faculty of computer science © 2006 cmput 229 representing information numbers, numbers, and numbers

Faculty of Computer Science

CMPUT 229 © 2006

Representing Information

Numbers, Numbers, and Numbers

© 2006

Department of Computing Science

CMPUT 229

Slide’s source

Yale N. Patt and Sanjay J. Patel, Introduction to Computing Systems: From bits & gates to C & Beyond, McGrawHill Press, 2001, Chapter2.

© 2006


CMPUT 229

Positional Number System

329 923

329

923

9 2 3

9 23

329 923

© 2006


CMPUT 229


329 923

3102 + 2101 + 9100 9102 + 2101 + 3100

3100 + 210 + 91 9100 + 210 + 31

300 + 20 + 9 900 + 20 + 3

923329

© 2006


CMPUT 229


32913 92313

3132 + 2131 + 9130 9132 + 2131 + 3130

316910 + 21310 + 9110 916910 + 21310 + 3110

50710 + 2610 + 910 152110 + 2610 + 310

The same positional system works with different basis:

15501054210

© 2006


CMPUT 229

Binary System

1102 92316

122 + 121 + 020 9162 + 2161 + 3160

1410 + 1210 + 0110 925610 + 21610 + 3110

410 + 210 + 010 230410 + 3210 + 310

In computers we are mostly interested on bases 2, 8, and 16.

233910610

© 2006


CMPUT 229

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)

© 2006


CMPUT 229

Sign-Magnitude Representation

Value 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 sign-magnitude representation weuse the first bit of the pattern to indicateif it is a positive or a negative number.

© 2006


CMPUT 229

Sign-Magnitude Represetation


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?

© 2006


CMPUT 229



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 sign-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

© 2006


CMPUT 229

Sign-Magnitude RepresentationHaving two patterns to represent0 is wasteful.


But does it have the arithmetic property?


1001 pattern(-1)+ 0001 pattern(1) 1010 = ??


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

© 2006


CMPUT 229



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.


But does it have the arithmetic property?


1001 pattern(-1)+ 0001 pattern(1) 1010 = pattern(-2)

PattPatel, pp. 20

© 2006


CMPUT 229

1’s-Complement Representation

Value 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 following-1 + 1 = ??-0 + 1 = ?? 0 + 1 = ??

PattPatel, pp. 20

© 2006


CMPUT 229

2’s-Complement Representation

Value 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

© 2006


CMPUT 229

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 has8 bits, the decimal value must be in the [-27; +(27-1)] =[-128;+127] interval.

PattPatel, pp. 23

© 2006


CMPUT 229

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

© 2006


CMPUT 229

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

© 2006


CMPUT 229

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

© 2006


CMPUT 229

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

© 2006


CMPUT 229

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

© 2006


CMPUT 229

Decimal to Binary Conversion (Another Method)

We can also use repeated long division:

105/2 = 52 remainder 1




6/2 = 3 remainder 0

3/2 = 1 remainder 1

1/2 = 0 remainder 1

© 2006


CMPUT 229

Decimal to Binary Conversion (Another Method)

We can also use repeated long division:





6/2 = 3 remainder 0

3/2 = 1 remainder 1

1/2 = 0 remainder 1

rightmost digit

10510 = 011010012

© 2006


CMPUT 229

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

© 2006


CMPUT 229

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 $FACE representsa 2’s complement binary number,

what is its decimal value?

In the 68K assembler, by convention, the character $

is printed in front of an hexadecimalnumber to indicate base 16.

First we need to look up the binaryrepresentation of F, which is 1111.

Therefore $FACE is a negative number, and we have to flip all the bits.


© 2006


CMPUT 229

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:

$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 = 0163 + 5162 + 3161 + 2160

= 0 + 5256 + 316 + 21

= 1280 + 48 + 2

= 133010

$FACE = -133010


© 2006


CMPUT 229

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

© 2006


CMPUT 229

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

© 2006


CMPUT 229

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

© 2006


CMPUT 229

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

© 2006


CMPUT 229

Sign-extension

What is the 8-bit representation of +510?

0000 0101


0000 0000 0000 0101

What is the 8-bit representation of -510?

1111 1011


1111 1111 1111 1011

PattPatel, pp. 27

© 2006


CMPUT 229

Sign-extension


0000 0101


0000 0000 0000 0101


1111 1011


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

© 2006


CMPUT 229

Some Useful Numbers to Remember

20 = 110 = $0001

21 = 210 = $0002

22 = 410 = $0004

23 = 810 = $0008

24 = 1610 = $0010

25 = 3210 = $0020

26 = 6410 = $0040

27 = 12810 = $0080

28 = 25610 = $0100

29 = 51210 = $0200

210 = 102410 = 0100 0000 0000 =$0400 = 1K

220 = 210 × 210 = 102410 × 102410 = 0001 0000 0000 0000 0000 0000 =

$0010 0000 = 1M

230 = 210 × 210 × 210 = 1 G

240 = 210 × 210 × 210 × 210 = 1 T

© 2006


CMPUT 229

Data and Addresses

“She is 104.”

“I am staying at 104.”

“It costs 104.”

“There is 104.”

“My office is 104.”

“Stop in front of 104.”

“My house is 104.”

In each case, how do we know what 104 is?

From the context in which it is used! The same is true fordata and addresses. The samenumber is data on one instanceand address on another.

“Write 100 at $0780.”

“Write 100 at 100.”

© 2006


CMPUT 229

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

Clements, pp. 152

© 2006


CMPUT 229

Fractions: Binary to Decimal

Another way to convert a binary fraction into

a decimal value is as follows:

0.101012 = 12-1 + 02-2 + 12-3 + 02-4 + 12-5

= 10.5 + 00.25 + 10.125 + 00.0625 +

10.03125

= 0.5 + 0.125 + 0.03125

= 0.65625

© 2006


CMPUT 229

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

faculty of computer science © 2006 cmput 229 representing information numbers, numbers, and numbers

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