QUES.1 WRITE A PROGRAM (USING FORK() AND EXEC()) WHERE PARENT AND CHILD EXECUTE:

A) SAME PROGRAM, SAME CODE

CODE:#include<stdio.h>int main(){int pid;pid=fork();if(pid==-1){printf("fork failed\n");exit(-1);}else{printf("my process identifier =%d \n",pid);exit(0);}}

Output:[user3@localhost user3]$ cc p1a.c[user3@localhost user3]$ ./a.outmy process identifier =0my process identifier =8217

B) SAME PROGRAM, DIFFERENT CODE

CODE:#include<stdio.h>int main(){ int pid; pid=fork(); if(pid==-1) { printf("fork failed\n"); exit(-1); } else if(pid==0) { printf("i am a child process\n"); exit(0); } else if(pid>0) { printf("i am a parent process\n"); exit(0); }}

OUTPUT:[user3@localhost user3]$ cc p1b.c[user3@localhost user3]$ ./a.outi am a child processi am a parent process

C) DIFFERENT PROGRAMS

CODE:#include<stdio.h>int main(){ int pid; pid=fork(); if(pid==-1) { printf("fork failed\n"); exit(-1); } else if(pid==0) { execlp("/bin/ls","ls",NULL); exit(0); } else if(pid>0) { printf("i am a parent process\n"); exit(0); }}

OUTPUT: [user3@localhost user3]$ cc p1c.c

[user3@localhost user3]$ ./a.out03 cient.c client2.out new.c richsan serv.c tanvi26 clent.c client.c p1a.c ser1.c server1.c5.c client1.c clnt1.out p1b.c ser1.out server1.outa.out client1.out clnt.c p1c.c ser.c server2.outi am a parent process

D) PARENT WILL FAIL FOR CHILD CHILD TO TERMINATE

CODE:#include<stdio.h>int main(){ int pid; pid=fork(); if(pid==-1) { printf("fork failed\n"); exit(-1); } else if(pid==0) { execlp("/bin/ls","ls",NULL); exit(0); } else if(pid>0) { wait(NULL); printf("i am a parent process\n"); printf("child terminated"); exit(0); }}

OUTPUT:[user3@localhost user3]$ cc p1d.c[user3@localhost user3]$ ./a.out03 cient.c client2.out new.c p1d.c ser.c server2.out26 clent.c client.c p1a.c richsan serv.c tanvi5.c client1.c clnt1.out p1b.c ser1.c server1.ca.out client1.out clnt.c p1c.c ser1.out server1.outi am a parent process

E) WHERE PARENT AND CHILD CAN EXECUTE DIFFERENT PROGRAM AND A CHILD PROCESS IS EXECUTING A USER CREATED FILE.

CODE:#include<stdio.h>int main(){ int pid; pid=fork(); if(pid==-1) { printf("fork failed\n"); exit(-1); } else if(pid==0) { execlp("/home/user3/square","square",NULL); exit(0); } else if(pid>0) { printf("i am a parent process\n"); wait(NULL); printf("child terminated"); exit(0); }}

Vi square.c#include<stdio.h>int main(){int x;printf("enter a no.\n");

scanf("%d",&x);printf("square=%d",x*x);exit(0);}

OUTPUT:[user3@localhost user3]$ cc p1e.c[user3@localhost user3]$ ./a.outenter a no.6square=36i am a parent processchild terminated

F) HIERARCHY OF PROCESS

CODE:#include<stdio.h>int main(){ int pid; pid=fork(); if(pid==-1) { printf("fork failed\n"); exit(-1); } else if(pid==0) { int cid; cid=fork(); if(cid==0) { printf("i am the grandchild\n"); exit(0); } else { printf("i am a child\n"); exit(0); } } else if(pid>0) { printf("i am a parent process\n"); exit(0); }}

OUTPUT:[user3@localhost user3]$ ./a.outi am the grandchildi am a childi am a parent process

QUES.2(a) WRITE A PROGRAM TO DEMONSTRATE INTER PROCESS COMMUNICATION BETWEEN PARENT AND CHILD.CODE:#include<stdio.h>int main(){ int fd[2],pid; int n; char line[20]; if(pipe(fd) < 0) { printf("Error in creating pipe"); } else{ pid=fork(); if(pid < 0) printf("Fork failed"); else if (pid > 0) { close(fd[0]); write(fd[1],"Hello World",12); } else if(pid==0) { close(fd[1]); n=read(fd[0],line,20); if(n<0) printf("Error in reading");write(1,line,n);}}return 0;}OUTPUT:[user3@localhost user3]$ cc prog2.c

[user3@localhost user3]$ ./a.out[user3@localhost user3]$ Hello World

QUES 2(b) : WRITE A PROGRAM TO DEMONSTRATE INTER PROCESS COMMUNICATION BETWEEN PARENT AND CHILD(FULL DUPLEX).CODE :#include<stdio.h>int main(){ int fd1[2],fd2[2],pid,n; char line[20]; if(pipe(fd1)<0) { printf("Error creating file"); } else { pid=fork(); if(pid<0) printf("Fork failed"); else if(pid>0) { close(fd1[0]); write(fd1[1],"Hello World",12); } else if(pid==0) { close(fd1[1]); n=read(fd1[0],line,20); if(n<0) printf("Error in reading"); write(1,line,n); } } if(pipe(fd2)<0) { printf("Error creating file"); } else { pid=fork(); if(pid<0) printf("Fork failed"); else if(pid==0) { close(fd2[0]);

write(fd2[1],"Good Bye",9); } else if(pid>0) { close(fd2[1]); n=read(fd2[0],line,20); if(n<0) printf("Error in reading"); write(1,line,n); } }return 0;}

OUTPUT :[user3@localhost user3]$ cc q2b.c[user3@localhost user3]$ ./a.outGood Bye[user3@localhost user3]$ Hello World Good Bye

QUES 3 : WRITE A PROGRAM TO REPORT A LINUX KERNEL BEHAVIOR INCLUDING CPU TYPE AND MODEL,KERNEL VERSION,INFORMATION ON CONFIGURED MEMORY,AMOUNT OF FREE AND USED MEMORY.CODE :#include<stdio.h>int main(){ system("echo Amount of time CPU has spend in User Mode : "); system("cat /proc/stat|grep -m1 'cpu'|cut -c 6-9"); system("echo"); system("echo Amount of time CPU has spend in System Mode : " ); system("cat /proc/stat|grep -m1 'cpu'|cut -c 11"); system("echo"); system("echo Amount of time CPU has spend in Idle Mode : "); system("cat /proc/stat|grep -m1 'cpu'|cut -c 13-16"); system("echo"); system("echo Number of context switches : "); system("cat /proc/stat|grep -m1 'ctxt'|cut -c 6-11"); system("echo"); system("echo Number of processes created sinc system was last booted : "); system("cat /proc/stat|grep -m1 'processes'|cut -c 11-14");

system("echo"); system("echo CPU type : "); system("cat /proc/cpuinfo|grep -m1 'cpu'|cut -c 13-15"); system("echo"); system("echo CPU model : "); system("cat /proc/cpuinfo|grep -m1 'model name'"); system("echo"); system("echo Kernel Version : "); system("cat /proc/sys/kernel/osrelease"); system("echo"); system("echo Total memory in KB : "); system("free -o |grep 'Mem'|cut -c 10-20"); system("echo"); system("echo Used memory in KB : "); system("free -o |grep 'Mem'|cut -c 20-30"); system("echo"); system("echo Free memory in KB : "); system("free -o |grep 'Mem'|cut -c 30-40"); system("echo");return 0;}

OUTPUT :

Amount of time CPU has spend in User Mode :8150

Amount of time CPU has spend in System Mode :1

Amount of time CPU has spend in Idle Mode :62 5

Number of context switches :113578

Number of processes created sinc system was last booted :8656

CPU type : 15

CPU model :model name : Intel(R) Pentium(R) 4 CPU 3.00GHz

Kernel Version :2.4.20-8smp

Total memory in KB : 504368

Used memory in KB : 481576

Free memory in KB : 22792

Ques 4.a) Write a program to print file details includeing owner access permissions, file access time, where filename is given an argument.

#include<stdio.h>#include<stdlib.h>#include<sys/stat.h>int main(int argc, char *argv[]){struct stat statbuf;int fd;fd = open(argv[1],0,0744);if(fd==-1){printf("file not opened");exit(0);}fstat(fd,&statbuf);close(fd);printf("\n owner id is %d", statbuf.st_uid);printf("\n group id is %d", statbuf.st_gid);printf("\n size is %d", statbuf.st_size);printf("\n no. of links is %d", statbuf.st_nlink);printf("\n last access time is %s", ctime(&statbuf.st_atime));

printf("\n last modification time is %s", ctime(&statbuf.st_mtime));return 0;}

OUTPUT:

[user3@localhost richsan]$ cc prog4a.c[user3@localhost richsan]$ ./a.out file.c owner id is 502

group id is 502

size is 11

no. of links is 1

last access time is Mon Feb 18 10:46:38 2013

last modification time is Mon Feb 18 10:46:38 2013

Ques 4.b) Write a print information about a file if it exists else create the file and then print information.

#include<stdio.h>#include<stdlib.h>#include<sys/stat.h>#include<fcntl.h>int main(int argc, char *argv[]){struct stat statbuf;int fd;fd = open(argv[1],O_RDONLY|O_CREAT,0744);if(fd==-1){printf("file not opened");exit(0);}fstat(fd,&statbuf);close(fd);printf("\n owner id is %d", statbuf.st_uid);printf("\n group id is %d", statbuf.st_gid);printf("\n size is %d", statbuf.st_size);printf("\n no. of links is %d", statbuf.st_nlink);

printf("\n last access time is %s", ctime(&statbuf.st_atime));printf("\n last modification time is %s", ctime(&statbuf.st_mtime));return 0;}

OUTPUT:[user3@localhost richsan]$ cc prog4b.c[user3@localhost richsan]$ ./a.out file1.cowner id is 502

group id is 502

size is 0

no. of links is 1

last access time is Mon Feb 18 10:58:58 2013

last modification time is Mon Feb 18 10:58:58 2013

QUES 5 : WRITE A PROGRAM TO COPY FILES USING SYSTEM CALLS.CODE :#include<stdio.h>int main(int argc,char *argv[]){ char buf[500]; int fd1,fd2,count; fd1=open(argv[1],0); if(fd1== -1) { printf("File cannot be opened"); exit(-1); } fd2=creat(argv[2],0744); if(fd2==-1) { printf("File cannot be created"); exit(-1); } count=read(fd1,buf,sizeof(buf));

write(fd2,buf,count); return 0;}

OUTPUT:user3@localhost user3]$ cc prog3.c [user3@localhost user3]$ ./a.out f1 f2[user3@localhost user3]$ cat f2hello....this is the lab of operating system..you are welcomed ...

QUES 6(a) : IMPLEMENT FCFS WITH ARRIVAL TIME 0 FOR ALL PROCESSES.CODE :#include<stdio.h>int main(){int P[50],B[50],C[50],W[50],T[50];int sum1=0,sum2=0;int i,n;printf("Enter the number of processes : ");scanf("%d",&n);for(i=0;i<n;i++){ P[i]=i+1; printf("Enter the burst time for process %d",i+1); scanf(" %d",&B[i]);}for(i=0;i<n;i++){ C[0]=B[0]; C[i]=C[i-1]+B[i];}for(i=1;i<n;i++){ W[i]=C[i-1];}for(i=0;i<n;i++){ T[i]=C[i];}printf("PROCESSES WAITING TIME TURNAROUND TIME");for(i=0;i<n;i++)

printf("\n%d\t\t%d\t\t%d",P[i],W[i],T[i]);\for(i=1;i<n;i++){ sum1=sum1+W[i];}for(i=0;i<n;i++){ sum2=sum2+T[i];}printf("The Average waiting time : %d",(sum1/n));printf("The Average turnaround time : %d",(sum2/n));return 0;}

OUTPUT :[user3@localhost user3]$ cc ques6a.c[user3@localhost user3]$ ./a.outEnter the number of processes : 4Enter the burst time for process 1 10Enter the burst time for process 2 12Enter the burst time for process 3 20Enter the burst time for process 4 05PROCESSES WAITING TIME TURNAROUND TIME1 0 102 10 223 22 424 42 47The Average waiting time : 18The Average turnaround time : 30

QUES 6(b) : IMPLEMENT FCFS WITH GIVEN ARRIVAL TIME FOR ALL PROCESSES.CODE : #include<stdio.h>int main()printf("Enter the number of processes : ");scanf("%d",&n);for(i=0;i<n;i++){ P[i]=i+1; printf("Enter the burst time for process %d",i+1); scanf(" %d",&B[i]);}for(i=0;i<n;i++){ A[i]=i+1; printf("Enter the arrival time for process %d",i+1); scanf(" %d",&A[i]);}for(i=0;i<n;i++){ C[0]=B[0]; C[i]=C[i-1]+B[i];}for(i=1;i<n;i++){ W[i]=C[i-1]-A[i];}for(i=0;i<n;i++){ T[i]=C[i]-A[i];}

printf("PROCESSES WAITING TIME TURNAROUND TIME");for(i=0;i<n;i++)printf("\n%d\t\t%d\t\t%d",P[i],W[i],T[i]);for(i=1;i<n;i++){ sum1=sum1+W[i];}for(i=0;i<n;i++){ sum2=sum2+T[i];}printf("The Average waiting time : %d",(sum1/n));printf("The Average turnaround time : %d",(sum2/n));return 0; }OUTPUT :[user3@localhost user3]$ cc ques6b.c[user3@localhost user3]$ ./a.outEnter the number of processes : 4Enter the burst time for process 1 10Enter the burst time for process 2 12Enter the burst time for process 3 20Enter the burst time for process 4 05Enter the arrival time for process 1 0Enter the arrival time for process 2 1Enter the arrival time for process 3 2Enter the arrival time for process 4 3PROCESSES WAITING TIME TURNAROUND TIME1 0 102 9 213 20 404 39 44The Average waiting time : 17The Average turnaround time : 28

QUES 6(c) : IMPLEMENT ROUND ROBIN SCHEDULING.CODE :#include<stdio.h>int main(){ int P[50],B[50],j=0,ttime=0,tslice,n,i; printf("Enter the number of processes : "); scanf("%d",&n); for(i=0;i<n;i++) { P[i]=i+1; printf("Enter the burst time for process %d",i+1); scanf(" %d",&B[i]); } printf("Enter the time slice : "); scanf(" %d",&tslice); printf("PROCESSES REMAINING TIME TOTAL TIME"); while(j<=n) { for(i=0;i<n;i++) { if(B[i]>0) { if(B[i]>=tslice) { ttime=ttime+tslice; B[i]=B[i]-tslice; } else { ttime=ttime+B[i]; B[i]=0; } printf("\n%d\t\t%d\t\t%d",P[i],B[i],ttime); } } j++; } return 0;}

OUTPUT:

[user3@localhost user3]$ cc ques6c.c[user3@localhost user3]$ ./a.outEnter the number of processes : 3Enter the burst time for process 1 : 4Enter the burst time for process 2 : 2Enter the burst time for process 3 : 3Enter the time slice : 2PROCESSES REMAINING TIME TOTAL TIME1 2 22 0 43 1 61 0 83 0 9

QUES 7(a) : USE PTHREADS TO CALCULATE THE SUM OF ‘N’ NON NEGATIVE NUMBERS.CODE :#include<pthread.h>#include<stdio.h>int sum;void *runner(void *param);int main(int argc,char *argv[]){ pthread_t tid; pthread_attr_t attr; if(argc!=2) { printf("Error : 2 arguments are needed"); exit(-1); } if(atoi(argv[1]) < 0) { printf("%d must be >= 0 \n",atoi(argv[1])); exit(-1); } pthread_attr_init(&attr); pthread_create(&tid,&attr,runner,argv[1]); pthread_join(tid,NULL); printf("Sum=%d",sum); } void *runner(void *param) { int i,upper=atoi(param);

sum=0; for(i=1;i<=upper;i++) sum=sum+i; pthread_exit(0); }

OUTPUT :[user3@localhost user3]$ cc -lpthread prog7a.c -o prog7a[user3@localhost user3]$ ./prog7a 5Sum=15

QUES 7(b) : USE PTHREADS TO CALCULATE THE FACTORIAL OF ‘N’ NUMBERS.CODE :#include<pthread.h>#include<stdio.h>int fact;void *factorial(void *param);int main(int argc,char *argv[]){ pthread_t tid; pthread_attr_t attr; if(argc!=2) { printf("Error : 2 arguments are needed"); exit(-1); } if(atoi(argv[1]) < 0) { printf("%d must be >= 0 \n",atoi(argv[1])); exit(-1); } pthread_attr_init(&attr); pthread_create(&tid,&attr,factorial,argv[1]); pthread_join(tid,NULL); printf("Factorial = %d",fact); } void *factorial(void *param)

{ int i,upper=atoi(param); fact=1; for(i=upper;i>=1;i--) fact=fact*i; pthread_exit(0); }

OUTPUT :[user3@localhost user3]$ cc -lpthread prog7b.c -o prog7b[user3@localhost user3]$ ./prog7b 5Factorial = 120

QUES 7(c) : USE PTHREADS TO PRINT THE FIBONACCI SERIES OF ‘N’ NUMBERS.CODE :#include<pthread.h>#include<stdio.h>int fib,x,y;int fib,x,y;void *fibonacci(void *param);int main(int argc,char *argv[]){ pthread_t tid; pthread_attr_t attr; if(argc!=2) { printf("Error : 2 arguments are needed"); exit(-1); } if(atoi(argv[1]) < 0) { printf("%d must be >= 0 \n",atoi(argv[1])); exit(-1); } pthread_attr_init(&attr); pthread_create(&tid,&attr,fibonacci,argv[1]); pthread_join(tid,NULL); } void *fibonacci(void *param)

{ int i,upper=atoi(param); x=1; y=1; printf("Fibonacci Series = "); printf(" %d",x); printf(" %d",y); for(i=3;i<=upper;i++) { fib=y+x; printf(" %d",fib); x=y; y=fib; } pthread_exit(0);}

OUTPUT : [user3@localhost user3]$ cc -lpthread prog7c.c -o prog7c[user3@localhost user3]$ ./prog7c 5Fibonacci Series = 1 1 2 3 5

QUES 7(d) : USE PTHREADS,CREATE 2 THREADS WHICH COMPUTE SUM OF ALL EVEN AND ODD NUMBERS UPTO ‘N’.CODE :#include<pthread.h>#include<stdio.h>int sum1,sum2;void *runner1(void *param);void *runner2(void *param);int main(int argc,char *argv[]){ pthread_t tid1,tid2; pthread_attr_t attr1,attr2; if(argc!=2) { printf("Error : 2 arguments are needed"); exit(-1); } if(atoi(argv[1]) < 0) { printf("The number must be >= 0 \n",atoi(argv[1])); exit(-1); } pthread_attr_init(&attr1); pthread_create(&tid1,&attr1,runner1,argv[1]); pthread_join(tid1,NULL); pthread_attr_init(&attr2); pthread_create(&tid2,&attr2,runner2,argv[1]);

pthread_join(tid2,NULL); printf("Sum of odd numbers = %d ",sum1); printf("Sum of even numbers = %d",sum2);} void *runner1(void *param) { int i,upper=atoi(param); sum1=0; for(i=1;i<=upper;i=i+2) sum1=sum1+i; pthread_exit(0); } void *runner2(void *param) { int i,upper=atoi(param); sum2=0; for(i=0;i<=upper;i=i+2) sum2=sum2+i; pthread_exit(0); }

OUTPUT :[user3@localhost user3]$ cc -lpthread prog7d.c -o prog7d[user3@localhost user3]$ ./prog7d 10Sum of odd numbers = 25 Sum of even numbers = 30[user3@localhost user3]$ ./prog7d -2The number must be >= 0