google.com, pub-4617457846989927, DIRECT, f08c47fec0942fa0 Learn to enjoy every minute of your life.Only I can change my life.: April 2014

Sunday, April 20, 2014

Solve more questions on networking -Computer Science

Networking

1. The network number plays what part in an IP address?
A. It specifies the network to which the host belongs.
B. It specifies the identity of the computer on the network.
C. It specifies which node on the sub-network is being addressed.
D. It specifies which networks the device can communicate with.

Answer :- A. It specifies the network to which the host belongs.



 2. What is the decimal equivalent to the binary number 101101?
A. 32
B. 35
C. 45
D. 44

Answer :-  C. 45



3. Convert the following decimal number to its binary form: 192.5.34. 11.

A. 11000000.00000101.00100010.00001011
B. 11000101.01010111.00011000.10111000
C. 01001011.10010011.00111001.00110111
D. 11000000.00001010.01000010.00001011


Answer:- A. 11000000.00000101.00100010.00001011



 4. Convert the following binary IP address to its decimal form: 11000000.00000101.00100010.00001011 A. 190.4.34.11
B. 192.4.34.10
C. 192.4.32.11
D. None of the above

Answer:- D. None of the above



 5. What portion of the following Class B address is the network address: 154.19.2.7?
A. 154
B. 154.19
C. 154.19.2
D. 154.19.2.7

Answer:- B. 154.19



6. Which portion of the IP address 129.219.51.18 represents the network?
A. 129.219
B. 129
C. 51.18
D. 18

Answer:-  A. 129.219



7. Which address is an example of a broadcast address on the network 123.10.0.0 with a subnet mask of 255.255.0.0?
A. 123.255.255.255
B. 123.10.255.255
C. 123.13.0.0
D. 123.1.1.1

Answer:-   B. 123.10.255.255



8. MAC address are _________ bits in length.
A. 12
B. 24
C. 48
D. 64


Answer :- C.  48



9. Where does the MAC address reside?

A. Transceiver
B. Computer BIOS
C. NIC
D. CMOS

Answer :- C.  NIC



10.Convert the decimal number 2989 to hex.
A. FDD1
B. BAD
C. TED
D. CAD

Answer  :- B.  BAD


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Solve questions on Networking


1. How many bits are in an IP address?

A. 16 bits
B. 32 bits
C. 64 bits
D. None of the above

Answer :- B.32 bits

2. What is the maximum value of each octet in an IP address?
A. 128
B. 255
C. 256
D. None of the above

Answer :- B. 255

3. How many bits are in a subnet mask?
A. 16 bits
B. 32 bits
C. 64 bits
D. None of the above

Answer :- B. 32 bits

4. A half-duplex circuit means
A. Only one side can talk at a time
B. The signal strength is cut in half
C. The signal strength is doubled
D. Two hosts can talk simultaneously

Answer :- A. Only one side can talk at a time

5. Attenuation means
A. Travel
B. Delay
C. A signal losing strength over distance
D. Loss of signal due to EMI

Answer :- C. A signal losing strength over distance

6. __________ means to convert binary data into a form that can travel on a physical communications link.
A. Encoding
B. Decoding
C. Encrypting
D. Decrypting

Answer :- A.Encoding

7. How many host addresses can be used in a Class C network?
A. 253
B. 254
C. 255
D. 256

Answer :- B. 254

8. What is the minimum number of bits that can be borrowed to form a subnet?
A. 1
B. 2
C. 4
D. None of the above

Answer :- B. 2

9. What is the primary reason for using subnets?
A. To reduce the size of the collision domain
B. To increase the number of host addresses
C. To reduce the size of the broadcast domain
D. None of the above

Answer :- C. To reduce the size of the broadcast domain

10.How many bits can be borrowed to create a subnet for a Class C network?
A. 2
B. 4
C. 6
D. None of the above

Answer :- C. 6

Study material of Computer Science :-


127.0.0.1 is is a special purpose IP address conventionally used as a computer's loopback address.

Definition Thrashing :-
In computer science, thrashing occurs when a computer's virtual memory subsystem is in a constant state of paging, rapidly exchanging data in memory for data on disk, to the exclusion of most application-level processing

Virtual Memory:-
In computing, virtual memory is a memory management technique that is implemented using both hardware and software. It maps memory addresses used by a program, called virtual addresses, into physical addresses in computer memory. Main storage as seen by a process or task appears as a contiguous address space or collection of contiguous segments.

The time taken to move the disk arm to the desired cylinder is called the :seek time

Class A
0    . 0   . 0   . 0      =  00000000.00000000.00000000.00000000
127.255.255.255   =  01111111.11111111.11111111.11111111
                                  0nnnnnnn.HHHHHHHH.HHHHHHHH.HHHHHHHH
Class B
128. 0. 0. 0          = 10000000.00000000.00000000.00000000
191.255.255.255 = 10111111.11111111.11111111.11111111
                               10nnnnnn.nnnnnnnn.HHHHHHHH.HHHHHHHH
Class C
192. 0. 0. 0          = 11000000.00000000.00000000.00000000
223.255.255.255 = 11011111.11111111.11111111.11111111
                               110nnnnn.nnnnnnnn.nnnnnnnn.HHHHHHHH
Class D
224. 0. 0. 0          = 11100000.00000000.00000000.00000000
239.255.255.255 = 11101111.11111111.11111111.11111111
                               1110XXXX.XXXXXXXX.XXXXXXXX.XXXXXXXX
Class E
240. 0. 0. 0          = 11110000.00000000.00000000.00000000
255.255.255.255 = 11111111.11111111.11111111.11111111
                               1111XXXX.XXXXXXXX.XXXXXXXX.XXXXXXXX
Where:
n indicates a binary slot used for network ID.
H indicates a binary slot used for host ID.
X indicates a binary slot (without specified purpose)

190.255.254.254 -class B

Friday, April 18, 2014

Solve Practice questions of Computer Science


1. Decryption & Encryption of data are the responsibility of the ___ layer.

(A) Physical
(B) Data link
(C) Presentation
(D) Session
(E) Application

Answer :- (C) Presentation


2. LANs can be connected by devices called ____ which operate in the data link layer?

(A) Hub
(B) Bridges
(C) HDLC
(D) Tunnel
(E) None of these

Answer :- (B) Bridges


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Thursday, April 17, 2014

Multiplexers

Multiplexer :-
                        It is known as MUX..
                        Multiplexing is the process of connecting many inputs to one output it is "many into one " process with appropriate logic control signals.
                        Multiplexer is a basic block or basic logic combinational circuit with
i) 'n' input lines
ii) 'm' control lines / selector lines
iii) one output line 'Y'

    It requires OR gate in the O/P.
  
   It cannot be used as a decoder.

   Example :- IC 74153,74157
 To calculate Select line  -  Formula :-         2m = n
where, n is the input lines
            m is the control lines

4:1 Multiplexer :
Truth table :-

S1        S0        Y =      
00D0
01D1
10D2
11D3

It has 4 input lines (D0 ,D1,D2,D3) two control lines (S1 S0) called  select lines and one output line by 'Y'
Working:-
1) When the control word is S1 and S0 is 0 0 then Y = D0
2) When the control word is S1 and S0 is 0 1 then Y = D1
3) When the control word is S1 and S0 is 1 0 then Y = D2
4) When the control word is S1 and S0 is 1 1 then Y = D3

Note that at any instant only one AND is enabled allowing only one data line transmission to the common output line.

Wednesday, April 16, 2014

Solve questions


1.In relational database row and column is called as ____and ____respectively.

(A)Tuple; Attribute
(B)Attribute; Tuple
(C)Tupple;Domain
(D)Attribute; Domain
(E)None of these

Answer :- (A)Tuple; Attribute

2.Arrange the following in increasing order of their size.

 (A) Database<File<Record<Field<Byte<Bit
 (B) Bit<Byte<Field<Record<File<Database
 (C) Bit>Byte>Record>Field>File>Database
 (D) Bit>Byte>File>Record>Field>Database


Answer :-   (B) Bit<Byte<Field<Record<File<Database

3. Round robin scheduling is essentially the preemptive version of

(A) FIFO
(B)Shortest job first
(C) Shortest remaining
(D)Longest time first
(E) None of these

Answer :- (A) FIFO

4. Which of the following identifies specific web page and its computer on the web page?

(A) Web site
(B) URL
(C) Web site address
(D) Domain Name
(E) None of these

Answer :- (B) URL

5.Communication handler is basically an ____ 16 bit micro controller?

(A) 8085
(B) 8086
(C) 8086A
(D) 80C186
(E) None of these

Answer :- (D) 80C186

6.GSM stands for ____

(A) General System for Mobile Communication
(B)Global System for Mobile Communication
(C) Guided System for Mobile Communication
(D) Growing System for Mobile Communication
(E) None of these

Answer :- (B) Global System for Mobile Communication

7.EDI refers to ----

(A) Electronic Data Interface
(B) Electric Device Interface
(C) Electronic Data Interchange
(D)Electronic Design Interface

Answer :- (C) Electronic Data Interchange

8.PAN refers to_____

(A) Peer Area Network
(B) Personnel Area Network
(C) Personal Area Network
(D) Payment Application Network

Answer :- (C) Personal Area Network

9.E-Commerce refers to ___

(A)Electrical Commerce
(B) Electronic Commerce
(C)Evolutionary Commerce
(D) Effective Commerce

Answer :- (B) Electronic Commerce

10.Which of the following is a type of E-commerce?

(A) B2B  (Business to Business)
(B) B2C  (Business to Consumer)
(C) C2C  (Consumer to Consumer)
(D) All the above

Answer :- (D) All the above


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Sunday, April 13, 2014

Solve question on C++

1) Object is ___
(A) A concept
(B) Abstraction
(C) Thing
(D) All of these

Answer :- (D) All of these

2) Each object is said to be ____ of its class.
(A) Attribute
(B) Instance
(C) Classification
(D) Operations

Answer :- Instance

3) Person is ____
(A) Object
(B) Class
(C) Attribute
(D) Structure

Answer :-  (B) Class

4) A class that serves only a base class from which classes are derived?
(A) Base class
(B) Class
(C) Abstract class
(D) Intermediate class

Answer :- (C) Abstract class

5) In C++ a function contained within a class is called a___
(A) In built function
(B) User defined function
(C) Member function
(D) None of these

Answer :- (C) Member function

6) In C++ , setw is ____
(A) Initialization operator
(B) Conditional operator
(C) Memory allocation operator
(D) Field width operator

Answer :- (C) Member function

7) Which of the following C++  statement is correct?
(A) int *p = new int
(B) int new = *p
(C) int *new = *p
(D) int *p = int new

Answer :- (A)  int *p = new int

8) In C++ by default ,the members of a class are ___
(A) Public
(B)  Private
(C) Both public and private
(D) None of these

Answer :- (B) Private

9) Which of the following operators can be overloaded in C++?
(A) Scope resolution operator
(B) Size of operator
(C) Conditional operator
(D) Arithmetic operator

Answer :- (D) Arithmetic operator

10) Which of the following is different from the group ?
(A) Private
(B) Protected
(C) Public
(D) Friend

Answer :- (D) Friend

11) Which of the following is a type of inheritance ?
(A) Multiple
(B) Multi level
(C) Hierarchical
(D) All of these

Answer : - (D) All of these

12) C++ was developed by ___
(A) Dennis Ritchie
(B) Bjarne Stroustrup
(C) Tannenbaum
(D) Milan Melkovinec

Answer :-  (B) Bjarne Stroustrup

13) A ____  variable provides an alias for a previously defined variable
(A) Global
(B) Local
(C) Refrence
(D) Dynamic

Answer :- (C) Refrence

14) Which statement is true in context of virtual functions?
(A) The virtual functions must be members of some class
(B) Virtual functions cannot be static members
(C) A virtual function can be friend of another class
(D) All of the above

Answer :- (D) All of the above

15) In C++ we can use same function name to create functions that perform a variety of different tasks.This is known as ____.
(A) Global function
(B) Function overloadiing
(C) Inheritance
(D) All of the above

Answer :- (B) Function overloading

16) A template can be used to create a family of ___
(A) Classes
(B) Funtions
(C) Class and function
(D) None of these

Answer :- (C) Class and function

17) Encapsulation is known as ___
(A) Information hiding
(B) Data abstraction
(C) Data about data
(D) None of these

Answer :- (A) Information hiding


18) The objects are __ if all their attribute values are identical.
(A) Identical
(B) Distinct
(C) Same
(D) None

Answer :- (B) Distinct

19) A type of inheritance that permits a class to have more than one super class and to inherit features from all ancestors ___
(A) Inheritance
(B) Multiple inheritance
(C) Subclasses
(D) Super classes

Answer :- (B) Multiple inheritance

20) Actions are performed in an object oriented systems by sending ______ to an object.
(A) Statements
(B) Messages
(C) Control
(D) None of these

Answer  :- (B) Messages

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Solve Computer Arithmetic -

1) The basic logic operation is _____.
(A) AND
(B) NOR
(C) NAND
(D) NOR

Answer :- (A) AND


2) Which of the following belongs to the logic families of integrated circuits?
(A) Transistor Transistor Logic (TTL)
(B) Metal Oxide Semiconductor (MOS)
(C) Complementary Metal Oxide Semiconductor (CMOS)
(D) All of the above

Answer :- (D) All of the above


3) _________ is preferable in system requiring low power consumption.
(A) Emitter Coupled Logic (ECL)
(B) Metal Oxide Semiconductor (MOS)
(C) Complementary Metal Oxide Semiconductor (CMOS)
(D) Transistor Transistor Logic (TTL)

Answer :- (C)  Complementary Metal Oxide Semiconductor (CMOS)


4) T flip-flop means _______
(A) Timed flip-flop
(B) Toggle flip-flop
(C) Tackle flip-flop
(D) Test flip-flop

Answer  :- (B) Toggle flip-flop


5) Half adder is an example of _____
(A) Combinational circuit
(B) Sequential circuit
(C) Asynchronous circuit
(D) None of these

Answer :- (A) Combinational circuit


6) Full adder is used to _____
(A) 2 bit addition
(B) 3 bit addition
(C) 4 bit addition
(D) Addition of unlimited number of bits

Answer :- (B) 3 bit addition

7) Which of the following flip-flop is different ,in context of their outputs?
(A) SR flip-flop
(B) D flip-flop
(C) JK flip-flop
(D) None of these

Answer :- (C) JK flip-flop


8) Which of the following is not a number system ?
(A) Hexadecimal
(B) Octal
(C) Radix
(D) Binary

Answer :- (C) Radix


9) Which of the following is not a combinational circuit ?
(A) Encoders
(B) Decoders
(C) Registers
(D) Multiplexers

Answer :- (C) Registers


10) Which of the following belongs to DeMorgan's theorems?
(A) x + x =1
(B) xy = yx
(C) (xy)' = x'  +  y'
(D) x . 0 = 0

Answer :- (C) (xy)' = x'  +  y'


11) An encoder has 2n input lines and ____ output lines
(A) 2
(B) n
(C) 2 * n
(D) n * n

Answer :- (B) n


12) There are occasions when it does not matter if the function produces 0 or 1 for a given minterm.
Minterms that may produce either 0 or 1 for the function are said to be ____
(A) Careless conditions
(B) Don't care condition
(C) Minterm condition
(D) Maxterm condition

Answer :- (B) Don't care condition


13) ASCII code for alphabetic character requires ____ bits
(A) 16
(B) 15
(C) 8
(D) 7

Answer :- (D) 7


14) The number that cannot be normalized, in floating point ___
(A) 1
(B) 0
(C) Both (A) and (B)
(D) None of these


Answer :- (B) 0


15) The 3-bit operation code for ADD operation is 001 and indirect memory address is 23 then 16-bit instruction code can be written as ____
(A) 00001000000010111
(B) 1001000000010111
(C) 10001000000010111
(D) None of the above

Answer :- (A) 00001000000010111


16) Which of following is/are application(s) of flip-flop?
(A) Bounce elimination switch
(B) Latch
(C) Counters
(D) All of the above

Answer :- (D) All of the above

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Wednesday, April 9, 2014

Solve

Solve


Directions: In each Q1 to Q3 of the following questions, there are five letter groups or words in each question. Four of these letter groups or words are alike in some way, while one is different. Find the one which is different.
Q.1.    (1) black    (2) red    (3) green    (4) paint    (5) yellow
Answer:- (4) paint

Q.2.    (1) BC    (2) MN    (3) PQ    (4) XZ    (5) ST
Answer :-(4)XZ

Q.3.    (1) Mango    (2)Apple    (3) Orange    (4) Guava    (5) Rose
Answer :- (5) Rose


Directions : In each of the following questions, there is a question mark in which only one of the five alternatives given under the question satisfies the same relationship as is found between the two terms to the left of the sign :: given in the question. Find the correct answer

Q.4.Foot : man : : hoof : ?
(1) leg    (2) dog    (3) horse    (4) boy    (5) shoe
Answer :- (3) horse

Q.5.Day : Night : : Kind : ?
(1) Dark    (2) Bright    (3) Cruel    (4) Generous    (5) Gratitude
Answer :-(3) Cruel

Q.6.Hut : mansion : : Rabbit : ?
(1) Hole    (2) Carrot    (3) Elephant    (4) Small    (5) Rat
Answer :- (3) Elephant


Now try the following questions.
Q.7.If the letters in the word TOPS can be rearranged to form a meaningful word beginning with O, the last
letter of that word is your answer. If more than one such word can be formed, M is the answer and if no
such word can be formed, X is the answer.
(1) T    (2) P    (3) M    (4) S    (5) X
Answer :- (4) S

Q.8.‘Some leaders are dishonest. Satyapriya is a leader.’
Which of the following inferences definitely
follows from these statements ?
(1) Satyapriya is honest
(2) Satyapriya is dishonest
(3) Some leaders are honest
(4) Leaders are generally dishonest
(5) Satyapriya is sometimes
Answer :- (3) Some leaders are honest


Directions :
In each of the following questions, you have to find out what will come in place of the question
mark (?).
Q.9. 42 + 73 + 137 = ?
(1) 352 (2) 252
(3) 242
(4) 142
(5) None of these
Answer :-(2) 252

Q.10.  20 ×1/2= ?
(1) 4
(2) 5
(3) 12
(4) 20
(5) None of these
Answer :-(5) None of these

Q.11.  0.7 x 0.5 = ?
(1) 35
(2) 0.35
(3) 0.0035
(4) 0.035
(5) None of these
Answer :-(2) 0.35

Q.12.At 10 rupees each, how many rupees will 6 lemons cost ?
(1) 6
(2) 10 (3) 60 (4) 61
(5) 610
Answer :- (3) 60

Q.13.Which of the following can be exact multiple of 4 ?
(1) 27114
(2) 58204
(3) 48402
(4) 32286
(5) None of these
Answer:- (2) 58204


Q.14.If the profit made by selling a pen for Rs.10 is as much as its cost, what is the cost price of the pen ?
(1) Rs.3/-
(2) Rs.5/-
(3) Rs.10/-
(4) Rs.20/-
(5) None of these
Answer:- (2) Rs.5/-


Directions :
In each of the following questions, select from amongst the five alternatives, the word nearest in meaning
to the word given in capitals.
Q.15.LETHAL
(1) light
(2) dangerous
(3) deadly
(4) cruel
(5) thoughtless
Answer:-(3) deadly

Q.16.CENTENARY
(1) a guard
(2) a hundred years
(3) a very old man
(4) hundred runs
(5) hundredth anniversary
Answer:-(5) hundredth anniversary

Q.17.TRIUMPH
(1) conquer
(2) smash
(3) earn
(4) brave
(5) capture
Answer:-(1) conquer


Directions :
In each of the following questions, select from amongst the five alternatives, the word most opposite in
meaning of the word given in capitals.

Q.18.LIVELY
(1) simple
(2) weak
(3) dull
(4) angry
(5) moron
Answer:-(3) dull

Q.19.INADVERTENT
(1) adequate
(2) available
(3) sluggish
(4) negligent
(5) intentional
Answer:-(5) intentional

Q.20.INEPT
(1) accurate
(2) skilful
(3) sensible
(4) artistic
(5) apt
Answer:-(2) skilful

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Tuesday, April 8, 2014

COCOMO

COCOMO stands for COnstructive COst MOdel

COCOMO ,COnstrustive COst MOdel is static single variable model .
Barry Bohem introduced COCOMO models.
There is a hierarchical of these models.

Model 1:
Basic COCOMO model is static single-valued model that computers software development effort (and cost)
as a function of program size expressed in estimate lines of code.

Model 2 :
Intermediate COCOMO model computers software development effort as a function of program size and a set of "cost drivers"  that include subjective assessments of product ,hardware ,personnel, project attributes.

Model 3:
Advanced COCOMO model incorporates  all characteristics of the intermediate version with an assessment of the cost driver's impact on each step,like analysis ,design ,etc.

COCOMO can be applied to the following software project's categories.
Organic mode
Embedded mode

 Problem :

KLOC =10.9
E = ab(KLOC)exp(bb)
    =2.4 (10.9)exp(1.05)
    =29.5 person-month
D = cb(E)exp(db)
     =2.5 (29.5) exp (.38)
    = 9.04 months
where ,
D is the chronological month
KLOC is the estimated number of delivered lines (expressed in thousands ) of code for project


The intermediate COCOMO model takes the following form ;

E = ai(LOC)exp(bi) X EAF
where ,
 E is the effort applied in person-month
LOC is the estimated number of delivered lines of code for the project.

The coefficient ai and the exponent bi are :
Software project      ai          bi
Organic                   3.2        1.05
Semi-detached        3.0       1.12
Embedded              2.8       1.20

(cocomo to study online)

for loop - syntax and explaination

The for loop :-

Syntax:
                   for (initialization;condition ;increase)
                          {         ---------;
                                    statement;
                           }

and its main function is to repeat statement while condition remains true , like the while loop .
But in addition ,for provides places to specify an initialization instruction and an increase instruction .
The three expression inside for loop have to be separated by semicolon. 
So this loop is specially designed to perform a repetitive action with a counter.

It works the following way :

1. Initialization is executed .Generally it is an initial value setting for a counter variable .This is executed only once.
2. Condition is checked ,if it is true the loop continues, otherwise the loop finishes and statement is skipped.
3. Statement is executed . As usual , it can be either single instruction or a block of instruction enclosed within curly brackets{}.

Different forms of for loop: -
1.  single statement:
for(i=0;i<10;i++)
statement;

2. compound statement:
for(i=0;i<10;i++)
{
         ----------;
         statement;
         ----------;
         ----------;
}

3. loop with no body

for(i=0;i<10;i++)
;
      or
for(i=0;i<10;i++);

4.Multiple initialization and multiple updates separated by comma :-
 for(i=0;j=0;i<10;i++;j++)
statement;

5.for (;i<10;i++)

6.for(;i<10;)

7.for(;;)
statement;

Nesting for statement :-
         One for statement can be written within another for statement .This is called nesting of for statement
for(i=0;i<25;i++)
{
    ------------;
    statement;
      for(j=0;j<10;j++)
         {
                statement;
          }
    ---------;
}
 here for every value of i the inner loop will be executed 10 times.

The for loop is very flexible powerful and most commonly used loop.
It is useful when the number of repetition is known in advanced.

Monday, April 7, 2014

ACID properties of transaction

Transaction :-
                      A transaction is collection of operations that performs a single logical function in a database application that may lead to success or failure in operation maintaining integrity.

ACID properties of transaction :-

Atomicity :-
Transaction must be either performed full or not at all.
It will run to completion as as individual unit ,at the end of which either no change have
occurred to database or database has been changed in consistent manner .
At the end of transaction update it will be accessible to all.

Consistency :-
A correct execution of transaction must take DB from one consistent state to another,
programmer is responsible for this.
This property implies that if database was in consistent state before start of transaction ,then
at any time of termination too it must be consistent.

Independence:-
Transaction must not make its update visible to other transactions.
All actions performed by it must be independent.
This property gives transaction a measure of relative independence.

Durability:-
Once changes are committed must never be lost .
And any updating in database may not loss updates made by transaction.

Sunday, April 6, 2014

SOLVE QUESTION ANSWERS ON OPERATING SYSTEM .


1.One can interface with operating system by means of -------
A) Operating system call in a program
B) Operating system commands
C) Operating system process
D) Both by operating system call and operating system commands

Answer :-  D) Both by operating system call and operating system commands


2. Which of the following is not type of processing ?
A) Serial
B) Network
C) Batch
D) Multiprogramming

Answer :- B) Network


3. Kernel is _____
A) A part of operating system
B) An operating system
C) A hardware
D) A register

Answer :-  A) A part of operating system



4. UNIX operating system is based on ______
A) Language structure
B) Kernel approach
C) Virtual machine
D) Time sharing

Answer :-   B) Kernel approach


5. A transition between two memory resident process in a memory resident process in amultiprograming system is called ______
A) Process switch
B) Mode switch
C) Transition switch
D) None of these

Answer :-    A) Process switch


6. Round robin scheduling algorithm falls under category of _____
A) Preemptive scheduling
B) Non- preemptive scheduling
C) Sometime preemptive sometime non-preemptive
D) None of tehse
Answer :-  A) Preemptive scheduling


7. Paging is  a _______
A) Virtual memory
B) Memory management scheme
C) Allocation of memory
D) Deadlock prevention scheme

Answer  :-    B) Memory management scheme


8. The first version of UNIX was written by _____
A) Dennis Ritchie
B) Andrew S
C) Ken Thompson
D) None of these


Answer :- C) Ken Thompson


9. Chaining and Indexing are the strategies of ______
A) Contiguous allocation
B) Non-contiguous allocation
C) Partition allocation
D) Static allocation

Answer :-  B) Non-contiguous allocation



10. Wasting of memory within a partition ,due to a difference in size of a partition and of the object resident within it is called _____
A) External fragmentation
B) Internal fragmentation
C) Compaction
D) Coalescing

Answer :-   B) Internal fragmentation


11.Which of the following is an operating system call ?
A) CREATE
B) LINK
C) SYSTEM
D) All of these

Answer :-    D) All of these


12.The primary job as the operating  system of a computer is to
A) command resources
B) manage resources
C) provide utilities
D) be use friendly

Answer : -  B) manage resources



13.Dirty bit for a page in a page table
A) helps avoid unnecessary writes on a paging device
B) helps maintain LRU information
C) allows only read on a page
D) none of these

Answer : -   A) helps avoid unnecessary writes on a paging device

14. The term "operating system " means
A) a set of programme which controls computer working
B) the way a computer operator works
C) conversion of high language into machine code
D)  the way a floppy disk drive operates

Answer :-  A) a set of programme which controls computer working


15. Windows is a/an
A) operating system
B) user interface
C) operating environment
D) programming platform

Answer :-  C) operating environment


16. When did IBM released teh first version of its Disk Operating System (DOS) version 1.0?
A) 1981
B) 1982
C) 1983
D) 1984


Answer :- A) 1981



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Network Topology

  • Compute network:-
Goals:-
1.Resource sharing .
2.Provide High Reliability
3.Distribution of workload
4.Saving Money
5.Expandability
6.Powerful communication Medium.
7.Protecting Information.
8.Preserving Information.

Applications:-
1) Business Applications
2) Home Applications

Network architecture describes how a computer network is configured and what strategies are employed .
Configurations include star, bus, ring, mesh and hierarchical. Strategies include terminal, peer-to-peer and  client-sever.

  • Network Hardware :-
Broadcasting :-
1.Addressing a packet to all destinations.
2.Received by every machine on network.
3.Multicasting

Point to point:-
1.Network connections between  individual pairs.
2.Packet may visit one more intermediate machines
3.Uni-casting.

A network can be arranged or configured in several different ways.This arrangement is called the network's topology.

Network topologies:-  Physical topologies -

MESH topology:-
1.Interconnected topology.
2.Point to point link.
3.n(n-1)/2 links for n devices.
4.Direct communication
5.Highest degree of fault tolerance


Advantages:
1.Traffic problems eliminated .
2.Robust
3.Fault identification and fault isolation easy.
4.Privacy.
5.Highest  reliability and security.

Disadvantages:
1Amount of cabling
2.Number of I/O ports.
3.Reconnection is difficult.
4.Installation of new devices is difficult.
5.Expensive.



STAR topology:-
1.Point to point link.
2.Hub central controller.
3.Communication through controller.
4.Controller exchange data.



Advantages :
1.Requires only one link and one I/O port
2.Less expensive.
3.Easy to install.
4.Robust.
5.Fault identification and isolation easy.

Disadvantages:
Performance depends on hub.
Therefore if hub fails.
So,Network stops
Since speed depends on hub connections.

In a star network,a number of small computers or peripheral devices are linked to a central unit.This central unit may be a host computer or a file server.
All communications pass through this central unit 
Control is maintained by polling. That is ,each connecting device is asked (polled) whether it has a message to send .Each device is then in turn allowed to send its message.
One particular advantage of the star form of network is that it can be used to provide a time-sharing system. That is , several users can share resources (time) on a central computer.
The star is a common arrangement for linking several microcomputers to a mainframe that allows access to an organization's database.

BUS Topology:-

1.Linked by single cable i.e. multi point topology.
2.Devices have Drop lines and taps.
3.Used in LAN.


Advantages:-
1.Easy t install.
2.Cabling requires less compare to mesh and star.

Disadvantages:
1.Limited number of devices support .
2.Difficult to identify and isolate fault.
3.Degradation in quality of signal.
4.Bus cable fault ,stops all transmission.

In bus network each device in the network handles its own communications control.There is no host computer.
All communications travel along a common connecting cable called a bus.
As the information passes along the bus,it is examined by each device to see if the information is intended for it.
The bus network is typically used when only a few microcomputers are to be linked together.This arrangement is common for sharing data stored on different microcomputers.
The bus network is not as efficient as the star network for sharing common resources.
However , a bus network is less expensive and is in very common use.

RING topology:-

1.Connected to two devices .
2.Point to point link dedicated .
3.Signal travels in one direction.
4.Device relay ,if not intended.



Advantages:
1.Easy to install or reconfigure.
2.Fault isolation is easy.
3.Token

Disadvantages:-
1.Ring break ,network disable.
2.Number of devices connected and ring length.
3.Unidirectional ,data transfer is slow.

In ring network, each device is connected to two other devices , forming a ring .
There is no central file or computer.Messages are passed around the ring until they reach the correct destination.
With microcomputers, the ring arrangement is the least frequently used of the other networks .
However ,it often is used to link mainframes ,especially over wide geographical areas. These mainframes tend to operate fairly autonomously.
They perform most or all of their  own processing and only occasionally share data and programs with other mainframes.



HYBRID topology:-

1.Combination of topologies,not basic topology.
2.Two different basic topologies connection.
3.WAN have it.


Computer network - OSI and TCP/IP

Comparison between OSI and TCP/IP :-

OSI TCP/IP
1. Open Systems Interconnection 1. Transfer Control Protocol / Internet protocaol
2.Number of layers are seven (7). 2.Number of layers are four (4).
3.Connection oriented and connectionless
in network and only connection oriented in transport.
3.Only connectionless in Internet and both in Transport.
4. Distinction
1.Services
2.Interfaces
3.Protocols
4.Does not distinction between them.
5.Model first developed.
Therefore, Not biased toward one protocol general model.
5.Protocol first developed
Therefore, does not fit any other protocol stack.
6.No thought was given to inter-networking 6.Main goal to handle inter-networking.
7.Model useful,Protocols not popular. 7.Model not widely accepted ,Protocols universally used.
8.S and P layers empty.
D and N layers overfull.
Many function addressing ,flow control and error control reappear.
8.No distinguish nor mention P and D layer even have functionalities.

Computer network - Service

 Service :-
Service
_______________________|___________________
|                                                                                 |
                   Connection Oriented                                                       Connectionless
                   _______|_________                            ___________________|__________
                   |                              |                         |                              |                        |
Reliable connection       Unreliable connection      Unreliable        Acknowledgement      Request-
oriented service                                                    data-gram            data-gram                   reply
      ______|__________
      |                               |
Message stream         Byte stream


Difference between Connection oriented and Connectionless

          Connection-oriented                                                    Connectionless
1. It is similar to telephone system.                     1. This is modeled after the postal system.

2. Packets received are in same order as sent.    2. Messages arrive out of order.

3.Each service can be characterized by quality    3.Not all applications require reliable service
   of service-reliable or unreliable.                           and not all require connection oriented
                                                                              communication.
4.It has services:                                                 4.It has services: -
   i) Reliable message stream.                                  i) Unreliable data-gram
   ii) Reliable byte stream                                         ii)  Acknowledged data-gram
   iii) Unreliable connection.                                     iii) Request -reply

5. It has example:-                                               5.It has example
   i) Sequence of pages                                            i) Electronic junk mail.
   ii) Remote login.                                                   ii) Registered mail.
   iii) Digitized voice                                                 iii) Database query.



Services primitive :-
1) Set of primitives (operations)
2)Available to user process to access service.
3)depend on service.


Service primitive for implementing a simple connection oriented service:
     Primitive                                                           Meaning
1) LISTEN                -      Block waiting for an incoming connection.
2) CONNECT          -      Establish a connection with a waiting peer
3) RECEIVE             -      Block waiting for an incoming message
4) SEND                   -      Send a message to the peer.
5) DISCONNECT    -      Terminate a connection.

Saturday, April 5, 2014

Program in C++ for play-game TIC TAC TOE

#include <iostream.h>
#include <conio.h>

char square[10] = {'0','1','2','3','4','5','6','7','8','9'};
int testwinner();
void playchart();

int main()
{
    int player = 1,i,choice;
    char mark;
    clrscr();
    do
    {
        playchart();
        player=(player%2)?1:2;
        cout << "Player " << player << ", enter a number ==>  \a ";
        cin >> choice;
        mark=(player == 1) ? 'X' : 'O';
        if (choice == 1 && square[1] == '1')
            square[1] = mark;
        else if (choice == 2 && square[2] == '2')
            square[2] = mark;
        else if (choice == 3 && square[3] == '3')
            square[3] = mark;
        else if (choice == 4 && square[4] == '4')
            square[4] = mark;
        else if (choice == 5 && square[5] == '5')
            square[5] = mark;
        else if (choice == 6 && square[6] == '6')
            square[6] = mark;
        else if (choice == 7 && square[7] == '7')
            square[7] = mark;
        else if (choice == 8 && square[8] == '8')
            square[8] = mark;
        else if (choice == 9 && square[9] == '9')
            square[9] = mark;
        else
        {
            cout<<"Invalid move ";
            player--;
            getch();
        }
        i=testwinner();
        player++;
    }while(i==-1);
    playchart();
    if(i==1)
        cout<<"==>\aPlayer "<<--player<<" win ";
    else
        cout<<"==>\aGame draw";
    getch();
    return 0;
}

int testwinner()
{
    if (square[1] == square[2] && square[2] == square[3])
        return 1;
    else if (square[4] == square[5] && square[5] == square[6])
        return 1;
    else if (square[7] == square[8] && square[8] == square[9])
        return 1;
    else if (square[1] == square[4] && square[4] == square[7])
        return 1;
    else if (square[2] == square[5] && square[5] == square[8])
        return 1;
    else if (square[3] == square[6] && square[6] == square[9])
        return 1;
    else if (square[1] == square[5] && square[5] == square[9])
        return 1;
    else if (square[3] == square[5] && square[5] == square[7])
        return 1;
    else if (square[1] != '1' && square[2] != '2' && square[3] != '3' &&
             square[4] != '4' && square[5] != '5' && square[6] != '6' &&
            square[7] != '7' && square[8] != '8' && square[9] != '9')
        return 0;
    else
        return -1;
}

void playchart()
{
    clrscr();
    cout << "\n\n\tTic Tac Toe\n\n";
    cout << "Player 1 (X)  -  Player 2 (O)" << endl << endl;
    cout << endl;
    cout << "     |     |     " << endl;
    cout << "  " << square[1] << "  |  " << square[2] << "  |  " << square[3] << endl;
    cout << "_____|_____|_____" << endl;
    cout << "     |     |     " << endl;
    cout << "  " << square[4] << "  |  " << square[5] << "  |  " << square[6] << endl;
    cout << "_____|_____|_____" << endl;
    cout << "     |     |     " << endl;
    cout << "  " << square[7] << "  |  " << square[8] << "  |  " << square[9] << endl;
    cout << "     |     |     " << endl << endl;
}


Click here for more programs on C++

Program in C to sort all words of text in alphabatical order.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include<conio.h>
void sort_string(char*);
 int main()
{
   char string[100];
   clrscr();
   printf("Enter some text\n");
   gets(string);

   sort_string(string);
   printf("%s\n", string);
   getch();
   return 0;

}

void sort_string(char *s)
{
   int c, d = 0, length;
   char *pointer, *result, ch;

   length = strlen(s);

   result = (char*)malloc(length+1);

   pointer = s;

   for(ch='a';ch<='z';ch++ )
   {
      for(c=0;c<length;c++ )
      {
         if(*pointer == ch )
         {
            *(result+d) = *pointer;
            d++;
         }
         pointer++;
      }
      pointer = s;
   }
   *(result+d) = '\0';
   strcpy(s, result);
   free(result);
}
Click here for more programs on C++

Friday, April 4, 2014

Software testing -White box testing,Black box testing, Grey box testing

Software testing  :
Software testing is the process to uncover requirement, design and coding errors in the program.

Three most prevalent and commonly used software testing techniques for detecting errors, they are:
1)White box testing,
2)Black box testing and
3)Grey box testing.

1) White Box Testing Technique:
It is the detailed investigation of internal logic and structure of the code. In white box testing it is necessary
for a tester to have full knowledge of source code.

White box testing techniques are:-
i) Control Flow Testing: It is a structural testing strategy that uses the program control flow as a model
control flow and favours more but simpler paths over fewer but complicated path.
ii) Branch Testing: Branch testing has the objective to test every option (true or false) on every control
statement which also includes compound decision.
iii) Basis Path Testing: Basis path testing allows the test case designer to produce a logical complexity
measure of procedural design and then uses this measure as an approach for outlining a basic set of
execution paths.
iv) Data Flow Testing: In this type of testing the control flow graph is annoted with the information about how
the program variables are define and used.
v) Loop Testing: It exclusively focuses on the validity of loop construct.

Advantages:-
a)Side effects are beneficial.
b)It reveals error in hidden code by removing extra lines of code.

Disadvantages:-
a)It is very expensive.
b)Some of the codes omitted in the code could be missed out.





2) Black Box Testing Technique:
It is a technique of testing without having any knowledge of the internal working of the application. It only
examines the fundamental aspects of the system and has no or little relevance with the internal logical structure of the system.

Black box testing techniques are
i) Equivalence Partitioning: It can reduce the number of test cases, as it divides the input data of a software
unit into partition of data from which test cases can be derived.
ii) Boundary Value Analysis: It focuses more on testing at boundaries, or where the extreme boundary values
are chosen. It includes minimum, maximum, just inside/outside boundaries, error values and typical values.
iii) Fuzzing: Fuzz testing is used for finding implementation bugs, using malformed/semi-malformed data
injection in an automated or semi-automated session.
iv) Cause-Effect Graph: It is a testing technique, in which testing begins by creating a graph and establishing
the relation between the effect and its causes. Identity, negation, logic OR
and logic AND are the four basic symbols which expresses the interdependency between cause and effect.
v) Orthogonal Array Testing: OAT can be applied to problems in which the input domain is relatively small,
but too large to accommodate exhaustive testing.
vi) All Pair Testing: In all pair testing technique, test cases are designs to execute all possible discrete
combinations of each pair of input parameters. Its main objective is to have a set of test cases that covers all
the pairs.
vii) State Transition Testing: This type of testing is useful for testing state machine and also for navigation of
graphical user interface.

Advantages:-
a)Efficient for large code segment.
b)Programmer and tester are independent of each other.

Disadvantages:-
a)Inefficient testing.
b)Without clear specification test cases are difficult to design.





3) Grey Box Testing Technique:
 White box + Black box = Grey box, it is a technique to test the application with limited knowledge of the
internal working of an application and also has the knowledge of fundamental aspects of the system
The other name of grey box testing is translucent testing.

Grey box testing techniques are :
i) Orthogonal Array Testing: This type of testing use as subset of all possible combinations.
ii) Matrix Testing: In matrix testing the status report of the project is stated.
iii) Regression Testing: If new changes are made in software, regression testing implies running of test cases.
iv) Pattern Testing: Pattern testing verifies the good application for its architecture and design.

Advantages:-
a)The test is done from the user’s point of view rather than designer’s point of view.
b)Grey box testing provides combined benefits of white box and black box testing techniques

Disadvantages:-
a)Many program paths remain untested.
b)Test coverage is limited as the access to source code is not available.

Thursday, April 3, 2014

PCM - Pulse Code Modulation


PCM :- Pulse Code Modulation
It is the digitization process in which analog signal is represented in digital or discrete form .
It use ordinary analog signal which is digitized at the end office with the help of codec (coder decoder) codec produces 7 or 8 bit numbers .

Codec is inverse of modem -True
It code the digital to analog.
CO  DEC -COder (Analog to Digital) DECoder  (Digital to analog)

PCM is based Nyquest theorem:-
               There are various system for digital transmission:-
1.T1 carrier system (handle 24 voice channel)
2.Encoding system

There are three methods of PCM:
1.Differential pulse code modulation ( DPCM)
2.Delta Modulation
3.Predictive encoding

1.DPCM :-
Definition :It does not transmit actual sample .It transmit difference between current and previous value .Errors can ignored for voice or speech.

2.Delta modulation :
It is special care of DPCM it uses only one bit per sample .Each sample differs from its predecessor by either +1 or -1 .This is called as a sampling interval .
Granular noise :- The schema which is used for slow changing signal which gives distortion called as granular noise.

3.Predictive encoding : this preemptive to DPCM .Some sampling values are used to protect the next value,for this transmitter and receiver must uses same prediction algorithm.


Data  communication
   characteristics :
1. Delivery     :  Correct destination
2. Accuracy  :  Correct data
3. Timeliness  :  Fast enough
4. Jitter          :  Uneven  delay


Data  representation
1Ttext        :  email
2.Numbers :  Direct  conversion
3. Images   :  pixels
4.Audio      : continous
5. Video     :  movie

Data  flow :
1.Simplex
2.Half Duplex
3.Full Duplex

Solve Questions on Computer Science Single choice answers


Multiple Choice Questions ( Single Choice) :-


1) In SQL, which of the following is not a data Manipulation Language Commands?

A) Delete
B) Select
C) Update
D) Create
E)None of the above

Answer :- E)None of the above



2) Which of the following is not a type of SQL statement?

A) Data Manipulation Language (DML)
B) Data Definition Language (DDL)
C)Data Control Language (DCL)
D)Data Communication Language (DCL)
E)None of these

Answer :- D) Data Communication Language



3)Which of the following is not included in DML (Data Manipulation Language)

A)INSERT
B)UPDATE
C) DELETE
D)CREATE
E)None of these

Answer :-D)CREATE



4)TRUNCATE statement in SQL is a -

A) DML statement
B) DDL statement
C) DCL statement
D)DSL statement
E)None of these

Answer:- B) DDL statement



5) In SQL, which command is used to add new rows to a table?

A)Alter Table
B)Add row
C)Insert
D)Append
E)None of the Above

Answer :- C)Insert



6) Stack is also called ____

A) First In First Out (LIFO)
B)Last In First Out (FIFO)
C)First In Last Out (FILO)
D)First Come First Served (FCFS)
E)None of the above

Answer :- C)First In Last Out (FILO)



7)The full form of DDL is

A) Dynamic Data Language
B)Detailed Data Language
C)Data Definition Language
D)Data Derivation Language
E)All of these

Answer :- C)Data Definition Language



8) The OSI model consists of ___layers.

A) Nine
B)Eight
C)Seven
D)Five
E)Eleven

Answer :- C)Seven


9)Assembly language is a _____

A) Low Level Language
B)Middle Level Language
C)High level Language
D)User Language
E)None of these


Answer :- A) Low Level Language


10) Which of the following is a type of translator?

A)Assembler
B)Compiler
C)Interpreter
D)All of the Above
E) None of these

Answer:-  C)  Interpreter



11) Which of the following term is related to the stack?

 A) TOP
 B)PUSH
 C)POP
D)Rear
E) A, B and C.

Answer:-E) A, B and C.


12) In Queues, the end from where items inserted is called

A) Rear
B) Front
C) Top
D) Base
E) None of these

Answer:- A) Rear



13)Which protocol is used for browsing website:

A) TCP
B) FITFP
C)FTP
D)TFTP
E)None of these

Answer:-A) TCP



13) Which of the following is a browser?

A) Netscape Navigator
B) Mosaic
C)Mozilla Firefox
D)Google chrome
E)All of these

Answer:- E)All of these



14) Black Box Testing sometime called -

A) Data flow testing
B) Loop testing
C) Behavioral testing
D)Graph based testing
E)None of these

Answer:-D)Graph based testing



15)The Objective of testing is

A)Debugging
B)To uncover errors
C) To gain modularity
D) To analyze system
E) None of these

Answer :-B) To uncover errors



16)Choose the right sequence of SDLC (Software development life cycle) steps

A) Design, Requirement Analysis, Coding, Testing
B) Requirement Analysis, Design, Coding, Testing
C) Requirement Analysis, Design, Testing, Coding
D) Requirement Analysis, Coding, Design, Testing
E) None of these

Answer :-B) Requirement Analysis, Design, Coding, Testing



17) ODBC is based on ___.

A) Structured Query Language.
B) C language
C) C++ language
D) .net
E)None of these

Answer:- A) Structured Query Language.




18)JVM is a virtual machine that can execute ___
 A)C language
B).net programming
 C)RDBMS
 D)C++ Language
E)Java byte Code

Answer : E) Java byte Code


19)Which of the following virus overtake computer system when it boots and destroy in-formation?

A)Trojan
B)System infectors
C)Boot infectors
D)Stealth virus
E) None of these

Answer :- C) Boot infectors



20)The JDBC-ODBC bridge is

A) Three tiered
B)Multithread
C) Best for any platform
D) All of these
E) None of these

Answer :-B) Multithread



Click here for more questions

Wednesday, April 2, 2014

DTE and DCE

Terms related to modem connectivity:-

DTE :-
Data Terminal Equipment, a device that controls data flowing to or from a computer. The term is most often used in reference to serial communications defined by the RS-232C standard. This standard defines the two ends of the communications channel as being a DTE.
      
              It converts user information into signals or reconverts received signals. These can also be called tail circuits. A DTE device communicates with the data circuit-terminating equipment (DCE). The DTE/DCE classification was introduced by IBM.

           
              A DTE is the functional unit of a data station that serves as a data source or a data sink and provides for the data communication control function to be performed in accordance with the link protocol.

DCE :-
Data Communications Equipment (DCE) device. In practical terms, the DCE is usually a modem and the DTE is the computer itself, or more precisely, the computer's UART chip. For internal modems, the DCE and DTE are part of the same device.


Modem -
A modem (modulator-demodulator) is a device that modulates an analog carrier signal to encode digital information and demodulates the signal to decode the transmitted information
     In modem for Broadband ,
                  ADSL (Asymmetric Digital Subscriber Line) modems, a more recent development,



SQL related terms DCL,DML,DDl,TCL


1)DDL 
 
Data Definition Language (DDL) statements are used to define the database structure or schema. Some examples: 
CREATE - to create objects in the database
ALTER - alters the structure of the database
DROP - delete objects from the database
TRUNCATE - remove all records from a table, including all spaces allocated for the records are removed
COMMENT - add comments to the data dictionary
RENAME - rename an object


 
2)  DML

Data Manipulation Language (DML) statements are used for managing data within schema objects. Some examples:
SELECT - retrieve data from the a database
INSERT - insert data into a table
UPDATE - updates existing data within a table
DELETE - deletes all records from a table, the space for the records remain
MERGE - UPSERT operation (insert or update)
CALL - call a PL/SQL or Java subprogram
EXPLAIN PLAN - explain access path to data
LOCK TABLE - control concurrency



 
3)  DCL

Data Control Language (DCL) statements. Some examples:
GRANT - gives user's access privileges to database
REVOKE - withdraw access privileges given with the GRANT command


 
4)  TCL

Transaction Control (TCL) statements are used to manage the changes made by DML statements. It allows statements to be grouped together into logical transactions.
COMMIT - save work done
SAVEPOINT - identify a point in a transaction to which you can later roll back
ROLLBACK - restore database to original since the last COMMIT
SET TRANSACTION - Change transaction options like isolation level and what rollback segment to use


Next

Mathematical Pre-requisities

Set:-
           Set is a collection of object without repetition and each object of set is called element of a set example D={x/x is a day of week}
Properties
1) Empty set  - It contains no elements denoted by ∅ {}.
2) Subset -  Set A is subset of B if every element in set A is in B denoted as A ⊂ B.
3) Equal sets -Two sets are equal if A ⊂ B and B ⊂ A..
4) Power set -A is a power set if 2A.Set of all subsets A
        For example if,
           A={1,2,3}
           2A={ {∅},{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3} }
          23=8 elements.
5) Compliment A' such that A'={x/x ∉ A}.
6) Union - A U B is all the elements in set A and B.
7) Intersection -A ∩ B is the element s both which is in A  and also in B.
8) Cardinality is the number of sets.


Relation :-  A relation R in set S is collection of ordered pair of elements in S.
Properties :-
1) Reflexive  -R is reflexive in S if xRx ∀ (for all) x ∈ S.
2) Symmetric -R is symmetric in x if xRy ⇒ yRx
3) Transitivie - R is transitive in x if xRy and yRz ⇒xRz
4)Equivalence relation -If relation is all three reflexive,symmetric and transitive.

Tuesday, April 1, 2014

Compile,Link and Run

Compile,Link and Run
        When you're ready to compile HELLOMSG, you can  select Build Hellomsg.exe from the Build menu,or press F7,or  select the Build   icon from the Build  toolbar.The appearance of this  icon is  shown in the Build menu.If the Build  toolbar  is not currently  displaned,you can choose Customize from the Tools menu and select the Toolbars tab.Pick Build or Build MiniBar.

      Alternatively you can select Execute Hellomsg.exe from the Build menu ,or press Ctrl+F5 or click the Execute program icon which looks like red exclamation point from the Build toolbar.You ''ll get a message box asking you if you want to build the program .

       As normal ,during the compile stage ,the compiler generates an .OBJ (object) file from the C source code file.During the link stage the linker combines the .OBJ file with .LLB (library ) files to create the .EXE(executable) file .You can see a list of these library files by selecting Settings from the Project tab and clicking the Link tab .In particular ,you'll notice KERNEL32.LIB,USER.LIB,and GDI.LIB.These are "import libraries " for the three major Windows sub-systems.They contain dynamic -link library names and reference information that is bound into the .EXE file .Windows uses this information to resolve calls from the program  to functions KERNEL32.DLL,USER.DLL,and GDI.DLL. dynamic0link libraries.

          In the Visual C++ Developer Studio ,you can compile and link the program in different configurations .By defaiult ,these are called Debug and Release .The executable files are stored in subdirectories of these names .In the Debug configuration ,information is added to the .EXE file assists in deubgging the program  and in tracing through the program source code.
              
         

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