os lab completed

8/3/2019 Os Lab Completed

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CS 2257 OPERATING SYSTEMS

LAB

#include<stdio.

1A. IMPLEMENTATION OF FORK

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LAB

OUTPUT :

Fork domain the parent ( Pid=4124)

I am the child (cid=4125)of(pid=4124)

0

Parent waiting here for child

1

2

3

4

5

6

7

8

9

Child finished parent quiting too

1B. IMPLEMENTATION OF STAT

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LAB

PROGRAM

#include<stdio.h>#include<sys/types.h>#include<sys/stat.h>

#include<unistd.h>#include<fcntl.h>main(){int fd1,fd2,n;char source[30],ch[5];struct stat s,t,w;fd1=creat("test.txt",0644); printf("Enter the file to be copied\n");scanf("%s",source);fd2=open(source,0);

if(fd2==-1){ perror("file doesnot exist");exit(0);}while((n=read(fd2,ch,1))>0)write(fd1,ch,n);close(fd2);stat(source,&s); printf("Source file size=%d\n",s.st_size);fstat(fd1,&t); printf("Destination File size before closing=%d\n",t.st_size);close(fd1);fstat(fd1,&w); printf("Destination File Size after closing=%d\n",w.st_size);}

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LAB

IMPLEMENTATION OF STAT

OUTPUT:

"soce.c" 30L, 614C written

[ivcsea@localhost karthik]$ cc soce.c

[ivcsea@localhost karthik]$ ./a.out

Enter the file to be copied

soce.c

Source file size=614

Destination File size before closing=614

Destination File Size after closing=3

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LAB

1C. IMPLEMENTATION OF READ DIRECTORIES

PROGRAM:

#include<stdio.h>

#include<sys/types.h>

#include<dirent.h>

#include<sys/stat.h>

#define REQUEST_DIR "/"

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

FILE *fp;

DIR *dirp;

struct dirent *dp;

struct stat buf;

dirp=opendir(REQUEST_DIR);

chdir(REQUEST_DIR);

while((dp = readdir(dirp))!=NULL){

if(stat(dp->d_name,&buf)==-1)

perror("stat\n");

if(S_ISDIR(buf.st_mode))

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

else if(S_ISREG(buf.st_mode))

printf("%s is a regular file \n",dp->d_name);

}

(void)closedir(dirp);

}

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LAB

IMPLEMENTATION OF READ DIRECTORIES

OUTPUT:

. is a directory

.. is a directory

lost+found is a directory

dev is a directory

proc is a directory

.autofsck is a regular file

var is a directory

tmp is a directory

etc is a directory

root is a directory

usr is a directory

boot is a directory

bin is a directory

home is a directory

initrd is a directory

lib is a directory

mnt is a directory

opt is a directory

sbin is a directory

misc is a directory

.automount is a directory

tftpboot is a directory

.fonts.cache-1 is a regular file

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LAB

2. IMPLEMENTATION OF I/O SYSTEM CALLS

PROGRAM:

#include<stdio.h>int main(){char c;FILE *f;f=fopen("a.txt","w"); printf("\nEnter the TEXT to File:\n");while((c=getchar())!='\n')fprintf(f,"%c",c);

fclose(f); printf("\n\nContent of the File:\n\n");f=fopen("a.txt","r");while(fscanf(f,"%c",&c)!=EOF) printf("%c",c);fclose(f);return(0);}

IMPLEMENTATION OF I/O SYSTEM CALL

OUTPUT :

Enter the TEXT to File:SBC Engineering College

Content of the File:SBC Engineering College

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LAB

3A. IMPLEMENTATION OF ‘LS’ COMMAND

PROGRAM:

#include<sys/wait.h>

#include<unistd.h>

int fatal(char str[]);

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

{

int pid;

switch(pid=fork())

{

case -1:

fatal("fork error 1");

case 0:

execl("/bin/ls","ls",argv[1],NULL);

fatal("exec call 1");

}

while(wait((int *)0) !=pid);

}

int fatal(char str[])

{

printf("%s\n");

exit(1);

}

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LAB

IMPLEMENTATION OF ‘LS’ COMMAND

OUTPUT:

"ls2.c" 22L, 331C written[ivcsea@localhost karthik]$ cc ls2.c

[ivcsea@localhost karthik]$ ./a.out

a.out fork.c grep.sh ls1.c music.txt start text3.txt

arasua.c gg.c index.c ls2.c new.c sys.c

a.txt gr.c io.c ls.c second test.txt

first grep.c link.c music.dat soce.c text1.txt

[ivcsea@localhost karthik]$

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LAB

3B. IMPLEMENTATION OF “GREP” COMMAND

PROGRAM:

#include<stdio.h>main()

{char names[5][10],temp[10],c;int i,j,k,n=0;char ch,cha; printf("Enter names to stop type keyword END \n");scanf("%s",names[n]);while(strcmp(names[n],"END")>0){n++;scanf("%s",names[n]);

}

printf("\n");for(i=0;i<n;i++) printf("%10s",names[i]); printf("\n Enter the name you want to search \n");scanf("%s",temp); printf("\n string found \n");for(i=0;i<n;i++){if(strcmp(names[i],temp)==0){printf("\n%s\n",names[i]);

}} printf("\n Enter the first character to search \n");ch=getchar();scanf("%c",&ch); printf("\n sentence started with \n");for(i=0;i<5;i++)if(names[i][0]==ch){printf("\n %s \n",names[i]);break;

}

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LAB

printf("\nEnter the last character to search \n");cha=getchar();scanf("%c",&cha); printf("\n sentence end with \n");for(i=0;i<5;i++)

for(j=0;j<10;j++)if(names[i][j]==cha){printf("\n %s \n",names[i]);

break;}

OUTPUT:

"gg1.c" 50L, 887C written[ivcsea@localhost karthik]$ cc gg1.c

[ivcsea@localhost karthik]$ ./a.outEnter names to stop type keyword ENDcar war tar nar war END

car war tar nar war Enter the name you want to searchwar

string foundwar war

Enter the first character to searchc

sentence started withcar

Enter the last character to searchr

sentence end withcar war tar nar war [ivcsea@localhost karthik]$

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LAB

4A. FIRST COME FIRST SERVE SCHEDULING

ALGORITHM

PROGRAM:

#include<stdio.h>int main(){

int n,i,j;float s,t[20],w[20],tar[20],avgw,avgtr;printf("\n Enter the number of process");scanf("%d",&n);for (i=1;i<=n;i++){

printf("\n Enter the time for %d process ",i);scanf("%f",&t[i]);

}printf("\n\n");w[1]=0;for(i=2;i<=n;i++)w[i]=w[i-1]+t[i-1];for(i=1;i<=n;i++)tar[i]=w[i]+t[i];printf("\n process no \t cpu time \t waiting time \t tar time \n");for(i=1;i<=n;i++)printf("\n\t %d\t %f\t%f\t%f\t\n",i,t[i],w[i],tar[i]);s=0;for(i=1;i<=n;i++)s+=w[i];avgw=s/n;printf("\n The avg waiting time is %f",avgw);s=0;for(i=1;i<=n;i++)s+=tar[i];avgtr=s/n;printf("\n The avg turn around time is %f\n",avgtr);

}

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LAB

FIRST COME FIRST SERVE

OUTPUT:

Enter the number of process: 3

Enter the time for 1 process: 3Enter the time for 2 process: 4Enter the time for 3 process: 5

PROCESS CPU TIME WAITING TIME TAR TIME

1 3.000000 0.000000 3.0000002 4.000000 3.000000 7.0000003 5.000000 7.000000 12.00000

The avg waiting time is:3.333333The avg turn around time is:7.333333

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LAB

4B. SHORTEST JOB FIRST SERVE SCHEDULING

PROGRAM:

#include<stdio.h>main(){

int n,i,j,p[20],temp1;float s,t[20],w[20],tar[20],avgw,avgtr,temp;printf("\n enter the no of process\t");scanf("%d",&n);for(i=1;i<=n;i++){

printf("\n enter the time for %d process\t",i);scanf("%f",&t[i]);p[i]=i;

}printf("\n\n");for(i=1;i<=n;i++)for(j=1;j<=n;j++){

if(t[i]<t[j]){

temp=t[i];temp1=p[i];t[i]=t[j];p[i]=p[j];t[j]=temp;p[j]=temp1;

}}w[1]=0;for(i=2;i<=n;i++)w[i]=w[i-1]+t[i-1];for(i=1;i<=n;i++)tar[i]=w[i]+t[i];printf("\n process no \t cpu time \t waiting time \t tar time \t\n");for(i=1;i<=n;i++)printf("\n\t%d\t%f\t%f\t%f\t\n",p[i],t[i],w[i],tar[i]);s=0;for(i=1;i<=n;i++)s+=w[i];avgw=s/n;

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LAB

printf("\n the average waiting time is %f",avgw);s=0;for(i=1;i<=n;i++)s+=tar[i];avgtr=s/n;

printf("\n the avg time around time %f,avgtr");}

OUTPUT:

Enter the number of process: 4

Enter the time for 1 process: 6Enter the time for 2 process: 3Enter the time for 3 process: 5Enter the time for 4 process: 7

PROCESS CPU TIME WAITING TIME TAR TIME

2 3.000000 0.000000 3.0000003 5.000000 3.000000 8.0000001 6.000000 8.000000 14.0000004 7.000000 14.000000 21.000000

The avg waiting time is:6.250000The avg turn around time is: 6.250000

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LAB

5A. PRIORITY SCHEDULING

PROGRAM:

#include<stdio.h>struct process{int pno,btime,prior;} p[100],temp;int main(){int i,j,wt=0,n;float awt=0,ata=0; printf("\n\n Enter the number of process\t");

scanf("%d",&n); printf("\nEnter thr process number,priority,Execution time:");for(i=0;i<n;i++){ printf("\n Process\t");scanf("%d",&p[i].pno); printf("\nPriorrity\t");scanf("%d",&p[i].prior); printf("\nTime\t");scanf("%d",&p[i].btime);}for(i=0;i<n-1;i++){for(j=i+1;j<n;j++){if(p[j].prior<p[i].prior){temp=p[i]; p[i]=p[j]; p[j]=temp;}}} printf("\n pno\tpriority\t execution_time\t waiting_time\tturnaround_time");for(i=0;i<n;i++){ printf("\n%10d\t%20d\t%10d\t%10d\t%10d",p[i].pno,p[i].prior,p[i].btime,wt,wt+p[i].btime);awt=awt+wt;ata=ata+wt+p[i].btime;

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LAB

wt=wt+p[i].btime;} printf("\n average waiting time %f",awt/n); printf("\n average turn around time is %f",ata/n);exit(0);

getch();return 0;}

OUTPUT:

-------------Enter the number of process 4Enter thr process number,priority,Execution time:Process 1Priority 3

Time 4

Process 2Priority 2Time 1

Process 3Priorrity 1Time 5

Process 4Priorrity 4Time 2

pno priority execution_time waiting_time turnaround_time3 1 5 0 52 2 1 5 61 3 4 6 104 4 2 10 12

average waiting time 5.250000average turn around time is 8.250000

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LAB

5B. ROUND ROBIN SCHEDULING

PROGRAM:

#include<stdio.h>

typedef struct{int pno,btm,sbtm,wtm,lst;}spcb;int main(){int pp=-1,ts,flag,count,pno,ptm=0,i,n,twt=0,tttm=0;spcb proc[100];printf("\n \t\t\t ROUND ROBIN SCHEDULING\t\t\t");printf("\n Enter the number of process:");scanf("%d",&n);

printf("\n Enter the time slice:");scanf("%d",&ts);printf("\n Enter the burst time for %d process \n",n);for(i=0;i<n;i++){scanf("%d",&proc[i].btm);proc[i].wtm=proc[i].lst=0;proc[i].pno=i+1;proc[i].sbtm=proc[i].btm;}printf("\n \t\t\t PROCESS SYCHRONISATION \t\t\t");do{flag=0;for(i=0;i<n;i++)if((count=proc[i].btm)>0){flag=1;count=(count>=ts)?ts:count;printf("\n process %d from %d",proc[i].pno,ptm);printf("t0 %d \n",(ptm+=count));proc[i].btm-=count;if(pp!=i){proc[i].wtm+=ptm-proc[i].lst-count;proc[i].lst=ptm;

}}

}while(flag);

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LAB

ROUND ROBIN SCHEDULING

printf("\t\t process no. \t waiting time \t turn around time\n");for(i=0;i<n;i++)

{twt+=proc[i].wtm;tttm+=proc[i].wtm+proc[i].sbtm;printf("\n \t\t%d\t\t%d\t\t",proc[i].pno,proc[i].wtm);printf("%d",proc[i].wtm+proc[i].sbtm);

}printf("\n Total average time =%d",twt);printf("\n average waiting time=%f",(float)twt/n);printf("\n Total turn around time =%d",tttm);printf("\n Average turn around time =%f",(float)tttm/n);}

"nimmi1.c" 56L, 1412C written[ivcsea@localhost ivcse]$ cc nimmi1.c[ivcsea@localhost ivcse]$ ./a.out

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LAB

OUTPUT:

ROUND ROBIN SCHEDULING

Enter the number of process:3

Enter the time slice:2

Enter the burst time for 3 process643

PROCESS SYCHRONISATIONprocess 1 from 0t0 2

process 2 from 2t0 4






process no. waiting time turn around time

1 7 132 6 103 8 11

Total average time =21average waiting time=7.000000Total turn around time =34Average turn around time =11.333333

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LAB

6A. INTERPROCESS COMMUNICATION USING

SHARED MEMORY

PROGRAM:

#include<stdio.h>

#include<sys/shm.h>

#include<sys/ipc.h>

int main()

{

int child,shmid,i;

char*shmptr;

child=fork();

if(!child)

{

shmid=shmget(2041, 32, 0666 | IPC_CREAT);

shmptr=shmat(shmid, 0, 0);

printf("\n\n SHARED MEMORY\n\n");

printf("\nPARENT WRITING\n");

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

{

shmptr[i]='a' + i;

putchar(shmptr[i]);

}

printf("\n\n%s",shmptr);

wait(NULL);

}

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LAB

INTERPROCESS COMMUNICATION USING

SHARED MEMORY

else

{

shmid=shmget(2041, 32, 0666);

shmptr=shmat(shmid, 0, 0);

printf("\nCHILD IS READING:\n");

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

putchar(shmptr[i]);

shmdt(NULL);

shmctl(shmid,IPC_RMID,NULL);

}

return 0;

}

OUTPUT:

SHARED MEMORY

PARENT WRITING:

abcdefghij

CHILD IS READING:

abcdefghij

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LAB

6B. INTERPROCESS COMMUNICATION USING

PIPES

PROGRAM:

#include<stdio.h>

int main()

{

int fd[2],child;

char a[10];

printf("\nPIPES:\n");

printf("\nEnter the srting to enter into the pipe:");scanf("%s",a);

pipe(fd);

child=fork();

if(!child)

{

close(fd[10]);

write(fd[1],a,5);

wait(0);

}

else

{

close(fd[1]);

read(fd[0],a,5);

printf("\nThe string retrieved from the pipe is %s\n",a);

}

return 0;

}

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LAB

INTERPROCESS COMMUNICATION USING

PIPES

OUTPUT:

PIPES:

Enter the string to enter into the pipe:computer

The string retrieved from the pipe is:computer

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LAB

7. PRODUCER-CONSUMER PROBLEM

PROGRAM

#include<stdio.h>

#define eof '\n'

int main()

{

int n;

char buf[10],item[10];

int i,full=0,empty=5,mutex=1;for(i=0;i<5;i++)

{

printf("\n produce an item in buffer");

while((item[i]=getchar())!=eof)

if((mutex==1)&&(empty!=0))

{

buf[i]=item[i];

full=full+1;

empty=empty-1;

}

printf("\n\t\tfull=%d",full);

printf("\n\t\tempty=%d",empty);

printf("\nthe item moved in the buffer \n");

}

if((mutex==1)&&(empty==5))

printf("\n the buffer is full \n");

for(i=0;i<5;i++)

{

if((mutex==1)&&(full!=0))

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LAB

PRODUCER-CONSUMER PROBLEM

{

printf("\n %c",buf[i]);

full=full-1;

empty=empty+1;

}

printf("full=%d",full);

printf("empty=%d",empty);

}

if((mutex==1)&&(empty==5)) printf("\n the buffer is empty");

}

OUTPUT:

produce an item in buffer5

full=1

empty=4

the item moved in the buffer


full=2

empty=3



full=3

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LAB

PRODUCER-CONSUMER PROBLEM

empty=2



full=4

empty=1



full=5

empty=0


5full=4empty=1

3full=3empty=2

4full=2empty=3

7full=1empty=4

2full=0empty=5

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LAB

8A. FIRST FIT ALGORITHM

PROGRAM:


{int a,b[50],i,s,c,n;printf("\n\t\t\t FIRST FIT\n");printf("\n Enter the no of memory process\t");scanf("\n %d",&a);printf("\n Enter the required no of slots \n");for(i=1;i<=a;i++){scanf("\n %d",&b[i]);}

do

{printf("\n Enter the job size \t");scanf("\n %d",&s);for(i=1;i<=a;i++){if(s<=b[i]){printf("\n %d is allocated to slot %d of size %d\n",s,i,b[i]);b[i]=b[i]-s;printf("\n Remaining space of slot %d is %d\n",i,b[i]);n=1;break;}elsecontinue;}if(n!=1)printf("\n %d is not allocated to any slot",s);n=0;printf("\n 1.Enter the job\n2.Exit\n");printf("\n Enter your choice\t");scanf("\n %d",&c);}while(c==1);}

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LAB

OUTPUT:

FIRST FIT

Enter the no of memory process 3Enter the required no of slots36755

Enter the job size 5555 is allocated to slot 2 of size 67

Remaining space of slot 2 is 121.Enter the job

2.Exit

Enter your choice 1

Enter the job size 2323 is allocated to slot 3 of size 55Remaining space of slot 3 is 321.Enter the job2.ExitEnter your choice 2

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LAB

8B. WORST FIT ALGORITHM

PROGRAM:

#include<stdio.h>

main()

{

int a[50],n,i,j,p,t;

printf("\n WORST FIT ALGORITHM");

printf("\n\n Enter Unallocated space \n");

scanf("\n %d",&n);

printf("\n Enter the memory space \n");

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

scanf("%d",&a[i]);printf("\n Enter the time memory required space");

scanf("%d",&p);

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

for(j=i+1;j<=n;j++)

if(a[i]>a[j])

{

t=a[i];

a[i]=a[j];

a[j]=t;

}

printf("\n THE WORST FIT PROCESS %d\n",a[i]);

}

30




LAB

OUTPUT:

WORST FIT ALGORITHM

Enter Unallocated space

3

Enter the memory space

34

43

23

Enter the time memory required space20

THE WORST FIT PROCESS 43

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LAB

8C. BEST FIT ALGORITHM

PROGRAM:


{int a,b[50],i,j,temp,s;printf("\n BEST FIT ALGORITHM");printf("\n Enter the number of allocated process");scanf("%d",&a);printf("\n Enter the unallocated space in KB");for(i=1;i<=a;i++)scanf("%d",&b[i]);for(j=j+1;j<=a;j++)if(b[i]>b[j]){

temp=b[j];b[j]=b[i];b[i]=temp;}printf("\n Enter the space for required process\n");scanf("%d",&s);for(i=1;i<a;i++){if(s<=b[i]){printf("\n \n The best fit for required space is %d \n",b[i]);break;}}}

32




LAB

OUTPUT:

BEST FIT ALGORITHM

Enter the number of allocated process3Enter the unallocated space in KB122334Enter the space for required process20

The best fit for required space is 23

33




LAB

9. MEMORY ARRANGEMENT-II

PAGE REPLACEMENT

PROGRAM:

#include<stdio.h>

int m,n,i,j,k,flag,count=0,refer[100],page_frame[100][2],fault=0,min,no_frames;

void replace(int z)

{

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

{

flag=1;

for(j=0;j<no_frames;j++)

if(refer[i]==page_frame[j][0])

{

m=j;

flag=0;

}

if(flag)

{

fault++;

min=32000;


if(page_frame[j][1]<min)

{

min=page_frame[j][1];

k=j;

}

page_frame[k][0]=refer[i];

page_frame[k][1]=++count;


printf("\n %d",page_frame[j][0]);

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LAB

printf("\n");

}

else

{

printf("No page fault\n");

if(z==2)

page_frame[m][1]=++count;

}

}

printf("Number of page fault is: %d \n",fault);

}

int main()

{

printf("\n Enter the number of reference: ");

scanf("%d",&n);

printf("\n Enter the number of frames: ");

scanf("%d",&no_frames);

printf("\n Enter the reference string: ");

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

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

printf("\n FIFO ALGORITHM\n");

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

{

page_frame[i][0]=-1;

page_frame[i][1]=count;

}

replace(1);

fault=0;

count=0;

printf("\n\t\t\tLRU ALGORITHM\n");

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

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LAB

{

page_frame[i][0]=-1;

page_frame[i][1]=count;

}

replace(2);

return 0;

}

OUTPUT:

Enter the number of reference: 20

Enter the number of frames: 3

Enter the reference string: 7,0,1,2,0,3,0,4,2,3,0,3,2,1,2,0,1,7,0,1

FIFO ALGORITHM

7 -1 -1

7 0 -1

7 0 1

2 0 1

NO page fault

2 3 -1

2 3 0

4 3 0

4 2 0

4 2 3

0 2 3

NO page fault

NO page fault

0 1 3

0 1 2

NO page fault

NO page fault

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LAB

7 1 2

7 0 2

7 0 1

Number of page fault is : 15

LRU ALGORITHM

7 -1 -1

7 0 -1

2 0 1

NO page fault

2 0 3

NO page fault

4 0 3

4 0 2

4 3 2

0 3 2

NO page fault

NO page fault

1 3 2

NO page fault

1 0 2

NO page fault

1 0 7

NO page fault

NO page fault

Number of page fault is: 12

37




LAB

IMPLEMENTATION OF FILE ALLOCATION TECHNIQUE USING INDEX

PROGRAM:

#include<stdio.h>

#include<stdlib.h>

typedef struct student

{

int rno;

char name[20];

int age;

int marks;

}ss;

typedef struct index

{

int irno;

long offset;

int flag;

}is;

FILE *ptr=NULL;

FILE *iptr=NULL;

void main()

{

int ch;

int status;

int rno;

char op1='y';

ptr=fopen("std.dat","wb");

if(ptr==NULL)

{

printf("file not found");

exit(0);

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LAB

}

iptr=fopen("std1.dat","wb");

if(iptr==NULL)

{

printf("file not found");

exit(0);

}

fclose(ptr);

fclose(iptr);

while((op1='y')||(op1='Y'))

{

printf("\n 1.add");

printf("\n 2.delete");

printf("\n 3.modify");

printf("\n 4.search");

printf("\n 5.display");

printf("\n 6.exit \n");

printf("enter your option (1-6):");

scanf("%d",&ch);

switch(ch)

{

case 1:

status=ADD_RECORD();

if(status==1)

printf("record successfully added");

else

printf("record could not be added");

break;

case 2:

printf("enter the rollno of the record to delete");

scanf("%d",&rno);

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LAB

status=DELETE_RECORD(rno);

if(status==1)

printf("record deleted successfully");

else

printf("record could not be deleted");

break;

case 3:

printf("enter rollno of the record to modify:");

scanf("%d",&rno);

status=MODIFY_RECORD(rno);

if(status==1)

printf("record modified successfully");

else

printf("record could not be modified");

break;

case 4:

printf("enter the rollno to search:");

scanf("%d",&rno);

status=SEARCH_RECORD(rno);

if(status==1)

printf("record found");

else

printf("record not found");

break;

case 5:

DISPLAY_ALL_RECORD();

break;

case 6:

exit(0);

}

fflush(stdin);

40




LAB

printf("\n do you want to continue(Y/N):");

scanf("%c",&op1);

if((op1=='n')||(op1=='N'))

break;

}

}

int ADD_RECORD()

{

ss stud;

is ind;

long offset;

ptr=fopen("std.dat","rb+");

if(ptr==NULL)

{

printf("file not opened");

return -1;

}

while(fread(&stud,sizeof(stud),1,ptr));

offset=ftell(ptr);

printf("rollno:");

scanf("%d",&stud.rno);

printf("name of the student:");

scanf("%s",&stud.name);

printf("age:");

scanf("%d",&stud.age);

printf("marks:");

scanf("%d",&stud.marks);

fwrite(&stud,sizeof(stud),1,ptr);

fclose(ptr);

iptr=fopen("std1.dat","ab+");

if(iptr==NULL)

41




LAB

{


return -1;

}

ind.irno=stud.rno;

ind.offset=offset;

ind.flag=1;

fwrite(&ind,sizeof(ind),1,ptr);

fclose(iptr);

SORT_FILE();

return 1;

}

int DELETE_RECORD(int rno)

{

is ind;

int flag=0;

iptr=fopen("std1.dat","rb+");

if(iptr==NULL)

{


return -1;

}

while(fread(&ind,sizeof(is),1,iptr));

{

if(ind.irno==rno)

{

flag=1;

}

}

if(flag==1)

{

42




LAB

ind.flag=-1;

fseek(iptr,-(long) sizeof(is),SEEK_CUR);

fwrite(&ind,sizeof(is),1,iptr);

fclose(iptr);

return 1;

}

fclose(iptr);

return -1;

}

int MODIFY_RECORD(int rno)

{

is ind;

ss stud;

int flag=0;


if(iptr==NULL)

{


return -1;

}

while(fread(&ind,sizeof(is),1,iptr))

{

if(ind.irno==rno)

{

flag=1;

break;

}

}

fclose(iptr);

if(flag==1)

{

43




LAB


if(ptr==NULL)

{


return -1;

}

fseek(ptr,ind.offset,SEEK_SET);

fread(&stud,sizeof(stud),1,ptr);

printf("enter name:");

scanf("%s",stud.name);

printf("age:");

scanf("%d",&stud.age);

printf("marks:");

scanf("%d",&stud.marks);

fseek(ptr,-(long) sizeof(stud),SEEK_CUR);

fwrite(&stud,sizeof(stud),1,ptr);

fclose(ptr);

return 1;

}

return -1;

}

int SEARCH_RECORD(int rno)

{

is ind;

ss stud;


if(iptr==NULL)

{


return -1;

}

44




LAB


{

if(ind.irno==rno && ind.flag==1)

{

fclose(iptr);

return 1;

}

}

fclose(iptr);

return -1;

}

int DISPLAY_ALL_RECORD()

{

is ind;

ss stud;


if(iptr==NULL)

{


return -1;

}


if(ptr==NULL)

{


return -1;

}


{

if(ind.flag==1)

{

45




LAB

fseek(ptr,ind.offset,SEEK_SET);

fread(&stud,sizeof(stud),1,ptr);

printf("\n\n rollno:%d",stud.rno);

printf("\n name:%s",stud.name);

printf("\n age:%d",stud.age);

printf("\n marks:%d",stud.marks);

}

}

fclose(iptr);

fclose(ptr);

return 1;

}

int SORT_FILE()

{

int size;

int i,j,flag=0;

is ind,ind_temp;

ss stud;


if(iptr==NULL)

{


return -1;

}


if(ptr==NULL)

{


return -1;

}

size=0;

46




LAB


size++;

fclose(iptr);

iptr=fopen("std.dat","rb+");

if(iptr==NULL)

{


return -1;

}

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

{

flag=0;

for(j=0;j<size-(i+1);j++)

{

fseek(iptr,j*sizeof(is),SEEK_SET);

fread(&ind,sizeof(is),1,iptr);

fread(&ind_temp,sizeof(is),2,iptr);

if(ind.irno>ind_temp.irno)

{

fseek(ptr,j*sizeof(is),SEEK_SET);

fwrite(&ind_temp,sizeof(is),1,ptr);

fseek(iptr,(j+1)*sizeof(is),SEEK_SET);

fwrite(&ind,sizeof(is),1,iptr);

flag=1;

}

if(flag==0)

break;

}

fclose(iptr);

return 1;

}

47




LAB

OUTPUT:

1. Add

2. Delete

3. Modify

4. Search

5. Display

6. Exit

Enter your option(1-6):1

Rollno:01

Name of the student:muni

Age:25

Marks:78

Record Successfully Added

Do You want to continue(Y/N):y

1. Add

2. Delete

3. Modify

4. Search

5. Display

6. Exit


Rollno:02

Name of the student:raj

Age:25

Marks:72

Record Successfully Added



Enter the rollno of the record to delete02

Record could not be deleted


48




LAB


Enter roll no of the record to modify:01

Enter name:muni

Age:30

Marks:80

Record modified successfully


1. Add

2. Delete

3. Modify

4. Search

5. Display

6. Exit


Rollno:01

Name:muni

Age:30

Marks:80

Do You want to continue(Y/N):6

49




LAB

IMPLEMENTATION OF FILE ALLOCATION TECHNIQUE USING LINKED

PROGRAM

#include<stdio.h>

#include<conio.h>

typedef struct list

{

int val;

struct list *next;

}item;

item *additem(item *entry, int value);

void printitems(void);

item *head=NULL;

item *tail=NULL;

main(int argc,char *argv[])

{

clrscr();

tail=additem(tail,5);



printitems();

}

item *additem(item *entry, int value)

{

item *newitem;

newitem=(item*)malloc(sizeof(item));

if(entry==NULL)

{

head=newitem;

printf("first list_item in list\n");

}

50




LAB

else

{

entry->next=newitem;

printf("adding %d to list. Last value was %d \n",value,entry->val);

}

newitem->val=value;

newitem->next=NULL;

return newitem;

}

void printitems(void)

{

item *ptritem;

for(ptritem=head;

ptritem!=NULL;

ptritem=ptritem->next){

printf("Value is %d\n",ptritem->val);

getch();

}

}

51




LAB

OUTPUT:

adding 5 to list. Last value was 2

adding 7 to list. Last value was 5

2

5

7

os lab completed

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