BITS-WIPRO:MS IN SOFTWARE ENGINEERING

BITS-WIPRO:MS IN SOFTWARE ENGINEERING
14-01-2007
Course number and course title
SEWP ZC221 Structured Programming
Instructer : Rajesh Kulkarni
hodcse@biet.ac.in
rkpv2002@gmail.com
http://children-off-lesser-gods.blogspot.com
http://360.yahoo.com/rkpv2005
week 1: Elementary computer organization; introduction to number
systems, Representation of Integers, Real numbers, Overflow,
Bit, Data types and operations
1.1.1 Computer OrganizationThe way hardware components are connected together in a computer system.Computer architecture is the structure and behavior of various functional modules and how they interact to provide the processing needs of the user.Computer design is the development of hardware according to given set of specifications. Computer organization is I. CPU Organization II. Memory Organization III. I/O OrganizationI. CPU Organization: ALU,CU, Registers which are memory elements for temporary storage Registers are memory element for temporary storage. Size of register indicates how much information the processor can operate at one time, 16, 32, 64 bit.Buses : data buses indicate information moving capability 16,32,64, address buses indicate ram size or memory handling capability; 8086 220= 1M.II.Memory Organization: MAR,MBR,CacheIII. I/O Organization: Programmed I/O, Interrupt driven I/O, DMA
1.2 Data Representation
Registers contain both data and control information for data manipulation. Data are numbers and other binary coded information. Common types of data indicate Data Types.
Data Types
1. Numbers2.Letters or alphabets3.symbols
Data are represented in registers in binary coded form. Registers are made of flip-flops have two states which are capable of storing one bit of information. This change with the arrival of a clock pulse.1.2.1Number systems
Radix or base: 2, 10, 8, 16, rBinary coded octal numbers
----TABLE OF BCD,HEXA,OCTAL CAN BE KEPT
Number system
ASCII
Character Code in microcomputers Communication
Seven Bit Scheme
27 = 128 Unique code
Extended ASCII = 2 8 = 256
Information is Stored in the form of Bits for two voltages or magnetic States
A à 01000001 à 65
B à 01000010 à 66
----- HERE ASCII TABLE CAN BE KEPT------ BINARY Machine language Two Digits 0 and 1 Compatible with digital electronic Circuits
BITSON - True - 0
Off – False – 1
Binary Notation ……… 16s 8s 4s 2s 1s
1. ( 1101)2 = ( ? ) 10
1 1 0 1
8 4 2 1 =
+ + + = 13
2. 1 0 1 0
8 4 2 1 = 10
+ + +
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 1
Binary Addition
1. 8 4 2 1
1 0 1 1 11
1 1 01 +13
------------- -------
24 ß 1 1 0 0 0 24
-----------------------------------
2. 1 1 0 1
1 1 1 0
3. 1 0 1 1
0 0 1 1
4. 1 1 1 0
1 1 1 1
Binary Multiply
1. 1 0 0 1 9
x 1 1 1 0 x 14
______________ _____________
0 0 0 0 ( 126)10
1 0 0 1
1 0 0 1
1 0 0 1
­­­­­­­­­­­­­­­­­_______________
( 1111110) 2 = (126)10
_________________
2. 1 0 1 1 11
1 1 0 0 12
_________
132
_____________
Decimal = 10 hexa = 6
Hexadecimal
6 + 10 = 16
0 1 2 3 4 5 6 7 8 9 A B C D E F
Integer Representation
1 0 1 0 1 0 1 0 = 170
Unsigned Representation
1 0 1 0 1 0 1 0 = -42
Signed
MS B = 1 à Negative
= 0 à Positive
complements
Simplifying the subtraction and logical manipulation
Allows to represent positive and negative numbers.
Deal with subtraction using addition
10s complement
9s complement
1s complement
2s Complement
base or radix ==== r
r’s complement
r-1’s complement
r=10
10’s complement
10-1= 9’s complement
9’s complement
N Base r-1 r – 1’s Complement of N
N r r-1 (rn - 1) - N
10 9 (10n – 1) – N where 10n – 1 = n 9s
n=4 , 10000 – 1 = 9999
546700 999999 – 546700= 453299
10’s complement is 9’s complement + 1
r=2
2’s complement
2-1=1’s complement
1’s complement
16 = 10000
24 - 1 = 1111 subtraction of binary is done thru
2’s complement is 1’s complement + 1
Base Conversion Table
======= ADDITION, MULTIPLICATION TABLE CAN BE KEPT HERE =====
Binary To Decimal
1. ( 1 0 0 1 0) 2 = ( ? )10
2 4 x 1 + 0 + 0 + 2 1 x 1 + 20 x 0 =
16 +2 + 0 = ( 18) 10
2. ( 1 0 1 1 0 ) 2 = ( ?) 10
2 0 x 0 + 21x1 +22 x 1 + 23 x 0 + 24 x 1
= 0 + 2 +4 + 0+ 16 = (22) 10
3. ( 1 1 0 1 0 )2 =
4. ( 1 1 1 1 0 )
5. ( 1 0 0 0 1 1 1 1) 2
6. ( 0 1 0 1 0 0 0 1 ) 2
7. ( 0 1 1 1 0 0 1 1) 2
Binary to Hexadecimal
( 1 1 0 11 0) 2 = ( ? ) 16
Group digits in set of four, begining at the right
11 0110
____ __________ = (36) 16
3 6
___________________
2. ( 1 0 11 11) 2 = 10 1 1 1 1
____ _____________
2 15
= ( 2 F) 16
3. ( 10 111 0) 2 = ( ?) 16
4. (11 0 0 11 0) 2
5. ( 11 00 10) 2
6. (11 01 11) 2
7. ( 00 1100) 2
8. ( 111010) 2
9. ( 111111) 2
10. (1001011)2
Decimal to Binary
Divide the number by 2
( 24 ) 10 = ( ?) 2
2 24 0
2 12 0
2 16 0
2 3 1
2 1 1
0 ( 1 1 0 0 0 ) 2
1 .( 48) 10 =
(113) 10 =
(127) 10 =
Decimal to Hexadecimal
Divide the number by the base Number
1. ( 1 6 9) 10
16 169
16 10 9
0 10
( A 9 ) 16
2. ( 95 ) 10 =
3. ( 46) 10 =
4. ( 3315.3) 10 = ( ? ) 16
Hexadecimal to Binary
Convert each digital to its four digit binary equivalent and record from left to right
1. ( A C ) 16 = A C
1010 1100
Hexa To Decimal
1. ( A9) 16 = ( ? ) 10
A X 16 1 + 16 0 x 9
= 16 x 10 +9
= (169) 10
2. ( A F) 16
3. (4 B3.3) 16
4. CF3.4CC
Octal to Decimal
Sol – B2D
1. ( 234.14) 8 = ( ? ) 10
3( 26) 10
( 30) 10
(143) 10
(81) 10
(114) 10
Sol – B2H
3. ( 2E) 16
4. (66) 16
5. (32) 16
6. (37)16
7. (OD) 16
8. (3A) 16
9. ( 7F) 16
10. (4B) 16
Sol – D2 B
1. ( 1 1 0 0 0 0) 2
2. ( 1 1 1 0 0 0 1 ) 2
3. ( 1 1 1 1 1 1 1) 2
2. ( 5F) 16
3. ( 2E) 16
H 2 D
2. ( 175 ) 10
3. 1203 .1875
Octal 2 D
1. 156 .1875