CMPUT 229 - Computer Organization and Architecture I
29
Overflow (another example)
Could overflow happen when we add two negative numbers?
Decimal
- 12+ -6-18
Binary
10100+ 11010
01110 = +14 ?
Again we can detect overflow by detecting that weadded two negative numbers and got a positive result.
Could we get overflow when adding a positive anda negative number?
PattPatel pp. 28
CMPUT 229 - Computer Organization and Architecture I
30
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
PattPatel pp. 27
CMPUT 229 - Computer Organization and Architecture I
31
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
To sign-extend anumber to a largerrepresentation, allwe have to do is toreplicate the sign bituntil we obtain the
new length.
PattPatel pp. 27
CMPUT 229 - Computer Organization and Architecture I
32
Storing characters
ASCII (American Standard Code for Information Interchange) One byte
UTF-8 (8-bit Unicode Transformation Format) One to four bytes Coincides with ASCII Be able to represent any languages
CMPUT 229 - Computer Organization and Architecture I
33
Endianess
0x100010000x100010010x100010020x100010030x100010040x100010050x100010060x10001007
0x000x000x000x130xFF0xFF0xFF0x97
Address Value
In the previous example, there are two ways to store +1910 atthe address 0x10001000 and -10510 at the address 0x10001004:
0x100010000x100010010x100010020x100010030x100010040x100010050x100010060x10001007
0x130x000x000x000x970xFF0xFF0xFF
Address Value
What is the difference?
CMPUT 229 - Computer Organization and Architecture I
34
Little-End and Big-End
+1910 = 0000 0000 0000 0000 0000 0000 0001 0011
+1910 = 0x0000 0013
Little end of +1910 Big end of +1910
(binary)
(hexadecimal)
-10510 = 1111 1111 1111 1111 1111 1111 1001 0111
+1910 = 0xFFFF FF97
Little end of +1910 Big end of +1910
(binary)
(hexadecimal)
CMPUT 229 - Computer Organization and Architecture I
35
Endianess
0x100010000x100010010x100010020x100010030x100010040x100010050x100010060x10001007
0x000x000x000x130xFF0xFF0xFF0x97
Address Value
The question is: which end of the integer do we store first in memory?
0x100010000x100010010x100010020x100010030x100010040x100010050x100010060x10001007
0x130x000x000x000x970xFF0xFF0xFF
Address Value
Big end of +1910 Little end of +1910
Little Endian Byte OrderBig Endian Byte Order
DECstations and Intel 80x86 are little-endians.
Sun SPARC and Macintosh are big-endians.
CMPUT 229 - Computer Organization and Architecture I
36
Floating Number
How to represent a very large number?
How about a number of the form 2.5006 x 10
Floating number IEEE standard
-8
CMPUT 229 - Computer Organization and Architecture I
37
IEEE 754 Floating Number
S exponent fraction (precision)
1 8 23
N = -1 x 1. Fraction x 2 S exponent – 127
CMPUT 229 - Computer Organization and Architecture I
38
Example
-65
8
Top Related