Oslab

ASSIGNMENT I Problem 1:Write a shell script that will assign the value Hi! to a variable called all. Then print the value of all with no quotes, with single quotes and with double quotes.

Solution: all=Hi! echo -e "With no quotes :$all\n" echo -e "With single quotes :'$all'\n" echo -e "with double quotes :\"$all\"\n" Output: [root@localhost ~]# ./prob1.sh With no quotes :Hi! With single quotes :'Hi!' with double quotes :"Hi!" Problem 2 :Write a shell script that has two user created variables, var1 and var2.Ask for values of the variable from the user and take in any values of the two variables and then print them as:i)values of var1 followed by values of var2 separated by a comma ii) values of var2 followed by values of var1 separated by the word “and”. Solution: echo -e "Enter variable 1 :\c" read var1 if [ -z var1 ] ; then echo -e "You did not entered variable. Exit\n" ; exit 1 fi echo -e "Enter variable 2 :\c" read var2 if [ -z var2 ] ; then echo -e "You did not entered variable. Exit\n" ; exit 2 fi echo -e "i) $var1 , $var2 " echo -e "ii)$var2 and $var1 " Output: [root@localhost ~]# ./prob2.sh Enter variable 1 :var1 Enter variable 2 :var2 i) var1 , var2

ii)var2 and var1

Problem 3:Write a shell script that reads command line arguments and prints the number of arguments. Given this sequence of words as command line arguments “To the OS Laboratory at Alpha Lab Welcome”, the script should echo back the information that there are 8 arguments. Additionally it should echo back some of the arguments in the following order: “Welcome To the alpha Lab Os Laboratory”. Solution: echo -e "Number of arguments :$#" echo -e "$8 $1 $2 $6 $7 $3 $4" Output: [root@localhost ~]# ./prob3.sh To the OS Laboratory at Alpha Lab Welcome Welcome To the alpha Lab Os Laboratory

Problem 4: Write a shell script that accepts a filename as an argument and performs the following tasks for that file: a. searches all possible directories and prints the pathname if the file exists otherwise prints an error message and exits. b. prints the access rights of the file. c. prints the size of the file. d. prints the date of creation and the last date of modification of the file. Solution: if [ $# -eq 0 -o $# -gt 1 ] ; then echo -e "This program takes only one argument" ; exit 1 fi if [ -z "$1" ] ; then echo -e "Filename Null. Exit\n" ; exit 2 fi echo -e "Searching all possible directories....\n" find . -name "$1" 2>/dev/null >temp if [ `cat temp | wc -l ` -eq 0 ] ; then echo -e "No such file exists.\n" ; exit 3 fi echo -e " Path Access Rights Size Modification Date\n"

for var in `cat temp` ; do aright=`ls -l $var | tr -s ' ' |cut -d" " -f1` size=`ls -l $var | tr -s ' ' |cut -d" " -f5` mdate=`ls -l $var | tr -s ' ' |cut -d" " -f6,7` printf "%19s %13s %6d %16s\n" $var $aright $size "$mdate" done Output: ./prob4.sh soumya Searching all possible directories.... Path

Access Rights Size Modification Date

./soumya

-rw-r--r--

45

Feb 7

Problem 5:- Write a shell script that accepts a file name as input .The program then asks for a string of characters(i.e. any word ) from the user .The file will be searched to find whether it contains the given word .If it does, the program will display the no. of occurrences of the word. The program should also display the line numbers where the word occurred .If the file does not contain the word, and then an appropriate error message will be displayed. Solution: echo Enter a File Name read fname if [ -f $fname ] then echo Enter a Pattern to be Recognized read pattern if ! grep $pattern -n $fname then echo Pattern NOT Exists fi else echo File Does NOT Exist fi Output: [root@localhost ~]# cat soumya chloroform aerobic olfactory caesious Euonia [root@localhost ~]# ./prob5.sh Enter a File Name soumya Enter a Pattern to be Recognized chloroform 1:chloroform

Problem 6:Extend the shell script written in problem 5 to perform the following task: user is asked to enter 2 different patterns or words. The first pattern will be matched with the contents of the file and replaced with the second pattern if a match occurs. If the first pattern does not occur in the file, an appropriate error message will be displayed. Solution: echo Enter a File Name read fname if [ -f $fname ] then echo Enter a Pattern to be Replaced read pattern1 echo Enter another Pattern to be Inserted read pattern2 if grep $pattern1 -c $fname then sed s/$pattern1/$pattern2/g $fname > aaa rm $fname mv aaa $fname else echo 1st Pattern NOT Exists fi else echo File Does NOT Exist fi Output: [root@localhost ~]# cat soumya chloroform aerobic olfactory caesious Euonia [root@localhost ~]# ./prob6.sh Enter a File Name soumya Enter a Pattern to be Replaced chloroform Enter another Pattern to be Inserted mercury 1 [root@localhost ~]# cat soumya mercury aerobic olfactory

caesious Euonia

Problem 7: Write a shell script factor_table that will generate a table of factors for an integer number given on the command line. Solution: if [ $# -eq 0 ] ; then echo -e "No arguments." ; exit 1 fi for no in "$@" ; do i=2 flag=prime num=`expr $no - 1` result=" " error=0 while [ $i -le $num ] ; do rem=`expr $no % $i` if [ $rem -eq 0 ] ;then flag=comp result=$result" "$i #appedning result fi i=`expr $i + 1` done if [ $error -eq 1 ] ; then echo -e "$no is not an integer\n" elif [ "$flag" != "comp" ] ; then echo -e "$no is a prime number\n" else echo -e "Factors of $no :$result\n" fi done Output: [root@localhost ~]# ./prob7.sh 105 Factors of 105 : 3 5 7 15 21 35

Problem 8:- Write a shell script that takes as command line argument the name of a C program and removes the comments from it. Solution: if [ $# -eq 0 -o $# -gt 1 ] ; then echo -e "Improper number of arguments.\n" ;exit 1 fi sed 's|\(^[^"]*\)\(/\*.*\*/\)\([^"]*$\)|\1\3|g' $1 > temp #include<stdio.h> /*r*/ sed 's|\(".*\)\(/\*.*\*/\)\([^"]*$\)|\1\3|g' temp >temp1 #/*r*/ printf("tt"); sed 's|\(^[^"]*\)\(/\*.*\*/\)\(.*"\)|\1\3|g' temp1 > temp # printf("tt"); /*r*/ sed 's|\(.*; *\)\(/\*.*\*/\)\(.*\)|\1\3|g' temp > temp1 #printf(";/*rr"); /*r*/ printf("rr"); sed 's|\(^[^"]*\)\(//\)\([^"]*$\)|\1|g' temp1 > temp #//comment sed 's|\(".*\)\(//\)\([^"]*$\)|\1|g' temp1 > temp #//comment print(""); sed 's|\(^[^"]*\)\(//\)\(.*"\)|\1|g' temp > temp1 #printf("tt"); //cc sed 's|\(.*; *\)\(//.*\)|\1|g' temp1 > temp #printf(";/*rr"); //comment out printf("rr"); mv temp $1 rm -f temp temp1 cat $1 Output: [root@localhost ~]# cat test.c int main() { int first,second,third; // declaration third=first+second ;//adding } [root@localhost ~]# ./prob8.sh test.c int main() { int first,second,third; third=first+second ; } Problem 9:Write a shell script that takes multiple filenames as command line arguments and prints the frequency of occurrence of vowels in each file. The output should clearly denote the frequency of occurrence of each vowel in each file. Solution: count_v(){ no=`grep -c $2 $1` y=0 while [ $no -ne 0 ] do y=`expr $y + $no`

sed 's/'$2'/""/' $1 > doc2.txt cp -f doc2.txt $1 no=`grep -c $2 $1` done echo $y rm -f doc2.txt rm -f doc1.txt } work_done(){ tr [aA] 'a' < $1 | tr [eE] 'e' | tr [iI] 'i' | tr [oO] 'o' | tr [uU] 'u' > doc3.txt echo $1 no=`count_v doc3.txt a` echo " freq of occurence of a: $no" no=`count_v doc3.txt e` echo " freq of occurence of e: $no" no=`count_v doc3.txt i` echo " freq of occurence of i: $no" no=`count_v doc3.txt o` echo " freq of occurence of o: $no" no=`count_v doc3.txt u` echo " freq of occurence of u: $no" rm doc3.txt } if [ $# -eq 0 ] ; then echo "Error:No files mentioned!" fi i=1 while [ $i -le $# ] do a=`echo $* | cut -d " " -f $i` work_done $a i=`expr $i + 1` done Output: [root@localhost ~]# cat soumya chloroform aerobic olfactory caesious Euonia [root@localhost ~]# ./prob9.sh soumya soumya freq of occurence of a: 4 freq of occurence of e: 3 freq of occurence of i: 3 freq of occurence of o: 8 freq of occurence of u: 2

ASSIGNMENT – II Problem 1:a) Find a linux command that will be able to show you the details of the currently running processes with real time update of CPU usage in the system. Keep the command executing in one window. b) Write a program in C that runs for 45 seconds, uses the CPU but does not perform any i/o. Compile and convert it into a process. Execute this in another window and see the output of window (a). Now stop this using ctrl- z (before it automatically stops) and see how the change is reflected in the other window. c) Find another Linux command that will bring it back to running state again. Linux command used to see the memory details : top C- program used:

#include <stdio.h> int main(){ int t1,t2; t1=time(); t2=time(); while(t2-t1<45000000){ t2=time(); }//45 sec delay return 0; } Command used to bring the program to ready state : fg Problem 2:Develop a small linux shell called BCSE. The shell will display a prompt ,accept user commands and execute them. The overview and functions of BCSE are as follows:• At start the shell will display the prompt string hi- BCSE!! • After displaying the prompt. it is ready to accept and execute a command. • A command line has the syntax command[arguments] • When a valid command is entered ,the shell will execute it.During excution ,the shell may accept other commands, ie, commands may run in the foreground and background. • The shell should display error message when command is invalid, or when there are problems with arguments or execution of the command. • Once the command is completed(or error message displayed) , trhe shell will redisplay the prompt and the whole cycle repeats.

The commands to be executed by the BCSE are as follows:i) ii)

iii) iv)

dirnew directory-name action : creates a directory fedit [filename] ([] denotes argument is optional) action: the file will be opened with vi editor, and if no argument is given, a new file will be opened by the editor. fcontent filename action: prints the contents of a file on the screen info [filename] action: displays essential information about the file ,comprising of the full pathname of the file the size of the file

v)

last modification adte name of creator exitbcse action : this will quit the shell

Solution: #include<stdio.h> #include<string.h> #include #include<sys/types.h> #include<wait.h> #define BUFSIZE 200 #define ARGVSIZE 40 #define DELIM "\n\t\r " int main() { char buf[BUFSIZE]; char *arg[ARGVSIZE]; char cmdline[BUFSIZE]; char cmd[10]; int i=0,argcount=0,returnval,bg=0,err; pid_t pid; for(;;) { for(i = 0; i < ARGVSIZE; i++) arg[i] = "\0"; for(i = 0; i <= BUFSIZE; i++) buf[i] = '\0'; for(i = 0; i <= BUFSIZE; i++) cmdline[i] = '\0'; for(i=0;i<10;i++) cmd[i]='\0'; bg=0; printf("Hi-BCSE >"); fgets(buf,BUFSIZE,stdin); arg[0]=strtok(buf,DELIM); for(i=1;arg[i-1]!=NULL;i++) { arg[i]=strtok(NULL,DELIM); } argcount=i-2; if(argcount>=0) { if(strcmp(arg[argcount],"&")==0) {bg=1; arg[argcount]=NULL;argcount--;} } if(arg[0]==NULL) continue; if(strcmp(arg[0],"exitbcse")==0) { if(argcount==0) exit(0); else { printf("exitbcse:No arguments required\n");continue; } }

} {

} {

}

else if(strcmp(arg[0],"dirnew")==0) { strcpy(cmd,"mkdir"); strcpy(cmdline,"mkdir"); if(argcount==0) { printf("dirnew :Error: Arguments required\n"); continue; } } else if(strcmp(arg[0],"fedit")==0) { strcpy(cmd,"vi"); strcpy(cmdline,"vi"); else if(strcmp(arg[0],"fcontent")==0) strcpy(cmd,"cat"); strcpy(cmdline,"cat"); if(argcount==0) { printf("fcontent :Error: Arguments required\n"); continue; } else if(strcmp(arg[0],"info")==0) strcpy(cmd,"ls"); for(i=argcount;i>=1;i--) arg[i+1]=arg[i]; arg[1]="-l"; argcount++; arg[argcount+1]=NULL; strcpy(cmdline,"ls -l"); else { printf("BCSE:Invalid Command\n"); continue; } pid=fork(); if(pid==0) { arg[0]=cmd; err=execvp(arg[0],arg); if(err==-1) {err=0;_exit(0);} } else if(pid>0) { if(bg==0) { wait(NULL);}

} } return(0);

}

Output: [root@localhost ~]# gcc -o bcse hibcse.c [root@localhost ~]# ./bcse Hi-BCSE > Hi-BCSE >fcontent soumya chloroform aerobic olfactory caesious Euonia Hi-BCSE >fedit soumya Hi-BCSE >info soumya -rw-r--r-- 1 root root 52 Apr 14 12:44 soumya Hi-BCSE >dirnew soumyava Hi-BCSE >dirnew dirnew :Error: Arguments required Hi-BCSE >exitbcse [root@localhost ~]#

ASSIGNMENT III Problem 1:- Consider the following scenario and develop a program to simulate it. Assume that each memory location is capable of holding only 1 byte. Five programs are currently executing and following information pertaining to each program is provided.

Name Program Program Program Program Program

Size (bytes) 300 155 66 120 400

1 2 3 4 5

Starting memory address 450 070 002 290 950

(i) You have to accommodate a new program (60 bytes) in the system. Use (a) first fit policy and (b) best fit policy. Show the original and modified memory map in each case. #include<stdio.h> #include<stdlib.h> #include<malloc.h> #define MemSize 2048 #define MaxProcess 100 int ProcessListIndex=0; char Memory[MemSize]; int ProcessList[MaxProcess][3];

int main(){ int int int int int int int

nProcesses; i,j; pid; size; start_loc; number_partitions; choice;

int int int int

blockSize; bestBlockSize;int flag=0; firstBlockSize; BestFitBlock;

for(j=0;j<MemSize;j++){ Memory[j]='0'; }

printf("Enter the initial number of processes"); scanf("%d",&nProcesses); for(i=0;i
printf("\n Enter process id: "); scanf("%d",&pid); printf("\n Enter program size: "); scanf("%d",&size); printf("\n Enter starting location: "); scanf("%d",&start_loc); ProcessList[ProcessListIndex][0]=pid; ProcessList[ProcessListIndex][1]=size; ProcessList[ProcessListIndex][2]=start_loc; ProcessListIndex++; number_partitions=size; for(j=0;j
} /***************ORIGINAL MEMORY MAP*****************************************/ printf("\n original memory map :\n "); for(j=0,blockSize=0;j<MemSize;j++){ if(Memory[j]=='0'){ if(blockSize!=0){ printf("End:%d\n ",j-1); } blockSize=0; } else if(Memory[j]=='1'){ if(blockSize==0){ printf("\n Start:%d -",j); } blockSize++; } } printf("\n Enter New process id: "); scanf("%d",&pid); printf("\n Enter program size: "); scanf("%d",&size); number_partitions=size; printf("Enter your choice : \n 1.First Fit \n 2.Best Fit\n choice :"); scanf("%d",&choice); switch(choice){

case 1: //FIRST FIT flag=0; for(i=0,blockSize=0;i<MemSize && flag==0;i++){ if(Memory[i]=='1') blockSize=0; if(Memory[i]=='0'){ blockSize++; if(number_partitions==blockSize){ flag=1; printf("Starting location : %d",(i-blockSize+1) ); }

}

}//end for break; case 2://BEST FIT flag=0; BestFitBlock=2048; for(i=0,blockSize=0;i<MemSize && flag==0;i++){ if(Memory[i]=='1') blockSize=0; if(Memory[i]=='0' ){ blockSize++; if(Memory[i+1]=='1' && (blockSize=number_partitions)){ BestFitBlock=blockSize; start_loc=(i-blockSize+1); } } }//end for printf("Starting location : %d",start_loc ); break; } return 0; }

Problem2:Write a program to simulate i. Least Recently Used (LRU) page replacement algorithm ii. First In First Out (FIFO) page replacement algorithm for the following memory reference string: 1 2 3 4 1 2 5 7 1 3 7 5 2 3 4 Assume there are 4 frames in main memory. Solution:

#include<stdio.h> #include #include<stdlib.h> #define NO_OF_FRAMES 4 int no_of_requests; void lru(int *,int ); void fifo(int *,int ); int * get_page_requests(); int main() { int *page_request; page_request=get_page_requests(); lru(page_request,no_of_requests); fifo(page_request,no_of_requests); return 1; } int * get_page_requests() { int *a,n,i;

printf("\n\n\tENTER NO. OF PAGE REQUESTS:"); scanf("%d",&n); a=(int *)malloc(n*sizeof(int)); printf("\n\n\tenter the page requests one by one:"); for(i=0;i
if(count<=3) {frame[count]=a[i]; count++; printf("\tmiss!"); } else {printf("\tmiss!"); for(j=0;j
void lru(int *a,int n) { int count,hit,miss; int *frame; int i,j,use_index; frame=(int*)malloc(NO_OF_FRAMES*sizeof(int)); hit=0; count=0; miss=1; for(i=0;i
for(i=0;i
printf("\n\ncurrent frame:"); for(j=0;j
{miss=0;use_index=j;break;} if(!miss) {printf("\t hit! "); hit++; for(j=use_index;j>0;j--) frame[j]=frame[j-1]; frame[0]=a[i]; } else { if(count<3) { for(j=count;j>0;j--) frame[j]=frame[j-1]; frame[0]=a[i]; count++; } else { if(frame[count]==-1) printf("\tmiss!"); else printf("\tmiss!"); for(j=count;j>0;j--) frame[j]=frame[j-1]; frame[0]=a[i]; } } miss=1; } printf("\n\n\t %d hits for %d requests",hit,n); }

Output: [root@localhost ~]# gcc -o fifolru fifolru.c [root@localhost ~]# ./fifolru ENTER NO. OF PAGE REQUESTS:15 enter the page requests one by one: request no. 1:1 request no. 2:2 request no. 3:3

request no. 4:4 request no. 5:1 request no. 6:2 request no. 7:5 request no. 8:7 request no. 9:1 request no. 10:3 request no. 11:7 request no. 12:5 request no. 13:2 request no. 14:3 request no. 15:4 LRU

current frame: current frame:1 current frame:2 1 current frame:3 2 1

page requested:1 page requested:2 page requested:3 page requested:4

miss! miss! miss! miss!

current frame:4 3 2 1 page requested:1

hit!

current frame:1 4 3 2 page requested:2

hit!

current frame:2 1 4 3 page requested:5

miss!

current frame:5 2 1 4 page requested:7

miss!

current frame:7 5 2 1 page requested:1

hit!

current frame:1 7 5 2 page requested:3

miss!

current frame:3 1 7 5 page requested:7

hit!

current frame:7 3 1 5 page requested:5

hit!

current frame:5 7 3 1 page requested:2

miss!

current frame:2 5 7 3 page requested:3

hit!

current frame:3 2 5 7 page requested:4

miss!

6 hits for 15 requests FIFO

current frame: current frame:1 current frame:1 2 current frame:1 2 3

page requested:1 page requested:2 page requested:3 page requested:4

miss! miss! miss! miss!

current frame:1 2 3 4 page requested:1

hit!

current frame:1 2 3 4 page requested:2

hit!

current frame:1 2 3 4 page requested:5

miss!

current frame:2 3 4 5 page requested:7

miss!

current frame:3 4 5 7 page requested:1

miss!

current frame:4 5 7 1 page requested:3

miss!

current frame:5 7 1 3 page requested:7

hit!

current frame:5 7 1 3 page requested:5

hit!

current frame:5 7 1 3 page requested:2

miss!

current frame:7 1 3 2 page requested:3

hit!

current frame:7 1 3 2 page requested:4

miss!

5 hits for 15 requests