OS Programs for Record

8/8/2019 OS Programs for Record

1/71

STUDY OF BASIC COMMANDS OF UNIX

COMMANDS:

CAL:

DESCRIPTION:

The cal command is used to display the calendar.

SYNTAX:

[$ CAL YEAR]

DATE:

DESCRIPTION:

The date command is used to display the day,month,year and time.

SYNTAX:

[$DATE]

The command can also be used within suitable format specified as argument .

Each format specified as argument .Each format is prescribed by a symbol followed by the %

Operator describing the format,

-m -> Display the month in number

-h -> Display the month in words

-d -> Display the day in words

-y -> Display last two digits of the year

OPTIONS:

U print or sets coordinate universal time

S set the current time to specified time

MAN:


2/71

DESCRIPTION:

Format and displays online manuals.

SYNTAX:

Man [option] name

CLEAR:

DESCRIPTION:

Clear the terminal screen .

WHO:

DESCRIPTION:

Displays the information of logged in user.

OPTIONS:

-a-> All

-b-> Time of the last symbol boot

-d->Prints dead processor

WHO AM I:

DESCRIPTION:

Show the login of the current user .

PASSWD:

DESCRIPTION:

Update on users authentication token.

SYNTAX:

Passwd [*][-1][-v][-f][-d][-n mindays][-m maxdays][username]

OPTIONS:

-k-> Used to lock an account

-n->Will set minimum password time

LOGOUT:


3/71

DESCRIPTION:

Exit off from current login.

FILE COMMANDS:

pwd:

DESCRIPTION:

Points the name of the current working directory

OPTIONS:

-help->displays help

mkdir:

DESCRIPTION:

Used to make/create the new directory.

SYNTAX:

mkdir [option] directory

OPTIONS:

-m ->sets permission mode]

cd:

DESCRIPTION:

Change the current directory to specified path.

SYNTAX:

cd [directory]

rmdir:

DESCRIPTION:

Used to remove/delete the empty directories.

SYNTAX:

rmdir [option]directory]


4/71

OPTIONS:

-output->diagnostics from each step

ls:

DESCRIPTION:

Used to list directory contents.

SYNTAX:

ls [option][file]

OPTIONS:

-a-> do not hide entries starting with author of each

-c-> list entries by columns

-d-> list directory entries

-f-> do not sort

-l-> use long history format

-r->recursive list

-s->sort by size

-v->sort by version

cat:

DESCRIPTION:

Concatenate files and prints on the standard output.

SYNTAX:

cat [option] [file]

OPTION:


5/71

-a->show all

-n->number of all output lines

-t->show tabs

cp:

DESCRIPTION:

copies files and directories.

SYNTAX:

cp [option] source text

OPTIONS:

-r->copy recursively

-u->copy only when the source is never

rm:

DESCRIPTION:

Removes files (or) directories.

SYNTAX:

rm[option]..file..

OPTION:

-f->never prompt

-r->recursive remove

-i->interactive operation

mv:

DESCRIPTION:

Move or remove files.

SYNTAX:


6/71

mv[options]..source text

OPTION:

-f->do not prompt

-u->update

-i->interactive operation

file:

DESCRIPTION:

Determines the file type.

SYNTAX:

file[-f file name]

wc:

DESCRIPTION:

Prints the number of bytes,words&lines.

SYNTAX:

wc[options][file]

OPTION:

-c-> prints the character count

-w->prints word count

-l-> print the line count

head:

DESCRIPTION:

Gives the first part of the file.

SYNTAX:head[options]..[file]


7/71

OPTIONS:

-c->print first N bytes

-n->print first n lines

tail:

DESCRIPTION:

Gives the last part of the file.

SYNTAX:

tail[option]..[file]

OPTIONS:

-c->print last N bytes

-n->print last N lines

GREATEST OF THREE NUMBERS

PROGRAM :

echo enter a value:read a

echo enter b value:

read b

echo enter c value:

read c

if[$a-gt $b-a $a-gt $c]

thenecho a=$a is greatest

elif[$b-gt$c]


8/71

then

echo b=$b is greatest

elseecho c=$c is greatest

fi

OUTPUT :

Enter a value:3

Enter a value:2

Enter c value:5

C=5 is greatest


9/71

ARITHMETIC OPERATION USING SWITCH CASE

PROGRAM :

echo 1.add

echo 2.sub

echo 3.mul

echo 4.div

echo enter your choice:

read n

echo enter a value:

read a

echo enter b value:

read b

case $ n in


10/71

1)c=expr $a+$b;;

2) c=expr $a-$b;;

3) c=expr $a*$b;;

4) c=expr $a/$b;;

esac

OUTPUT :

1.add

2.sub

3.mul

4.div

Enter your choice:3

Enter a value:3

Enter b value:5

The result is 15


11/71

STUDENT DETAILS USING WHILE LOOP

PROGRAM :

echo enter the no of students:

read n

count=1

while[$count-le$n]

do

echo enter the name:

read name

echo enter the no:

read no

echo enter the marks m1,m2,m3:

read m1

read m2

read m3


12/71

tot=explr $m1+$m2+sm3

avg=expr $ tot/3

echo name is:$ name

echo no is : $ no

echo avg is: $ avg

echo total is: $ tot

count=expr $ count+1

OUTPUT :

enter the name: student

enter the no:1

enter the marks m1,m2,m3:100 100 100

total is:300

avg is:100


13/71

SYSTEM CALLS USING FORK()

PROGRAM :

#include#include#include#includemain(){int pid;int a,b,t;

printf("\nEnter a and b :");scanf("%d%d",&a,&b);

pid=fork();if(pid


14/71

printf("\nParent process");wait(2);}

printf("pid value as %d\n",getpid());printf("A as:%d,B as:%d\n",a,b);}

OUTPUT :

Enter a and b : 4

5

Child process to do swapping

pid value as 6651

A as:5,B as:4

Parent process pid value as 6649

A as:4,B as:5


15/71

SYSTEM CALLS USING EXEC()

PROGRAM :

#include#include#include#includemain(int age ,char *args[]){int pid;pid=fork();

if(pid


16/71

}

OUTPUT :

Child Process

Child Completed


17/71

FILE CREATION

PROGRAM :

#include#include#include#include#includemain(){int fd;char buf1[100],buf2[]="hello everybody",fname[30];

printf("enter filename\n");scanf("%s",fname);fd = creat(fname,S_IRWXU);if(fd < 0){

printf("error in file creation");}else{write(fd,buf2,30);

printf("contents written successfully\n");

fd=open(fname,O_RDONLY);read(fd,buf1,30);

printf("contents is:\n%s\n",buf1);close(fd);


18/71

}}

OUTPUT :

jec@jec-desktop:~/uma$ cc systemcall3.c

jec@jec-desktop:~/uma$ ./a.out

enter filename

devi

contents written successfully

contents is:

hello everybody

jec@jec-desktop:~/uma$ cat devi

hello everybody


19/71

FILE ATTRIBUTES

PROGRAM:

#include

#include

#include

#include

int main()

{

struct stat statv;

if(stat("/home/staff/rajkumar/ww",&statv)==-1)

perror("stat");

else

{

printf("\n the file user id is: %d",statv.st_uid);

printf("\n the file group id is: %d",statv.st_gid);

printf("\n the file size is :%d",statv.st_size);

printf("\n the file inode number is :%d",statv.st_ino);

}

return (0);

}


20/71

OUTPUT:

the file user id is: 920

the file group id is: 921

the file size is :33

the file inode number is :8230320


21/71

CHANGE THE CONTEXT OF A PROCESS

PROGRAM:

#include

#include

#include

#include

#include

#include

int system (const char *cmd)

{

pid_t pid;

int status;

switch(pid=fork())

{

case -1:

return -1;

case 0:

execl ("/bin/sh","sh","-c",cmd,0);

perror ("execl");

exit(errno);

}

if(waitpid(pid,&status,0)==pid && WIFEXITED(status))

return WEXITSTATUS(status);

return -1;


22/71

}

int main()

{

int rc=0;

char buf[256];

do

{

printf("\n sh>");

if(!gets(buf))

break;

rc=system(buf);

}

while(!rc);

exit(rc);

}


23/71

OUTPUT:

sh>ls

a.out Desktop gg m3.txt odd.sh p2.c prim.sh sal.sh small.sh

bb fact.sh ha menu.sh p1o.c p2c.c raj simcd.c user.sh

big.sh ff hai mm.c p1pc.c p3.c rajkumar simls.c ww

chat1.c ff.sh io4.c mul.sh p1w.c p4.c rrr simmv.c

chat2.c fib.sh m2.txt mylist p1z.c pn.sh rr.sh small1.sh

sh>cal January 2009

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


24/71

SCANNING A DIRECTORY

PROGRAM:

#include

#include

//#include

#include

#include

#include

#if defined(BSD)&&I_POSIX_SOURCE

#include

typedef struct dirent Dirent;

#else

#include

typedef struct dirent Dirent;

#endif

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

{

Dirent* dp;

int i;

int cnt;

DIR *dir_fdesc;

while(--argc>0)

{

dir_fdesc=opendir(*++argv);

for(i=0;i


25/71

if(i)

printf("%s\n",dp->d_name);

//if(strcmp(dp->d_name,".")&&strcmp(dp->d_name,".."))

cnt++;

}

if(!cnt)

{

rmdir(*argv);

break;

}

rewinddir(dir_fdesc);

}

closedir(dir_fdesc);

}

}


26/71

OUTPUT:

/a.out /home/staff/rajkumar/ggpp.sh

hh.txt

rajkumar

gg


27/71

IMPLEMENTATION OF I/O SYSTEM CALLS

PROGRAM:

#include

#include

#include

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

{

char buf[256];

int fdesc, len;

while(--argc>0)

{

if(access(*++argv,F_OK)){

fdesc=open(*argv,O_WRONLY|O_CREAT,0744);

write(fdesc, "HELLO WORLD\n",12);

}

else

{

fdesc=open(*argv,O_RDONLY);while(len=read(fdesc,buf,256))

write(1,buf,len);

}

close(fdesc);

}

}

OUTPUT:


28/71

./a.out ram.c

[cse@localhost ~]$ vi ram.c

HELLO WORLD

SIMULATION OF UNIX COMMANDSPROGRAM :

#include


29/71

int main(){char c;int n;

printf("1.List of files in the directory\n2.List the lines\n3.Copy\n4.Remove\n");scanf("%d",&n);switch(n){case 1:system("ls");

break;case 2:system("grep argc john.c");

break;case 3:system("cp z c");

printf("the file is copied");

break;case 4:system("rm z c");

printf("the file is deleted");break;}}

OUTPUT :

1.List of files in the directory

2.List the lines

3.Copy

4.Remove

1

Ric.c


30/71

Sab.c

Sat.c


2.List the lines

3.Copy

4.Remove

2

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


2.List the lines

3.Copy

4.Remove

3

The file is copied


2.List the lines

3.Copy

4.Remove

4

The file is deleted

FIRST COME FIRST SERVE SCHEDULING

PROGRAM :

#include

int main()

{

int n,i,b[10],w[10],t[10];

float ae,aw,at;

printf("\n\tFIRST COME FIRST SERVE");

printf("\n\t-------------------------------------");

printf("\n\nENTER THE NUMBER OF JOBS : ");


31/71

scanf("%d",&n);

for(i=0;i


32/71

ENTER THE BURST TIME FOR THE JOB3 : 30

ENTER THE BURST TIME FOR THE JOB4 : 40

JOB BURSTTIME WAITING TIME TURNAROUNDTIME

-------------------------------------------------------------------------------------------------

1 10 0 10

2 20 10 30

3 30 30 60

4 40 60 100

AVERAGE BURSTTIME IS = 25.0000000

AVERAGE WAITING TIME IS = 24.9999992

AVERAGE TURNAROUNDTIME IS = 50.00000

SHORTEST JOB FIRST SCHEDULING

PROGRAM :

#include

int main()

{

int n,b[12],b1[12],j[12],w[12],e[12];

int i,k,t,t1;

float aw=0,ae=0;

printf("\n\tSHORTEST JOB FIRST\n");

printf("\t-----------------------------------");

printf("\n\nENTER THE NUMBER OF JOBS : ");


33/71

scanf("%d",&n);

for(i=0;i


34/71

OUTPUT :

SHORTEST JOB FIRST

-------------------------------

ENTER THE NUMBER OF JOBS: 3

ENTER THE BURST TIME FOR THE JOB1 = 30



JOB BURSTTIME WAITING TIME TURNAROUNDTIME


35/71

-----------------------------------------------------------------------------------------------

1 10 0 10

2 20 10 30

3 30 30 60

AVERAGE WAITING TIME IS = 13.333333

AVERAGE TURNAROUNDTIME IS = 33.333332

PRIORITY SCHEDULING

PROGRAM :

#include

int main(){

int i,j,n,temp,start[10],pri[10],finish[10],burst[10],sw=0,st=0;

printf("\nPRIORITY SCHEDULING\n");

printf("\nENTER THE NUMBER OF JOBS : ");

scanf("%d",&n);

start[0]=0;

for(i=0;i


36/71

for(j=i+1;j


37/71

4 2 6 10

8 3 10 18

AVERAGE TURNAROUND TIME IS 11.333333

AVERAGE WAITING TIME IS 5.333333

ROUND ROBIN SCHEDULING

PROGRAM :

#include

main()

{

int i, wa=0,cum[30],b[30],count=0,f=0,wait[30],b1[30];

int k,x[30],xx[30],a,t=0,w,c,n,max=0,wt=0,s,j;printf("\nROUND ROBIN SCHEDULING");

printf("\nENTER THE NUMBER OF JOBS : ");

scanf("%d",&n);

for(i=0;i


38/71

printf("\nENTER THE TIME SLICE FOR JOBS : ");

scanf("%d",&s);

k=1;

for(j=0;j


39/71

}

}

}

printf("\n WAITING TIME");

for(i=0;i


40/71

2 4 6 8 10 12

WAITING TIME

4

2

6

AVERAGE WAITING TIME IS 4.000000

AVERAGE EXECUTION TIME IS 8.00000

AVERAGE TURN AROUND TIME IS 12.000000

CHATTING PROGRAM

PROGRAM:

#include

#include

#include

#include

#include

#include

int main()

{

int mid;

char mess[20];

printf("\n\n INTER PROCESS COMMUNICATION USING MESSAGE QUEUE...");

mid=msgget(27,IPC_CREAT|0777);

if(mid==-1)

{

printf("\n\t invalid message id..");


41/71

exit(1);

}

printf("\n enter the message text:");

scanf("%s",mess);

msgsnd(mid,mess,strlen(mess),0);

printf("\n\n message has been sent..\n");

printf("\n the sent message is:%s\n",mess);

}

OUTPUT:

INTER PROCESS COMMUNICATION USING MESSAGE QUEUE...

enter the message text:haimessage has been sent..

the sent message is:hai


42/71

PROGRAM:

#include

#include

#include#include

#include

#include

int main()

{

int i,mid,len;

char mess[20];

printf("\n\n INTER PROCESS COMMUNICATION USING MESSAGE QUEUE...");

mid==msgget(27,IPC_CREAT|0777);

if(mid==-1)

{

printf("\n\t invalid message id..");

exit(1);

}


43/71

msgrcv(mid,mess,100,0,0);

printf("\n\n message has been recieved..\n");

printf("\n\n the recieved message is:%s\n",mess);

}

OUTPUT:

INTER PROCESS COMMUNICATION USING MESSAGE QUEUE...

message has been recieved..

the recieved message is:hai


44/71

PRODUCER CONSUMER PROBLEM

PROGRAM :

#include

#include

#include

#define bz 10

void producer();

void consumer(int y[bz]);

int wait(int);

int signal(int);

int i,n,in=0,out=0,s,c,w,sig,mutex=1,np[bz],buffer[bz],ne[bz];

main()

{

do

{

printf("\n 1.produce the items\n2.consumer the items\n");

scanf("%d",&s);

if(s==1)

{

producer();

}


45/71

if(s==2)

{

consumer(buffer);

}

printf("\nto continue ,press 1 or else press 0");

scanf("%d",&c);

}

while(c!=0);

}

void producer()

{

w=wait(mutex);

printf("\nthe semaphore value of wait is =%d\n",w);

printf("\nEnter the no of items:\n");

scanf("%d",&n);

printf("\nEnter the items");

for(i=0;i


46/71

{

buffer[i]=np[i];

printf("%d\n",buffer[in]);

in=(in+1)%bz;

}

sig=signal(mutex);

printf("the semaphore value of signal is %d\n",sig);

}

void consumer(int y[bz])

{

if(mutex)

printf("\n the semsphore value of wait is=%d\n",w);

if(in==out)

{

printf("\n do nothing\n");

}

printf("\nthe consumer consumes the items:\n");

for(i=0;i


47/71

int wait(int n)

{

n++;

return(n);

}

OUTPUT :

1.produce the items

2.consumer the items

1

The semaphore value of wait is =2

Enter the no of items:

3

Enter the items2

3

4

1

given items are:2

3

4

1

the producer produce the items:

2

3

4

1

the semaphore value of signal is 2


48/71

to continue ,press 1 or else press 01

1.produce the items

2.consumer the items

2

the semsphore value of wait is=2

the consumer consumes the items:

2

3

4

1

the semaphore value of signal is 2

to continue ,press 1 or else press 0


49/71

DINING PHILOSOPHERS PROBLEM

PROGRAM :

#includechar state[10],self[10],spoon[10];void test(int k){

if((state[(k+4)%5]!='e')&&(state[(k+1)]!='e')){state[k]='e';self[k]='s';spoon[k]='n';spoon[(k+4)%5]='n';}

}void pickup(int i){state[i]='h';

test(i);if(state[i]=='h')self[i]='w';

}void putdown(int i){state[i]='t';spoon[i]='s';spoon[i-1]='s';test((i+4)%5);test((i+1)%5);

}main(){int ch,a,n,i;printf("\nDINING PHILOSOPHERS PROBLEM :");for(i=0;i


50/71

printf("\n%d\t\t%c\t\t%c\t\t%c",i+1,state[i],self[i],spoon[i]);

printf("\n1.EXIT\n2.HUNGRY\n3.THINKING");

printf("\nEnter your choice :");

scanf("%d",&ch);

while(ch!=1)

{

switch(ch)

{

case 1:

return 0;

case 2:

printf("\nEnter which philosopher is hungry :");

scanf("%d",&n);

pickup(n-1);

break;

case 3:

printf("\nEnter which philosopher is thinking :");

scanf("%d",&n);

putdown(n-1);

break;

}

printf("State of each philosopher :");

printf("\nPHIL NO:\tTHINK/EAT\tSTATUS\t\tSPOON");

for(i=0;i


51/71

}

OUTPUT :

DINING PHILOSOPHERS PROBLEM :

Initial state of each philosopher :PHIL NO: THINK STATUS SPOON1 t s s2 t s s3 t s s4 t s s5 t s s

1.EXIT

2.HUNGRY3.THINKING

Enter your choice :2

Enter which philosopher is hungry :2State of each philosopher :PHIL NO: THINK/EAT STATUS SPOON1 t s n2 e s n3 t s s

4 t s s5 t s s

1.EXIT2.HUNGRY3.THINKING


Enter which philosopher is thinking :4State of each philosopher :

PHIL NO: THINK/EAT STATUS SPOON1 t s n2 e s n3 t s s4 t s n5 e s n

1.EXIT

2.HUNGRY

3.THINKING



52/71

PAGING

PROGRAM :

#include

main(){

int i,j,arr[100],pt[20],val,pgno,offset,phymem,fs,nf;

printf("Memory management -Paging\n");

printf("\n*****************************");

printf("\nEnter size of physical memory");

scanf("%d",&phymem);

for(i=20,j=0;i


53/71

OUTPUT :

Memory management -Paging

*****************************

Enter size of physical memory

64

Enter size of frame or page

8

No of frames available are 8

Enter page table

1

2

3

4

5

6

7

8

Enter page no

5

Enter offset

3

The physical address is :31


54/71

BEST FIT

PROGRAM :

#include

main()

{

int i,n,m,ps[100],bs[150],j,pi[50],bi[50],t;

printf("enter the no of blocks\n");

scanf("%d",&n);

printf("enter the block size\n");

for(i=1;i


55/71

}

for(i=1;i


56/71

225033004400enter no of process3enter process 1 size:225enter process 2 size:275enter process 3 size:325

process 0 is allocated to block of size 200process 1 is allocated to block of size 250process 2 is allocated to block of size 300

FIRST FIT

PROGRAM :

#include

main()

{

int i,n,m,ps[100],bs[150],j;


57/71

printf("enter the no of blocks\n");

scanf("%d",&n);

printf("enter the block size\n");

for(i=1;i


58/71

400

2

500

3

300

4

600

enter no of process

4

enter process 1 size:

200


250


350


550

process 1 is allocated to block of size 400



process 4 is not allowed

ORPHAN PROCESS

PROGRAM:

main()

{

if(fork()==0)

{


59/71

system("clear");

printf("CHILD:i am child & my number is :%d \n",getpid());

printf("CHILD: my parent is :%d \n",getppid());

}

else

{

printf("PARENT:i am parent:%d \n",getpid());

printf("PARENT: my parent is :%d \n",getppid());

}

}

OUTPUT:

CHILD:i am child & my number is :2663

CHILD: my parent is :1


60/71

PARENT-CHILD PROCESS

PROGRAM:

main()

{

if(fork()==0)

{

system("clear");

printf("CHILD:i am child & my number is :%d \n",getpid());

printf("CHILD: my parent is :%d \n",getppid());

}

else


61/71

{

printf("PARENT:i am parent:%d \n",getpid());

sleep(2);

printf("PARENT: my parent is :%d \n",getppid());

}

}

OUTPUT:

PARENT:i am parent:2682

CHILD:i am child & my number is :2683

CHILD: my parent is :2682

PARENT: my parent is :2434


62/71


63/71

{

dip=fork();

if(dip==0)

{

printf("2st Child Process Id is %d\n",getpid());

printf("2st Child Process TERMINATING FROM THE MEMORY\n");

}

else

{

cpid=wait(0);

printf("Child with pid:%d is died\n",cpid);

cpid=wait(0);

printf("Child with pid:%d is died\n",cpid);

}

}

}

OUTPUT:

1st Child Process Id is 2699

1st Child Process TERMINATING FROM THE MEMORY

2st Child Process Id is 2700

2st Child Process TERMINATING FROM THE MEMORY

Child with pid:2699 is died

Child with pid:2700 is died


64/71

ZOMBIE PROCESS

PROGRAM:

main()

{

int pid=fork();

system("clear");

printf("my number is %d\n",getpid());

if(pid>0)

{

printf("My parent number is %d\n",getpid());

sleep(05);

}


65/71

}

OUTPUT:

my number is 2716

My parent number is 2716

my number is 2717


66/71

SINGLY LINKED LIST

PROGRAM:

#include

#include

struct queue

{

int no;

struct queue *next;

}

*p,*q,*temp,*head=NULL;

int main()

{

int ch;


67/71

void enqueue();

void dequeue();

void list();

while(1)

{

printf("\n 1...Insert");

printf("\n 2...Delete");

printf("\n 3...Display");

printf("\n 4...Quit");

printf("Enter your choice:");

scanf("%d",&ch);

switch(ch)

{

case 1:

enqueue();

break;

case 2:

dequeue();

break;

case 3:

list();

break;

default:

exit(0);

}

}

return 0;

}

void enqueue()


68/71

{

temp=(struct queue*) malloc(sizeof(struct queue));

printf("\nEnter number to be inserted:");

scanf("%d",&temp->no);

if(head==NULL)

{

head=temp;

temp->next=NULL;

}

else

{

q=head;

while(q->next!=NULL)

{

q=q->next;

}

q->next=temp;

temp->next=NULL;

}

printf("\n\nElement %d is pushed into queue",temp->no);

}

void dequeue()

{

p=head;

if(head==NULL)

{

printf("\n\nQueue is empty");

}

else


69/71

{

head=head->next;

printf("\n\nElement %d is dequeued from queue",p->no);

free(p);

}

}

void list()

{

if(head==NULL)

{

printf("\n\nQueue is empty");

}

else

{

p=head;

printf("\n\nElement present in queue\n");

while(p!=NULL)

{

printf("%d\t",p->no);

p=p->next;

}

}

}


70/71

OUTPUT:

1...Insert

2...Delete

3...Display

4...Quit

Enter your choice:1

Enter number to be inserted:12

Element 12 is pushed into queue

1...Insert

2...Delete

3...Display

4...Quit

Enter your choice:1



1...Insert

2...Delete

3...Display


71/71

4...Quit

Enter your choice:1



1...Insert

2...Delete

3...Display

4...Quit

Enter your choice:2

Element 12 is dequeued from queue

1...Insert

2...Delete

3...Display

4...Quit

Enter your choice:3

Element present in queue

13 14

1...Insert

2...Delete

3...Display

4...

Quit

Documents

OS Programs for Record