CPL 2 Lab

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INTRODUCTION OF LINUX

LINUX is now a widely used operating system.

For a general user there can many reasons to prefer LINUX OS for everyday use.

It is free, it comes with many programs including an pc suit,

Security in LINUX is high,

It is an open source operating system, you can change the system as your wish; even

create your own private LINUX distribution.

For a scientific user we may add three more.

It is a very suitable system – LINUX can easily handle multiple tasks and run for a long

time without crashing.

It allows parallel programming, it turns ordinary PC to a LINUX Workstation.

BASIC OF LINUX

1. USER RIGHTS AND FILE PERMISSIONS:-

Linux is a multi-tasking operating system.

Thus the system can be used by more than one user in the same time and can perform

multiple operations simultaneously.

The user of the system has limited rights to use the system and these rights are given

by the system administrator who has access to the root account 1.

Each user has a private directory called home, in the system.

This is the main working directory of the user.

The users cannot reach and modify the others directories and files except for shared

ones.

On the other hand the root has unlimited rights.

It can reach and modify all the directories and files.

There three types of permissions:

Read permission(r),


Write permission (w),

Execute permission(x)

Permission determine:

Who can read the file or the directory?

Who can write the file or the directory?

Or who can execute the file

Should be noted that in LINUX any file can be executable and it is not necessary to

have an extension like,

To see the file permissions in the current directory.

The 1s command can be used with the -1 flag.

2. CREATING A USER ACCOUNT:-

The easiest way to create a user account is to use the graphical interface (GUI).

If the default home directory should not be changed.

Management of the user accounts and file permissions can only be done by the

root.

The most used shell and also the default is the bash shell, so it should also be left

unchanged.

A shell is a program that interprets the command you have given to the system.

A group can be added to the system with a in similar way to adding a user.

Several users can be grouped together to give them specific group rights.

LINUX COMMANDS

GENERAL PURPOSE COMMANDS

DIRECTORY HANDLING COMMANDS

FILE HANDLING COMMANDS

PATTERN SEARCHING COMMANDS


INPUT AND OUTPUT REDIRECTION, WILD CARD PATTERNS AND

SECURITY COMMANDS

PIPES AND FILTERS

Ex No: 8

Date : LINUX COMMANDS

1.To display the date as mm/dd/yy.

[admin@localhost ~]$date “+%d/%m/%y”

OUTPUT:

08/05/2011

2.To display the time as HH:MM:SS(A.M/P.M)

[admin@localhost ~]$date “+%r”

OUTPUT:

11:13:41 PM

3.To display the current day in abbreviated form.

[admin@localhost ~] $date ”+%A”

OUTPUT:

Friday

4.To display the calendar May 2009..

[admin@localhost ~]$cal jan 2006

OUTPUT:

January 2006

Su Mo Tu We Th Fr Sa

1 2 3 4 5 6 7

8 9 10 11 12 13 14

15 16 17 18 19 20 21

22 23 24 25 26 27 28

29 30 31

5.To display the complete name of the terminal.

[admin@localhost ~]$tput longname


OUTPUT:

[admin@localhost ~]

6.To display about your operating system.

[admin@localhost ~]$ uname –a

OUTPUT:

Linux Localhost.localdomain 2.6.18 – 53.e15

RESULT:

Thus the Linux command for general purpose command has been executed successfully.

II.DIRECTORY HANDLING COMMANDS

7.To view the name of the current directory

[admin@localhost ~]$pwd

OUTPUT:

/home/admin

8.To display all the files including hidden files.

[admin@localhost ~]$ls –a

OUTPUT:

area.sh a.txt com.sh fahren.sh lines powers.sh vowel.sh

arith.sh basic.sh echo leap.sh lines.sh stu.sh w.sh

ari.txt big.sh echo.sh libdata odd sumeven.sh

a.tct b.txt fact.sh lib.sh odd.sh three.sh

9.To display all the files with its inode number.

[admin@localhost ~]$ls –i

OUTPUT:

11633005 5as.c 11632958s.sh

10.To display all the files including files in subdirectories.

[admin@localhost ~]$ls –R

OUTPUT:

5as.c Ajjj.c

A.c a.out

11. To create a directory called stud. Change to the stud. Directory

[admin@localhost ~]$mkdir stud


[admin@localhost ~]$cd stud

OUTPUT:

[admin@localhost ~stud]

12.To verify whether you have actually changed to the stud directory, return to your original

directory

[admin@localhost ~]$pwd

[admin@localhost ~]$cd..

OUTPUT:

/home/admin/stud

[admin@localhost ~]

RESULT:

Thus the Linux command for directory handling commands has been executed successfully.

III. FILE HANDLING COMMANDS

12. To create a directory called marks inside the stud director

[admin@localhost ~]$cd stud

[admin@localhost ~]$mkdir marks

OUTPUT:

[admin@localhost ~stud]

13. To create a file in the stud directory called slaec and enter the data for the slaec file

[admin@localhost ~stud]$cat >slaec

1. Arun cse

2. Anna ece

3. Anbu eee

Note : [ after entering the data in the files press ctrl + z key ]

14. To change to the marks directory and copy the files slaec from stud directory to marks

directory. To verify whether the files has been copied to the mark directory.


[admin@localhost ~stud]$cd marks

[admin@localhost ~marks]$cp/admin/home/stud/slaec slaec2

[admin@localhost ~marks]$1s

Slaec1

15.To rename the file slaec to slaeccse. To verify the file name has been changed.

[admin@localhost ~stud]$ln slaec1 slaeccse

[admin@localhost ~stud]$ls

OUTPUT:

Slaeccse

16. To move the file slaec1 from marks directory to stud directory. To verify whether the file has

been moved to the stud directory.

[admin@localhost ~]$mv slaec1/ramdisk/home/knoppix/stud

[admin@localhost ~]$cd ..

[admin@localhost ~]$1s

[admin@localhost ~]$cd marks

[admin@localhost ~]$1s

OUTPUT:

/home/admin/stud/marks/

17. To display the total numbers of lines, words and characters in a file

[admin@localhost ~]$wc slaeccse

OUTPUT:

4 12 45 slaeccse

18. To display the list of files starting from our current directory

[admin@localhost ~]$find.-print

OUTPUT:


./slaeccse

RESULT :

Thus the linux file handling commands has been executed successfully.

IV. PATTERN SEARCHING COMMANDS

18. To display students under “cse”department in the file named slaeccse

[admin@localhost ~]$cat slaeccse

1. Arun cse

2. Anna ece

3. Anbu eee

[admin@localhost ~]$grep “cse” slaeccse

OUTPUT:

Arun cse

19. To display the words ending with a or b or c in file

[admin@localhost ~]$cat >raj

This is a new file.

There are two types of files newa, newb and newc.

The second is new1, new2, new3.

[admin@localhost ~]$grep “new[a-c]” raj

OUTPUT:

There are two types of files newa, newb, and newc.

20. To display the students details whose names are arun, anna and anbu in the file named

[admin@localhost ~]Slaeccse

[admin@localhost ~]$egrep “arun|anna|anbu” slaeccse

OUTPUT:

1. Arun cse

2. Anna ece


3. Anbu eee

21. To display the lines in the file raj to which has exact string “there are”

[admin@localhost ~]$fgrep “there are” raj

OUTPUT:

There are two types of files newa, newb, and newc.

RESULT:

Thus the Linux commands pattern searching commands has been executed successfully.

V. INPUT AND OUTPUT REDIRECTION, WILD CARD PATTERNS AND

SECURITY COMMANDS

22. To redirect the system calendar contents to the file a1.

[admin@localhost ~]$ cal > a1

[admin@localhost ~]$ cat a1

OUTPUT:

January 2006


1 2 3 4 5 6 7

8 9 10 11 12 13 14

15 16 17 18 19 20 21

22 23 24 25 26 27 28

29 30 31

23. To move the contents of the named a1 to a new file by using redirection operator

[admin@localhost ~]$ cat < a1 > a2

[admin@localhost ~]$cat a2

OUTPUT:

January 2006


1 2 3 4 5 6 7

8 9 10 11 12 13 14

15 16 17 18 19 20 21

22 23 24 25 26 27 28

29 30 31

24. List the files starting with „a‟

[admin@localhost ~]$ ls a*


OUTPUT:

a1 a2 a1

25. To find the file names ending with “1”

[admin@localhost ~]$ ls * 1

OUTPUT:

a1 raj1

26. To find the file starting with “vignesh” and then followed by any one character and with any

extension of one character

[admin@localhost ~]$ls vignesh.?

OUTPUT:

Vignesh2.c

27. To revoke the execute function for the file name vignesh and granting all permissions to file

named raj

[admin@localhost ~]$chmode 777 vignesh

[admin@localhost ~]$ls -1

OUTPUT:

1localhost admin 168 may 1 03:47 vignesh

28. To revoke the execute permission for the user of the file vignesh

[admin@localhost ~]$ -rwxrwxrwx 1 localhost localhost 168 jan 1 03:47 vignesh

[admin@localhost ~]$ chmod u-x vignesh

[admin@localhost ~]$ -rw-rwxrwx 1 localhost localhost 168 jan 1 03:47 vignesh


RESULT:

Thus the linux commands for input and output redirection, wildcard patterns and security

commands has been executed successfully.

VI PIPES AND FILTERS

29. To display the permissions of the group for the files whose name being withv.

[admin@localhost ~]$ ls –l v *|tr –s “”|cut –d “” –c5-7

OUTPUT:

r - -

r - -

r - -

r - -

30. To create a file department with some student details

[admin@localhost ~]$ cat > dept

OUTPUT:

4001 mohan 80

4002 raj 65

4003 mani 75

31. To display the names of the students in the file department in alphabetical order

[admin@localhost ~]$sort + 1f -2 dept

OUTPUT:

4003 mani 75

4001 mohan 80

4002 raj 65

32. To display the list of students who have scored marks between 50 and 80

[admin@localhost ~]$cat dept | grep “[5-7][0-9]

OUTPUT:

4002 raj 65

4003 mani 75

33. To display the first line of the file raj

[admin@localhost ~]$head -1 raj

OUTPUT:


This is a new file.

RESULT:

Thus the unix pipes and filters commands has been executed successfully.

SHELL

PROGRAMMING

SHELL INTRODUCTION

A shell is a command language interpreter. It is possible to write and run your own shell.

As shells are no different than any others programs as far as the system is concerned.

This manual deals with the details of one particular shell, called shell script.

There are various graphical interfaces available for linux the still is a very neat tool.

The shell is not just a collection of commands but a really good programming language.

It is often possible to perform simple tasks using these scripts without writing a program

in a language such as c, by using the shell to selectively run others programs.

Many shell variables have special meaning to the shell.

There three mainly shells with UNIX

BOURN SHELL,

KORN SHELL,

C SHELL.

The shell is very good for system administrator tasks.

FUNCTION OF SHELL

It accepts commands from the user and interpreter it.

It starts executing the appropriate executable file.


It request the kernel to carry out 1\0 transfer. Thus the shell acts middle man between the

kernel and the user.

The shell implements a number of substitutions where sequences indicated by Meta

character.

The shell function is unlike any other than the wildcard function in that it communication

with of the world outside make.

The shell function performs the same function that back quotes (“ ”) perform in most

shells : it does command expansion.

This means that it takes as an argument a shell command and evaluates to the output of

the command expansion.

This means that it takes as an argument a shell command and evaluates to the output of

the command.

The only processing make does on the result is to convert each newline to a single space.

If there is a trailing newline it will simply be removed.

The command run by calls to the shell function are run when the function calls are

expanded.

SHELL PROGRAMMING

Even though there are various graphical interfaces available for linux the shell still is

a very neat tool.

The shell is not just a collection of command but a really good programming

language.

You can automate a lot of task with it, the shell is very good for system administrator

task.

You can quickly try out if your ideas work which makes it very useful for simple

prototyping.

WORKING WITH SHELL PROGRAM

The shell may be used to read and execute commands contained in a file.


The shell to read commands to read commands from file. Such a file is called a command

procedure of shell procedure.

When you login, the shell is started by the system in your home directory and begins by

reading commands from a file.

A login shell executed after you login to the system. Login also in your home directory.

The file contains several commands to be executed by linux each time two login.

The first is set command which is interpreted directly by the shell.

Ex No: 2

Date : Factorial Of a Given Numbers

AIM:

To find the factorial of numbers using shell script

ALGORITHM:

STEP1: Start the program

STEP2: Give the numbers is real

STEP3: Fact =1 ; i =1

STEP4: Do the while loop [ $i – u $n ]

STEP5: Give the condition do fact = `expr $i \* $fact`

STEP6: Stop the program.


PROGRAM

echo "enter the number"

read n

fact=1

i=1

while [ $i -le $n ]

do

fact=`expr $i \* $fact`

i=`expr $i + 1`

done

echo "the factorial number of $n is $fact”

OUTPUT

[administrator@localhost ~] $ sh fact.sh

enter the number 4

the factorial number of 4 is 24

RESULT:

Thus the given shell script has been executed successfully.

Ex No: 3

Date : Leap Year Or Not

AIM:

To find the given year is leap or not using shell script

ALGORITHM:



STEP2: Give the year

STEP3: Read the year

STEP4: If loop [expr $y % 4 –eq 0]

STEP5: If the remainder is zero the tear is leap or not leap year.


PROGRAM

echo "Enter The Year"

read y

if [ `expr $y % 2` -eq 0 ]

then

echo "the year is leap"

else

echo "the year is not leap"

fi

OUTPUT

[admin@localhost ~] $ sh leap.sh

Enter The Year 2008

the year is leap

RESULT:


Ex No: 4

Date : SUM OF EVEN NUMBERS


AIM:

To write a program to find the sum at even numbers upto n.

ALGORITHM:


STEP2: Enter the upper limit

STEP3: Give the limit and real n.

STEP4: Do the while loop [ $i – u $h ]

STEP5: Give the condition as run.


PROGRAM

echo " ENTER THE UPPER LIMIT "

read n

i=2

while [ $i -lt $n ]

do

sum=`expr $sum + $i `

i=`expr $i + 2 `

done

echo " SUM OF EVEN NUMBERS UPTO $n IS $sum"

OUTPUT

[administrator@localhost ~] $ sh sumeven.sh

ENTER THE UPPER LIMIT 10

SUM OF EVEN NUMBERS UPTO 10 IS 20

RESULT:


Ex No: 5


Date : SUM OF N NUMBERS

AIM:

To write a shell script to find the sum of n numbers

ALGORITHM:

STEP1: Start the program.

STEP2: Enter the upper limit.

STEP3: read n

STEP4: Do the while condition

Give condition [ $i –le $h ]

STEP5: Sum = expr $sum + $i

i = expr $i + 1


PROGRAM

echo "enter the upper limit "

read n

i=1

sum=0

while [ $i -le $n ]

do

sum=`expr $sum + $i`

i=` expr $i + 1`

done

echo "the sum of numbers is $sum"

OUTPUT

[administrator@localhost ~] $ sh upperlim.sh

enter the upper limit 5

the sum of numbers is 15


RESULT:


Ex No: 6

Date : MENU DRIVEN PROGRAM WITH UNIX COMMANDS

AIM:

To write a shell program to generate menu driver program with unix commands.

ALGORITHM:

STEP 1: Start the program.

STEP2: Read and display the (1) list of files (2) today‟s date (3) process status

(4) logged in users (5) display the current working directory (6) quit.

STEP 3: Read ch.

STEP 4:condition case “$ch”

STEP 5:condition satisfied to display the choice.

STEP 6:* invalid choice.

STEP 7: execute the program

STEP 8: stop the program.

PROGRAM

echo "menu"

echo "1.list of files"

echo "2.today's date"

echo "3.process status"

echo "4.logged in users"

echo "5.display the current working directory"


echo "6.Quit"

echo -n "enter our choice"

read ch

case "$ch" in

1) ls -1 ;;

2) date ;;

3) ps ;;

4) who ;;

5) pwd ;;

6) exit ;;

*) echo "invalid choice"

exit ;;

esac

OUTPUT:

Menu

1.List of files

2.Todays date.

3.Process status

4.Logged in users

5.Display the current working directory.

6.Quit

Enter your choice: 2

22/11/2010

Enter your choice: 5

localhost/admin

RESULT:



Ex No: 7

Date : FIBONACCI SERIES

AIM:

To write a program to display the Fibonacci series.

ALGORITHM:


STEP2: Read limit after enter limit.

STEP3: Do the while loop [$var –ie èxpr $lim-2`]

STEP4: Give the condition.

do

n3= èxpr $n1+$n2`

n1= èxpr $n2`

n2= èxpr $n3`

var= èxpr $var+1`


PROGRAM

echo " ENTER THE LIMIT FOR FIBONACCI SERIES"

read

lim

n1=0

n2=1

var=0


echo "FIBONACCI SERIES IS "

echo "$n1"

echo "$n2"

while [ $var -lt èxpr $lim - 2` ]

do

n3=èxpr $n1 + $n2 `

n1=èxpr $n2 `

n2=èxpr $n3 `

var=èxpr $var + 1

` echo "$n2"

done

OUTPUT

[administrator@localhost ~]$ sh fibonacci.sh

ENTER THE LIMIT FOR FIBONACCI SERIES 5

FIBONACCI SERIES IS

0

1

1

2

3

RESULT:


Ex No: 8

Date : SUM AND PRODUCT FOR THE GIVEN TWO NUMBERS


AIM :

To write a shell program to find the sum and products for the given two numbers

ALGORITHM :


STEP2: Enter the value of a and b.

STEP3: read a, b;

STEP4: To find sum = product write sentences

Sum = ‟expr $a +$b‟

Product = „expr $a -$b‟

STEP5: To give the result write this

Sum = $sum\ +product -$prod


PROGRAM:

echo -n "enter the values for a & b"

read a b

sum=`expr $a + $b`

prod=`expr $a \* $b`

echo -n "sum=$sum, product=$prod"

OUTPUT

Enter the values for a and b

2 3

Sum = 5

Product = 6


RESULT:


Ex No: 9

Date : TEMPERATURE TO FAHRENHEIT.

AIM

To write a program to convert temperature in Fahrenheit to degree centigrade

ALGORITHM:


STEP2: Enter the temperature in Fahrenheit

STEP3: read f

STEP4: expr $f -32

expr $t /*5/9


PROGRAM

echo " ENTER THE TEMPERATURE IN FAHRENHEIT "

read f

t=`expr $f - 32 `

c=`expr $t \* 5 / 9`

echo "$c"

OUTPUT

[administrator@localhost ~] $ sh temp2far.sh


ENTER THE TEMPERATURE IN FARENHEIT 39

3

RESULT:


Ex No: 10

Date : VOWELS OF A STRING.

AIM:

To write a program to find the value of a string

ALGORITHM:


STEP2: read the string

STEP3: Do the while loop[$1 –gto]

STEP4: Give the condition

Do

Temp = „expr $str act –c$‟

Case $ temp in

STEP5: else condition is bring out

1 = „expr $l –I”


PROGRAM:


echo "enter the string"

read str

l=èxpr length $str`

vowel=0

while [ $l -gt 0 ]

do

temp=èxpr $str | cut -c $l`

case $temp in

a|A) vowel=èxpr $vowel + 1`;;

e|E) vowel=èxpr $vowel + 1`;;

i|I) vowel=èxpr $vowel + 1`;;

o|O) vowel=èxpr $vowel + 1`;;

u|U) vowel=èxpr $vowel + 1`;;

esac

l=èxpr $l - 1`

done

echo "the string has $vowel vowel"

OUTPUT

Enter the string

Operating System

The string has 5 vowels

RESULT:


Ex No: 11


Date : SUM OF DIGITS FOR GIVEN INTEGER

AIM:

To write a shell program to find the sum of digit for given integer.

ALGORITHM:


STEP2: Read n

STEP3: Assign sol = 0

STEP4: Condition while[$n –gto]

STEP5 : Compute a = „expr $n%10‟

Sd = „expr $sd + $a‟ and

n= „expr $n/10‟

STEP6: Print the sum of the digits in the given number is d.


PROGRAM:

echo "enter the number"

read n

sd=0

while [ $n -gt 0 ]

do

a=èxpr $n % 10`

sd=èxpr $sd + $a`

n=èxpr $n / 10`

done

echo "the sum of digits in the given number is $sd"

OUTPUT


Enter the number: 1 2 3 4 5

The sum of digits for the given number is: 15

RESULT:


Ex No: 12

Date: BASIC SALARY

AIM:

To write the program to find the basic salary.

ALGORITHM:


STEP2: Enter the basic salary.

STEP3: Real b.

STEP4: Give the condition.

da=èxpr $b \* 10/100`

hr=èxpr $b \* 20/100`

gr=èxpr $b + $da + $h`

STEP5: Exit the program salary=$gr.


PROGRAM

echo " ENTER THE BASIC SALARY OF THE EMPLOYEE "

read b

da=èxpr $b \* 10 / 100 `


hr=`expr $b \* 20 / 100 `

gr=`expr $b + $da + $hr `

echo " GROSS SALARY = $gr "

OUTPUT

[administrator@localhost ~] $ sh bsal.sh

ENTER THE BASIC SALARY OF THE EMPLOYEE 900

GROSS SALARY = 1170

RESULT:


Ex No: 13

Date: COMBINATION OF 123

AIM:

To write a shell script program to print the combination of 123.

ALGORITHM:


STEP2: Give the three condition for i&j&k in 123.

STEP3: Echo “$i $j $k”


PROGRAM

for i in 1 2 3


do

for j in 1 2 3

do

for k in 1 2 3

do

echo $i $j $k

done

done

done

OUTPUT

[administrator@localhost ~] $ sh comb.sh

1 1 1

1 1 2

1 1 3

1 2 1

1 2 2

1 3 3

2 1 1

2 1 2

2 1 3

2 2 1

2 2 2

2 2 3

2 3 1

2 3 2

2 3 3

3 1 1

3 1 2

3 1 3

3 2 1


3 2 2

3 2 3

3 3 1

3 3 2

3 3 3

RESULT:


Ex No: 14

Date: POWER OF A NUMBER

AIM:

To write a shell script to find the power of a number.

AIGORITHM:

STEP1: Stort the program.

STEP2: To find xn enter X=N value.

STEP3: Do the while loop [$i-u $h].

STEP4: Give the condition

X=expr $ x\* $ temp

I=expr $ i+1


PROGRAM

echo "Enter the value of X & N to find X^N "

read x n

temp=`expr $x` i=1

while [ $i -lt $n ]


do

x=`expr $x \* $temp`

i=`expr $i + 1`

done

echo "RESULT IS $x"

OUTPUT

[administrator@localhost ~] $ sh power.sh

Enter the value of X & N to find X^N 8 2

RESULT IS 64

RESULT:


Ex No: 15

Date: NUMBER OF CHARACTER LINES IN A FILE

AIM:

To write a program to find the number of character lines,words in a file.

ALGORITHA:

STEP1: Start program.

STEP2: Enter the file name.

STEP3: Give the if condition

If (file $file) then

STEP4: Bring the else condition.



PROGRAM:

echo "enter the file name"

read file

if ( file $file )then

echo "file exists"

echo "no of lines"

wc -l $file

echo "no of characters"

wc -c $file

echo "no of words"

wc -w $file

else echo "file does not exists"

fi

OUTPUT

[administrator@localhost ~] $ sh check.sh

enter the file name area.sh

file exists.

no of lines 11

no of characters 36

no of words 150

[administrator@localhost ~] $ sh check.sh

enter the file name focus.sh

file does not exists.

RESULT:


Ex No: 16


Date : CHECK WHETHER THE NUMBER IS ZERO OR

POSITIVE OR NEGATIVE

AIM:

To write a program to check the given number is positive or negative.

ALGORITHA:

STEP1: Start program.

STEP2: Enter the integer.

STEP3: Give the if condition

if [ $n -eq 0 ] then

STEP4: Condition satisfied the given number is zero

STEP5: Bring the else if condition

elif [ $n -gt 0 ]

STEP6:If the given numer is greater than zero print positive or else negative


PROGRAM

echo "enter any number"

read n

if [ $n -eq 0 ]

then

echo "number is zero "

elif [ $n -gt 0 ]

then

echo "number is positive"

else

echo "number is negative"

fi


OUTPUT

[administrator@localhost ~]$ sh posneg.sh

enter any number 4

number is positive


enter any number -98

number is negative


enter any number 0

number is zero

RESULT:


C

PROGRAMMING

Ex.No: 17

DATE: DISPLAY THE CONTENTS OF A FILE

AIM:

To write the C program to display the contents of a file

ALGORITHM:


STEP 1:start

STEP 2:write number of command line arguments

STEP 3:open a file in a read mode

STEP 4: display the content of the file

STEP 5: stop

PROGRAM:

#include<stdio.h>

int main(int argc,char *argv[])

{

FILE *fptr;

char ch;

int i;

printf("no of arguments %d \n",argc);

for(i=1;i<argc;i++)

{

fptr=fopen(argv[i],"r");

ch=getc(fptr);

while(ch!=EOF)

{

putchar(ch);

ch=getc(fptr);

}

}

}


OUTPUT

[admin@localhost ~]$ cc disp.c

[admin@localhost ~]$ ./a.out one.txt

no of arguments 2

This is linux laboratory.

RESULT:

Thus the program has been executed successfully and the output is verified.

Ex.No: 18

DATE: COPY THE CONTENTS OF ONE FILE TO ANOTHER

FILE

AIM:

To write the C program to copy the contents of a file to another file.

ALGORITHM:

STEP 1:start

STEP 2: check no of command line arguments is equal to 3 if not exit

STEP 3:if so, open two files one in read mode and other in write mode

STEP 4:copy the contents of file in read mode to file in write mode

STEP 5:stop

PROGRAM


#include<stdio.h>

int main( int argc,char *argv[ ] )

{

FILE *fptr1,*fptr2;

char ch;

int i;

if(argc>3)

{

printf("too many arguments");

return(0);

}

else if(argc<2)

{

printf("\n parameter required");

return(0);

}

else

{

fptr1=fopen(argv[1],"r+");

fptr2=fopen(argv[2],"w+");

ch=getc(fptr1);

while(ch!=EOF)

{

putc(ch,fptr2);

ch=getc(fptr1);

}

printf(“the contents of %s is successfully copied to %s \n”,argv[1],argv[2]);

}

}


OUTPUT

[admin@localhost ~]$ cc copy.c

[admin@localhost ~]$ ./a.out one.c two.c

the contents of one.c is successfully copied to two.c

RESULT:


Ex.No: 19

DATE: TO COUNT THE CHARACTERS AND LINES IN A

FILE

AIM:

To write the C program to count the characters and lines in a file.

ALGORITHM:

STEP 1: Start the program

STEP 2: Declare a file pointer fp.

STEP 3: Assign nlines=0, and nc=0.

STEP 4: Read the filename.

STEP 5: Open the file –filename under read mode assigned to fp.

STEP 6: If (fp1 == NULL) then print cannot open filename for reading and

then go to step 15.


STEP 7: Assign c=getc(fp)

STEP 8: Repeat the steps 9 to 10 until while (c!=EOF)

STEP 9: if (c ==‟\n‟) then nlines=nlines+1 else nc=nc+1.

STEP 10: Assign c=getc(fp)

STEP 11: Close the file-fp

STEP 12: If (nc!=0) then step 13 else step 14.

STEP 13: Print the total no of characters and total of lines in the file then goto

step 15.

STEP 14: Print the file is empty

STEP 15: Stop

PROGRAM

#include<stdio.h>

#include<conio.h>

#include<stdlib.h>

main()

{

FILE *fp;

int c,nc,nlines;

char filename[40];

nlines=0;

nc=0;

clrscr();

printf("\n\n\t\t To count the characters and lines in a given file \n");

printf("\n\n Enter a file name: ");

gets(filename);

fp=fopen(filename,"r");


if(fp==NULL)

{

printf("\n\n Cannot open %s for reading \n", filename);

getch();

exit(1);

}

c=getc(fp);

while(c!=EOF)

{

if(c=='\n')

nlines++;

else

nc++;

c=getc(fp);

}

fclose(fp);

if(nc!=0)

{

printf("\n\n There are %d characters in %s \n",nc,filename);

printf("\n\n There are %d lines %n",nlines);

}

else

printf("\n\n The File %s is empty \n",filename);

getch();

}

OUTPUT

[admin@localhost ~]$ cc count.c


[admin@localhost ~]$ ./a.out

To count the characters and lines in a given file

Enter the file name:strcopy.c

There are 345 characters in strcopy.c

There are 25 lines

RESULT:


Ex.No: 20

DATE: FACTORIAL OF A GIVEN NUMBER

AIM:

To write a C program to find the factorial of a given number using recursive

function.

ALGORITHM

Main() program


STEP 2: Read the value of num.

STEP 3: Assign fact1=call fact(num) function

STEP 4: print fact1

STEP 5: Stop

Fact() function



STEP 2: if(n==1||n==0) then step 3 else step 4

STEP 3: return (1)

STEP 4: return (n*fact(n-1))

PROGARM

#include<stdio.h>

long fact(int n);

main()

{

int num;

long fact1;

printf("\n Enter a number");

scanf("%d",&num);

fact1=fact(num);

printf("The funtion of %d is %d",num,fact1);

}

long fact(int n)

{

if(n==1||n==0)

return(1);

else

return(n*fact(n-1));

getch();

}

OUTPUT


[admin@localhost ~]$ cc fact.c


Enter a number: 4

The factorial of 4 is 24

RESULT:


Ex.No: 21

DATE: FIBONACCI SERIES

AIM:

To write a C program to print Fibonacci Series using recursive function.

ALGORITHM

main() program


STEP 2: Read the value of n.

STEP 3: call fib(0, 1, n) function

STEP 4: Stop

fib() function



STEP 2: if(num!=0) then step 3 else step 5

STEP 3: print n1

STEP 4: call fib(n2, n2+n1, num-1)

STEP 5: return

PROGRAM

#include<stdio.h>

void fib(unsigned int n1,unsigned int n2,unsigned int num);

main()

{

int n;

printf("\n Enter the limit :");

scanf("%d",&n); fib(0,1,n);

}

void fib(unsigned int n1,unsigned int n2,unsigned int num)

{

if(num!=0)

{

printf("%u \n",n1); fib(n2,n2+n1,num-1);

}

}

OUTPUT

[admin@localhost ~]$ cc fibo.c


Enter the limit: 8

0


1

1

2

3

5

8

13

RESULT:


Ex.No: 22

DATE: IMPLEMENTATION OF LS COMMAND

AIM:

To write a C program to implement ls Command

ALGORITHM:

STEP 1: Read the file to be displayed through command Line

STEP 2: Check whether the argument matches with file available in the current directory

STEP 3: If no matches display error

STEP 4: Else display the list of files which matches the given input by utilizing the

system

command


PROGRAM

#include<stdio.h>

main(int argc,char *argv[])

{

FILE *fp;

int i;

for (i=0;i<argc;i++)

{

if((fp=fopen(argv[i],"r"))==(FILE*)NULL)

{

perror(argv[i]);

}

else

printf("%s file is present",argv[i]);

}

exit(0);

}

OUTPUT

[admin@localhost ~]$ cc ls.c

[admin@localhost ~]$ ./a.out process.c

file is present

process.c file is present

[admin@localhost ~]$ cc ls.c

[admin@localhost ~]$ ./a.out area.sh

a.sh: No such file or directory

RESULT:



Ex.No: 23

DATE: LINEAR SEARCH

AIM:

To write a C program to perform linear search using function.

ALGORITHM:

main() program

STEP 1 : Start the program.

STEP 2 : Read the value of n.

STEP 3 : Assign i=0.

STEP 4 : repeat the step 5 to 6 until i < n.

STEP 5 : Read a[i].

STEP 6 : Compute i = i + 1.

STEP 7 : Read the value of x.

STEP 8 : Call linear(a,n,x) function.

STEP 9 : Stop.

linear() function


STEP 2 : Assign i=0 and flag=0.

STEP 3 : Repeat the step 4 to 5 until i < n.

STEP 4 : if(x==a[i]) then flag=1 and exit from loop.



STEP 6 : if ( flag==1) then print element found else print element not found.

STEP 7 : Return to the main program.

PROGRAM

#include<stdio.h>

void linear(int [], int, int);

main()

{

int i,n,x,a[20];

printf("\n Enter the limit : ");

scanf("%d",&n);

printf("\n Enter the Elements :");

for(i=0;i<n;i++)

{

scanf("%d",&a[i]);

}

printf("\n Enter the elements to be searched :");

scanf("%d",&x);

linear(a,n,x);

}

void linear(int a[],int n,int x)

{

int i,flag=0;

for(i=0;i<n;i++)

{

if(x==a[i])

{

flag=1; break;

}

}


if(flag==1)

printf("\n The element %d is found in the position %d",a[i],i+1);

else

printf("\n The element is not found in the list");

}

OUTPUT

[admin@localhost ~]$ cc linear.c


Enter the limit : 5

Enter the elements : 27 39 63 18 12

Enter the elements to be searched : 63

The element 14 is found in the position 2

RESULT:


Ex.No: 24

DATE: BINARY SEARCH

AIM:

To write a C program to perform binary search using function.

ALGORITHM:


STEP 2 : Read the value of n.


STEP 3 : Assign i =0.

STEP 4 : Repeat the step 5 to 6 until i < n.

STEP 5 : Read a[i].


STEP 7 : Read the value of s.

STEP 8 : Call binary(q,n,s) function.

STEP 9 : Stop.

binary() function


STEP 2 : Assign f=0, l=n-1 and flag=0.

STEP 3 : Repeat the step 4 to 6 until f <= n.

STEP 4 : Compute m=(f+1)/2.

STEP 5 : if(s < a[m]) then l=m-1 else step 6.

STEP 6 : if(s > a[m]) then f=m+1 else flag=1 exit from loop.

STEP 7 : if ( flag==1) then print element found else print element not found.

PROGRAM

#include<stdio.h>

void binary(int [], int, int);

main()

{

int s,i,n,a[100];


printf("\n\t\t To search for an element using binary search \n");

printf("\n Enter the no of elements : ");

scanf("%d",&n);

printf("\n Enter the Elements :");

for(i=0;i<n;i++)

{

scanf("%d",&a[i]);

}

printf("\n Enter the element to be searched :");

scanf("%d",&s);

binary(a,n,s);

}

void binary(int a[],int n,int s)

{

int f,l,m,flag=0;

f=0;

l=n-1;

while(f<=1)

{

m=(f+l)/2;

if(s<a[m])

l=m-1;

else if(s>a[m])

f=m+1; else

{

flag=1;

break;

}

}

if(flag==1)


printf("\n\n %d is found \n",s);

else

printf("\n\n %d is not found \n",s);

}

OUTPUT

[admin@localhost ~]$ cc binary.c


To search for an element using binary search

Enter the no. of elements : 5

Enter the elements : 65 85 25 95 45

Enter the element to be searched: 95

95 is found.

RESULT:


Ex.No: 25

DATE: STRING CONCATENATION

AIM:

To write a C program to concatenate two strings.

ALGORITHM:


STEP 2 : Declare the character pointer variables as s1, s2, and s3.


STEP 3 : Allocate memory for s1, s2, and s3 character pointer variables using

malloc function.

STEP 4 : Read the string s1 and s2.


STEP 6 : Repeat the steps 7 and 8 until ((c=*(s1+i))!=‟\0‟).

STEP 7 : Assign s3[i]=0.

STEP 8 : Compute i=i+1.

Step 9 : Assign k=0.

STEP 10 : Repeat the steps 11 to 12 until ((c=*(s1+i))!=‟\0‟).

STEP 11 : Assign s3[i+k]=0.

STEP 12 : Compute k=k+1.

STEP 13 : Assign s3[i+k]=‟\0‟.

STEP 14 : Print the concatenated string s3.

STEP 15 : Stop.

PROGRAM

#include<stdio.h>

#include<malloc.h>

#define length 20

main()

{

char *s1,*s2,*s3,c;


int i,k;

s1=(char *)malloc(length *sizeof(char));


s3=(char *)malloc(2*length *sizeof(char));

printf("\n Enter the String1 : ");

scanf("%s",s1);


scanf("%s",s2);

i=0;

while((c=*(s1+i)) != '\0')

{

s3[i]=c; i++;

}

k=0;

while((c=*(s2+k)) != '\0')

{

s3[i+k]=c;

k++;

}

s3[i+k]='\0';

printf("\n Concatenated String is %s ",s3);

getch();

}

OUTPUT

[admin@localhost ~]$ cc strcon.c



Enter the String1 : Raja

Enter the String2 : Sekaran

Concatenated String is : RajaSekaran

RESULT:


Ex.No: 26

DATE: STRING COPY

AIM:

To write a C program to copy the contents of one string to another.

ALGORITHM:


STEP 2 : Declare the character pointer variables as s1 and s2.

STEP 3 : Allocate memory allocation for s1 and s2 using malloc() function.

STEP 4 : Read s1.


STEP 6 : Repeat the steps 7 and 8 until ((c=*(s1+i))!=‟\0‟).

STEP 7 : Assign s2[i]=0.


STEP 8 : Compute i=i+1.

STEP 9 : Assign s2[i]=‟\0‟.

STEP 10 : Print s2.

STEP 11 : Stop.

PROGRAM

#include<stdio.h>

#include<malloc.h>

#define length 30

main()

{

char *s1,*s2,c;

int i;




scanf("%s",s1);

printf("\n The entered String1 is %s",s1);

i=0;

while((c=*(s1+i)) != '\0')

{

s2[i]=c;

i++;

}

s2[i]='\0';

printf("\n The Copied String1 in String2 is %s ",s2);

getch();

}


OUTPUT

[admin@localhost ~]$ cc strcpy.c


Enter the string1 : ASHOK

The entered string1 is ASHOK

The copied String1 in String2 is ASHOK

RESULT:


Ex.No: 27

DATE: FORK SYSTEM CALL

AIM:

To write a C program to view the main value using the fork function.

ALGORITHM:

STEP 1:start

STEP 2:initialize the variable fork, value.

STEP 3:when the value is equal to the fork, print the main value as value.

STEP 4:stop

PROGRAM

#include<stdio.h>


int main(void)

{

int fork(void),value;

value=fork();

printf("main:value =%d\n",value);

return 0;

}

OUTPUT

[admin@localhost ~]$ cc fork.c


main:value =0

main:value =5545

RESULT:


Ex.No: 28

DATE: WAIT SYSTEM CALL

AIM:

To write a C program to display the waiting process state.

ALGORITHM:

STEP 1: start

STEP 2: initialize the variable pid,status,exitch


STEP 3: if the pid is equal to fork and it satisfy the condition that it equal

to -1,then print the error message

STEP 4: when pid is equal to other print the child process else print the

parent process

STEP 5: when exit is equal to status and it is equal to -1 then print perror is in

wait else parent existing

STEP 6:stop

PROGRAM

#include<stdio.h>

#include<unistd.h>

int main(void)

{

int pid,status,exitch;

if((pid=fork())==-1)

{

perror("error");

return (0);

}

if(pid==0)

{

sleep(1);

printf("child process");

return (0);

}

else

{

printf("parent process\n");

if((exitch=wait(&status))==-1)

{


perror("during wait()");

return(0);

}

printf("parent existing\n");

return (0);

}

}

OUTPUT

[admin@localhost ~]$ cc wait.c


parent process

child process

parent existing

RESULT:


CPL 2 Lab

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