Download presentation source

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

page
1
page
2
page
3
page
4
page
5
page
6
page
7
page
8
page
9
page
10
page
11
page
12
page
13
page
14
page
15
page
16
page
17
page
18
page
19
page
20
page
21
page
22
page
23
page
24
page
25
page
26
page
27
page
28
page
29
page
30
page
31
page
32
page
33
Document related concepts
no text concepts found
Transcript 
CMPUT229 - Fall 2002
Topic1: Numbers, Numbers,
and Numbers
José Nelson Amaral
CMPUT 229 - Computer
Organization and Architecture I
1
Reading Assignment
Patt and Patel textbook:
Chapter2
CMPUT 229 - Computer
Organization and Architecture I
2
Positional Number System
329
3
2
9
923
9
2
3
9
9
2
2
3
3
923
329
CMPUT 229 - Computer
Organization and Architecture I
3
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 - Computer
Organization and Architecture I
4
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169 + 213 + 91
50710 + 2610 + 910
9169 + 213 + 31
152110 + 2610 + 310
54210
155010
CMPUT 229 - Computer
Organization and Architecture I
5
Binary System
In computers we are mostly interested on bases 2, 8, and 16.
1102
92316
122 + 121 + 020
9162 + 2161 + 3160
14 + 12 + 01
410 + 210 + 010
9256 + 216 + 31
230410 + 3210 + 310
610
233910
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)
CMPUT 229 - Computer
Organization and Architecture I
7
Signed Magnitude
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 signed magnitude representation we
use the first bit of the pattern to indicate
if it is a positive or a negative number.
Signed Magnitude
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?
Signed Magnitude
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 signed 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
??
pattern(-1)
pattern(1)
PattPatel pp. 20
Signed Magnitude
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 signed 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 = ??
PattPatel pp. 20
Signed Magnitude
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 signed 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)
PattPatel pp. 20
1’s Complement
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 folllowing
-1 + 1 = ??
-0 + 1 = ??
0 + 1 = ??
PattPatel pp. 20
2’s Complement
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.
PattPatel pp. 20
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 - Computer
Organization and Architecture I
15 23
PattPatel pp.
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 - Computer
Organization and Architecture I
16 24
PattPatel pp.
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 - Computer
Organization and Architecture I
17 24
PattPatel pp.
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 - Computer
Organization and Architecture I
18 24
PattPatel pp.
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 - Computer
Organization and Architecture I
19 24
PattPatel pp.
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
a6 = 1
Thus we got:
a6 = 1
a5 = 1
a4 = 0
a3 = 1
a2 = 0
a1 = 0
a0 = 1
10510 = 011010012
CMPUT 229 - Computer
Organization and Architecture I
20 25
PattPatel pp.
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
21
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 - Computer
Organization and Architecture I
22
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
Thus:
-10510 =100101112
CMPUT 229 - Computer
Organization and Architecture I
PattPatel pp.2322-23
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
By convention, the characters 0x
are printed in front of an hexadecimal
number to indicate base 16.
If the number 0xFACE 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 0xFACE is a negative number,
and we have to flip all the bits.
CMPUT 229 - Computer
Organization and Architecture I
PattPatel pp.2426-27
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
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 = 0163 + 5162 + 3161 + 2160
= 0 + 5256 + 316 + 21
= 1280 + 48 + 2
= 133010
0xFACE
= -133010
CMPUT
229 - Computer
Organization and Architecture I
PattPatel pp.2526-27
Binary Arithmetic
Decimal
Binary
19
+ 3
22
010011
+ 000011
010110
Binary
Decimal
14
- 9
5
910 = 0010012
-910 = 1101112
CMPUT 229 - Computer
Organization and Architecture I
001110
+ 110111
000101
26 25
PattPatel pp.
Overflow
What happens if we try to add +9 with +11 in a 5-bit 2-complement
representation?
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 representation
we say that the result has OVERFLOWed the
capacity of the representation?
CMPUT 229 - Computer
Organization and Architecture I
27 27
PattPatel pp.
Overflow Detection
What happens if we try to add +9 with +11 in a 5-bit 2-complement
representation?
Decimal
9
+ 11
20
Binary
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 - Computer
Organization and Architecture I
28 28
PattPatel pp.
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 we
added two negative numbers and got a positive result.
Could we get overflow when adding a positive and
a negative number?
CMPUT 229 - Computer
Organization and Architecture I
29 28
PattPatel pp.
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 - Computer
Organization and Architecture I
30 27
PattPatel pp.
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 - Computer
Organization and Architecture I
31 27
PattPatel pp.
Some Useful Numbers to
Remember
20 = 110 = 0x0001
21 = 210 = 0x0002
210 = 102410 = 0100 0000 0000 =
0x0400 = 1K
22 = 410 = 0x0004
23 = 810 = 0x0008
24 = 1610 = 0x0010
220 = 210 × 210 = 102410 × 102410 =
0001 0000 0000 0000 0000 0000 =
0x0010 0000 = 1M
25 = 3210 = 0x0020
26 = 6410 = 0x0040
230 = 210 × 210 × 210 = 1 G
27 = 12810 = 0x0080
28
= 25610 = 0x0100
240 = 210 × 210 × 210 × 210 = 1 T
29 = 51210 = 0x0200
CMPUT 229 - Computer
Organization and Architecture I
32
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 0x0780.”
“Stop in front of 104.”
“My house is 104.”
“Write 100 at 100.”
CMPUT 229 - Computer
Organization and Architecture I
33
Related documents