j. michael moore computer organization cpsc 110. j. michael moore high level view of a computer...
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J. Michael Moore
Unit of Storage
• ___–___________
– Smallest unitof measurement
Memory
Storage
Two possiblevalues
on offOR
J. Michael Moore
How Data Is Stored
• ___: a group of 8 bits; 28=256 possibilities– 00000000, 00000001, 00000010, 00000011,
… , 11111111
• ___________: long sequence of locations, each large enough to hold one byte, numbered 0, 1, 2, 3, …
• ___________: The number of the location
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
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
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
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
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
J. Michael Moore
Other common number representations
• _____________ Numbers– characters
• 0 1 2 3 4 5 6 7 8
– 7820 is 7x83 + 8x82 + 2x81 + 0x80
• _____________ Numbers– characters
• 0 1 2 3 4 5 6 7 8 9 A B C D E F
– 2FD6 is 2x163 + Fx162 + Dx161 + 6x160
J. Michael Moore
__________ Numbers
• Can we store a __________ sign?
• What can we do?– Use a ____
• Most common is ____________________
J. Michael Moore
Representing ________ Numbers
• ____________________________– flip all the bits
• change 0 to 1 and 1 to zero
– add 1– if the leftmost bit is __, the number is __ or
_______________– if the leftmost bit is __, the number is
_______________
J. Michael Moore
______________________
• What is -9?– 9 is 00001001 in binary– flip the bits → 11110110– add 1 → 11110111
• Addition and Subtraction are easy– always addition
J. Michael Moore
_______________________
• 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
J. Michael Moore
Real (____________) numbers
• Break the bits used into parts
0110101000000011
________ ________
Sign bits
J. Michael Moore
Limitations of Finite Data Encodings
• __________ - number is too large– suppose 1 byte stores integers in base 2,
from 0 (00000000) to 255 (11111111) (note: this is not ___________________ although it would have the same problem)
– if the byte holds 255, then adding 1 to it results in __, not _____
J. Michael Moore
Limitations of Finite Data Exchange
• ___________ Error– Insufficient ___________ (size of word)
• ex. Try to store 1/8, which is 0.001 in binary, with only two bits
– ______________ expansions in current base• ex. Try to store 1/3 in base 10, which is 0.3333…
– ______________ expansions in every base• ex. Irrational numbers such as