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J. Michael Moore Computer Organization CSCE 110

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Page 1: J. Michael Moore Computer Organization CSCE 110. J. Michael Moore High Level View Of A Computer ProcessorInputOutput Memory Storage

J. Michael Moore

Computer Organization

CSCE 110

Page 2: J. Michael Moore Computer Organization CSCE 110. J. Michael Moore High Level View Of A Computer ProcessorInputOutput Memory Storage

J. Michael Moore

High Level View Of A Computer

ProcessorInput Output

Memory

Storage

Page 3: J. Michael Moore Computer Organization CSCE 110. J. Michael Moore High Level View Of A Computer ProcessorInputOutput Memory Storage

J. Michael Moore

Getting Data In

Input

Others?

Page 4: J. Michael Moore Computer Organization CSCE 110. J. Michael Moore High Level View Of A Computer ProcessorInputOutput Memory Storage

J. Michael Moore

ProcessorInput Output

Memory

Storage

Page 5: J. Michael Moore Computer Organization CSCE 110. J. Michael Moore High Level View Of A Computer ProcessorInputOutput Memory Storage

J. Michael Moore

Getting Data Out

Output

Others?

Page 6: J. Michael Moore Computer Organization CSCE 110. J. Michael Moore High Level View Of A Computer ProcessorInputOutput Memory Storage

J. Michael Moore

ProcessorInput Output

Memory

Storage

Page 7: J. Michael Moore Computer Organization CSCE 110. J. Michael Moore High Level View Of A Computer ProcessorInputOutput Memory Storage

J. Michael Moore

Unit of Storage

• Bit–Binary digit

– Smallest unitof measurement

Memory

Storage

Two possiblevalues

on offOR

Page 8: J. Michael Moore Computer Organization CSCE 110. J. Michael Moore High Level View Of A Computer ProcessorInputOutput Memory Storage

J. Michael Moore

How Data Is Stored

• Byte: a group of 8 bits; 28=256 possibilities– 00000000, 00000001, 00000010, 00000011,

… , 11111111

• Memory: long sequence of locations, each large enough to hold one byte, numbered 0, 1, 2, 3, …

• Address: The number of the location

Page 9: J. Michael Moore Computer Organization CSCE 110. J. Michael Moore High Level View Of A Computer ProcessorInputOutput Memory Storage

J. Michael Moore

How Data Is Stored

• Contents of a location can change– e.g. 01011010 can become 11100001

• Use consecutive locations to store longer sequences– e.g. 4 bytes = 1 word

1 1 0 1 1 0 1 0 0 0 1 1 1 0 1 1 0 0 1 0 0 1 00

3...

bytes

0 1 2

bits

Page 10: J. Michael Moore Computer Organization CSCE 110. J. Michael Moore High Level View Of A Computer ProcessorInputOutput Memory Storage

J. Michael Moore

Binary Numbers

• Base Ten Numbers (Integers)– characters

• 0 1 2 3 4 5 6 7 8 9

– 5401 is 5x103 + 4x102 + 0x101 + 1x100

• Binary numbers are the same– characters

• 0 1

– 1011 is 1x23 + 0x22 + 1x21 + 1x20

Page 11: J. Michael Moore Computer Organization CSCE 110. J. Michael Moore High Level View Of A Computer ProcessorInputOutput Memory Storage

J. Michael Moore

Converting Binary to Base 10

• 23 = 8

• 22 = 4

• 21 = 2

• 20 = 1

1. 10012 = ____10 =

2. 1x23 + 0x22 + 0x21+ 1x20 =

3. 1x8 + 0x4 + 0x2 + 1x1 =

4. 8 + 0 + 0 + 1 =

5. 910

• 01102 = ____10 (Try yourself)

• 01102 = 610

Page 12: J. Michael Moore Computer Organization CSCE 110. J. Michael Moore High Level View Of A Computer ProcessorInputOutput Memory Storage

J. Michael Moore

Converting Base 10 to Binary

• 28 = 256• 27 = 128• 26 = 64• 25 = 32• 24 = 16• 23 = 8• 22 = 4• 21 = 2• 20 = 1

• 38810 = ____2

28 27 26 2425 2223 2021

1 1 10 00 0 00

• 388 - 256 (28) = 132

• 132 - 128 (27) = 4

• 4 - 4 (22) = 0

Page 13: J. Michael Moore Computer Organization CSCE 110. J. Michael Moore High Level View Of A Computer ProcessorInputOutput Memory Storage

J. Michael Moore

Converting Base 10 to Binary• 38810 = ____2 • 38810 / 2 = 19410 Remainder 0• 19410 / 2 = 9710 Remainder 0• 9710 / 2 = 4810 Remainder 1• 4810 / 2 = 2410 Remainder 0• 2410 / 2 = 1210 Remainder 0• 1210 / 2 = 610 Remainder 0• 610 / 2 = 310 Remainder 0• 310 / 2 = 110 Remainder 1• 110 / 2 = 010 Remainder 1

28 27 26 2425 2223 2021

1 1 10 00 0 00

Page 14: J. Michael Moore Computer Organization CSCE 110. J. Michael Moore High Level View Of A Computer ProcessorInputOutput Memory Storage

J. Michael Moore

Other common number representations

• Octal Numbers– characters

• 0 1 2 3 4 5 6 7 8

– 7820 is 7x83 + 8x82 + 2x81 + 0x80

• Hexadecimal Numbers– characters

• 0 1 2 3 4 5 6 7 8 9 A B C D E F

– 2FD6 is 2x163 + Fx162 + Dx161 + 6x160

http://fac-staff.seattleu.edu/quinnm/web/education/JavaApplets/applets/NumConv.html

Page 15: J. Michael Moore Computer Organization CSCE 110. J. Michael Moore High Level View Of A Computer ProcessorInputOutput Memory Storage

J. Michael Moore

Negative Numbers

• Can we store a negative sign?

• What can we do?– Use a bit

• Most common is two’s complement

Page 16: J. Michael Moore Computer Organization CSCE 110. J. Michael Moore High Level View Of A Computer ProcessorInputOutput Memory Storage

J. Michael Moore

Representing Negative Numbers

• Two’s Complement– flip all the bits

• change 0 to 1 and 1 to zero

– add 1– if the leftmost bit is 0, the number is 0 or

positive– if the leftmost bit is 1, the number is negative

Page 17: J. Michael Moore Computer Organization CSCE 110. J. Michael Moore High Level View Of A Computer ProcessorInputOutput Memory Storage

J. Michael Moore

Two’s Complement

• What is -9?– 9 is 00001001 in binary– flip the bits - 11110110– add 1 - 11110111

• Addition and Subtraction are easy– always addition

Page 18: J. Michael Moore Computer Organization CSCE 110. J. Michael Moore High Level View Of A Computer ProcessorInputOutput Memory Storage

J. Michael Moore

Two’s Complement

• Addition– 13 - 9 = 4– 13 + (-9) = 4– 00001101 + 11110111 = ?

0

0 100 0 0 11

1

1

0

1

01 1 1 11 1 1

1

0

11 1 1

0000 4=

1

This bit is lost

But that doesn’t matter since we get the

correct answer anyway

Page 19: J. Michael Moore Computer Organization CSCE 110. J. Michael Moore High Level View Of A Computer ProcessorInputOutput Memory Storage

J. Michael Moore

Real (Floating point) numbers

• Break the bits used into parts

0110101000000011

Mantissa Exponent

Sign bits

Page 20: J. Michael Moore Computer Organization CSCE 110. J. Michael Moore High Level View Of A Computer ProcessorInputOutput Memory Storage

J. Michael Moore

Limitations of Finite Data Encodings

• Overflow - number is too large– suppose 1 byte stores integers in base 2,

from 0 (00000000) to 255 (11111111) (note: this is not two’s complement although it would have the same problem)

– if the byte holds 255, then adding 1 to it results in 0, not 256

– http://www.artima.com/insidejvm/applets/InnerInt.html– http://classes.engr.oregonstate.edu/eecs/fall2007/cs160/applets/

TwosComplement.html

Page 21: J. Michael Moore Computer Organization CSCE 110. J. Michael Moore High Level View Of A Computer ProcessorInputOutput Memory Storage

J. Michael Moore

Limitations of Finite Data Exchange

• Roundoff Error– Insufficient precision (size of word)

• ex. Try to store 1/8, which is 0.001 in binary, with only two bits

– Nonterminating expansions in current base• ex. Try to store 1/3 in base 10, which is 0.3333…

– Nonterminating expansions in every base• ex. Irrational numbers such as

Page 22: J. Michael Moore Computer Organization CSCE 110. J. Michael Moore High Level View Of A Computer ProcessorInputOutput Memory Storage

J. Michael Moore

ProcessorInput Output

Memory

Storage