1.a) Create an input text file in few lines which describes about your aim in life and develop
a LEX program to count the number of characters, words, spaces & no. of lines in the
input file.
%{
#include<stdio.h>
int wc=0,uc=0,sc=0,lc=0,tc=0,splc=0,dc=0,lnc=0;
%}
%%
[\n] {lnc++;wc++;}
[‘ ‘\t] {sc++;wc++;}
[A-Z] {uc++;}
[a-z] {lc++;}
[0-9] {dc++;}
. {splc++;}
%%
int main()
{
FILE *fp;
char filenm[20];
printf(“\n Enter text File name:”);
scanf(“%s”,filenm);
fp=fopen(filenm,”r”);
if(fp==NULL)
printf(“\nError in opening !!”);
else
{
yyin=fp;
yylex();
printf(“\n num of words = %d\n”,wc);
printf(“\n num of space count=%d\n”,sc);
printf(“\n num of uppercase = %d\n”,uc);
printf(“\n num of lowercase = %d\n”,lc);
printf(“\n num of spl char =%d\n”,splc);
printf(“\n num of digit count =%d\n”,dc);
printf(“\n num of line count =%d\n”,lnc);
tc=uc+lc+sc+splc+dc;
printf(“\n total num of chars =%d\n”,tc);
}
}
1b) Create an input file which consists of a C program to find the area of a circle with
comment lines and develop a LEX program to count the number of comment lines in the
input file and also eliminate them and copy that program into separate file.
%{
#include<stdio.h>
int fl=0,c=0,fla=0;
%}
%%
\/\* {if(!fl) {fl=1;c++;} else REJECT;}
\*\/ {if(fl) fl=0; else REJECT;}
\/\/ {if(!fla) {fla=1;c++;} else REJECT;}
[\n] {if(fla) fla=0; else REJECT;}
. {if(!fl&&!fla) REJECT;}
%%
int main(int argc,char *argv[])
{
if(argc!=3)
{
printf(“Improper arguments.\nUsage is $./a.out <srcfilename><destfilename>\n”);
exit(0);
}
yyin=fopen(argv[1],”r”);
yyout=fopen(argv[2],”w”); yylex();
printf(“Number of comment statemensts is %d\n”,c);
}
2. Develop a LEX program to recognize valid arithmetic expression in the list given below and recognize the identifiers and operators present in the expression and print them separately.
|
List of Arithmetic expressions |
||
|
2a+5b-4c*5d |
C=a+b-c*d/e |
Z=x+y-w* |
|
P*q=r |
2m=7n+8q-3r |
d+c*a- |
%{
#include<stdio.h>
int id=0,op=0,a=0,s=0,d=0,m=0,c=0,ob=0,cb=0,flag=1;
%}
%%
[a-zA-z][a-zA-Z0-9]* {id++; printf(“%s is an identifier\n”,yytext);}
“+” {op++;a++;}
“*” {op++;m++;}
“-” {op++;s++;}
“/” {op++;d++;}
“=” {op++;c++}
“(” {ob++;}
“)” {if(ob>cb)
cb++;
else
flag=0;
}
. ;
%%
main()
{
printf(“\n\nEnter the Arithmetic Expression:”);
yylex();
int b=ob+cb;
if((id==op+1)&&(b%2==0)&&(ob==cb)&&(flag==1))
{
printf(“The Expression is a Valid Expression”);
printf(“\n\nNo: of identifiers are\n”);
printf(“Identifiers=%d\n”,id);
printf(“\nNo.of operators\n”);
if(a>0)
printf(“+\n”);
if(m>0)
printf(“*\n”);
if(s>0)
printf(“-\n”);
if(d>0)
printf(“/\n”);
if(c>0)
printf(“\n”);
printf(“addition op=%d\n subtraction=%d\n multiplication=%d\n division=%d\n”,a,s,m,d);
}
else
{
printf(“invalid expression\n”); exit(0);
}
}
3.a) Create an input file which consists of C program to swap two numbers without using temporary variables and for this file develop LEX program to recognize and count the number of identifiers in a given input file.
%{
#include<stdio.h>
int fc=0,def=0,id=0; char a[10000];
%}
%%
\( {fc=1;}
\) {fc=0;}
[\ \t]*int[\ ] {if(fc) def=1; else def=2;}
[\ \t]*char[\ ] {if(fc) def=1; else def=2;}
[\ \t]*float[\ ] {if(fc) def=1; else def=2;}
[\ \t]*double[\ ] {if(fc) def=1; else def=2;}
\*?[a-zA-Z]+[a-zA-Z0-9_]*;
{if(def){id++;def=0;strncat(a,yytext,yyleng);strcat(a,” “);}}
\*?[a-zA-Z]+[a-zA-Z0-9_]*[,=]
{if(def){id++;strncat(a,yytext,yyleng);strcat(a,” “);}}
\*?[a-zA-Z]+[a-zA-Z0-9_]*\[[0-9]*\][,=]? {if(def)
{id++;strncat(a,yytext,yyleng-1);strcat(a,” “);}};{if(def) def=0;}
.|\n { ;}
%%
int main(int argc,char *argv[])
{
if(argc!=2)
{
printf(“Improper Arguments specified.\n”);
printf(“Usage is $./a.out <filename>\n”);
exit(0);
}
yyin=fopen(argv[1],”r”); yylex();
printf(“Number of identifiers are %d.\nThey are : %s\n”,id,a);
}
int yywrap()
{
return 1;
}
3b) Write a short note on Computer Science and Engineering and develop LEX program to find and replace the word Computer Science as CS.
%{
#include<stdio.h> #include<string.h> FILE *ff,*fr;
char p[20],q[20],r[20],fname[20];
%}
%%
[a-zA-Z]+ {if(strcmp(p,yytext)==0)
fprintf(fr,q); else
fprintf(fr,yytext);
}
\n {fprintf(fr,yytext);}
. {fprintf(fr,yytext);}
%%
int main(int argc,char *argv[])
{
strcpy(fname,argv[1]);
strcpy(p,argv[2]);
strcpy(q,argv[3]);
ff=fopen(fname,”r+”);
fr=fopen(“rep.txt”,”w+”);
yyin=ff;
yylex();
return(0);
}
4.a) Create a file which lists the latest open source application useful for Computer Science and Engineering and develop a LEX program to copy the content of this file with line numbers to another file.
%{
#include<stdio.h>
#include<string.h>
char line[20];
int count=0,i=0;
FILE *out;
%}
%%
[‘\n’] {fprintf(yyout,”%d %s\n”,++count,line);}
(.*) {strcpy(line,yytext);}
%%
{
if(argc!=3)
{
printf(“Improper arguments.\nUsage is $./a.out <srcfilename><destfilename>\n”);
exit(0);
}
yyin=fopen(argv[1],”r”);
yyout=fopen(argv[2],”w”);
yylex();
}
4b) From the list of students admission dates given below, check whether the admission dates are in valid format [DD/MM/YYYY] by developing LEX program.
|
Student admission Dates |
|
|
Student 1 |
24/05/2019 |
|
Student 2 |
30-02-2020 |
|
Student 3 |
23.05.2019 |
|
Student 4 |
24/6/2019 |
|
Student 5 |
28/04/20 |
%{
#include<stdio.h>
int i=0,yr=0,valid=0;
%}
%%
([0-2][0-9]|[3][0-1])\/((0(1|3|5|7|8))|(10|12))\/([1-2][0-9][0-9][0-9]) {valid=1;}
([0-2][0-9]|30)\/((0(4|6|9))|11)\/([1-2][0-9][0-9][0-9]) {valid=1;}
([0-1][0-9]|2[0-8])\/02\/([1-2][0-9][0-9][0-9]) {valid=1;}
29\/02\/([1-2][0-9][0-9][0-9]) {while(yytext[i]!=’/’) i++;i++;
while(yytext[i]!=’/’)i++;i++; while(i<yyleng) yr=(10*yr)+(yytext[i++]-‘0’); if(yr%4==0||(yr%100==0&&yr%400!=0)) valid=1;}
.;
%%
int main()
{
printf(“enter the date\n”);
yylex();
if(valid==1)
printf(“It is a valid date\n”);
else
printf(“It is invalid date\n”);
}
5. Write and Execute YACC program to convert infix to postfix.
Lex Part
%{
#include<stdio.h>
#include “y.tab.h”
extern int yylval;
%}
%%
[0-9]+ {yylval=yytext[0];
return NUMBER;
}
[\t] ;
\n {return 0;}
. {return yytext[0];}
%%
Yacc Part
%{
#include<stdio.h>
int k=0;
int i;
char sym[26];
FILE *fp;
%}
%token NUMBER
%left’+”-‘
%left’*”/’
%nonassoc UMINUS
%%
state : exp {
printf(“converted postfix expression is =>”); fp=fopen(“postfix.txt”,”w”);
for(i=0;i<k;i++)
{
fprintf(fp,”%c”,sym[i]);
printf(“%c”,sym[i]);
}
fclose(fp);
}
;
exp:NUMBER {$$=$1;sym[k]=(char)$$;k++;}
|exp’+’exp {sym[k]=’+’;k++;}
|exp’-‘exp {sym[k]=’-‘;k++;}
|exp’*’exp {sym[k]=’*’;k++;}
|exp’/’exp {sym[k]=’/’;k++;}
;
%%
int yyerror(char *str)
{
printf(“invalid character”);
}
int main()
{
printf(“enter expression=>”); yyparse();
return(0);
}
6. Develop YACC program to evaluate an arithmetic expression involving operators +, -, * and /. to find the solutions for the problems given below; If
|
1 |
a=5,b=6,c=8 ,then a+b*c=? |
|
2 |
a=10,b=8,c=2 ,then a-b*c=? |
|
3 |
a=10,b=2,c=3, then a/b+c=? |
|
4 |
a=5,b=3,c=3,d=4,then a*b-c+d=? |
Lex Part
%{
#include “y.tab.h” extern int yylval;
%}
%%
[0-9]+ {yylval=atoi(yytext);return NUMBER;} [\t] ;
\n return 0;
. return yytext[0];
%%
Yacc Part
%{
#include<stdio.h>
#include<stdlib.h>
%}
%token NUMBER
%left’-”+’
%left’*”/’
%nonassoc UMINUS
%%
statement:expression{printf(“value=%d\n”,$$);}
;
expression:expression’+’expression {$$=$1+$3;}
|expression’-‘expression {$$=$1-$3;}
|expression’*’expression {$$=$1*$3;}
|expression’/’expression
{if($3==0)
{
yyerror(“\n divide by 0 error\n”); exit(0);
}
else
$$=$1/$3;
}
|’-‘expression%prec UMINUS {$$=-$2;}
|'(‘expression’)’ {$$=$2;}
|NUMBER
;
%%
main()
{
printf(“\n enter expression\n”); yyparse();
}
int yyerror(char *s)
{
printf(“%s\n”,s);
}
7 a) Write and Execute YACC program to recognize the grammar (anb, n >= 10).
Lex Part
%{
#include “y.tab.h”
%}
%%
[aA] {return A;}
[bB] {return B;}
. {return yytext[0];}
%%
Yacc Part
%{
#include<stdio.h> #include<stdlib.h>
%}
%token A B;
%%
str:A A A A A A A A A A S
;
S:A S
|B
;
%%
int main()
{
printf(“\n enter a string\n”); yyparse();
printf(“\n valid string\n”); return;
}
yyerror()
{
printf(“\n invalid string\n”); exit(0);
}
7.b) Write and Execute YACC Program to recognize the grammar (anbmck, m,n,k>=0 and m=n+k).
Lex Part
%{
#include “y.tab.h”
%}
%%
[aA] {return A;}
[bB] {return B;}
[cC] {return C;}
. {return yytext[0];}
%%
Yacc Part
%{
#include<stdio.h> #include<stdlib.h>
%}
%token A B C;
%%
str:S Z
|S
|Z
;
S:A S B
|A B Z:B Z C
|B C
;
%%
main()
{
printf(“\n enter a string of the form (a^n)(b^m)(c^k) and m=n+k\n”);
yyparse();
printf(“\n string recognized\n”);
return;
}
yyerror()
{
printf(“\n string not recognized\n”); exit(0);
}
8. Write a C/C++ program which demonstrates interprocesscommunication between a reader process and a writerprocess. Use mkfifo, open, read, write and close APIs in your program.
Writer process
#include <stdio.h>
#include <fcntl.h>
#include <sys/stat.h>
#include <sys/types.h>
#include <unistd.h>
int main()
{
int fd;
char buf[1024];
/* create the FIFO (named pipe) */
char * myfifo = “/tmp/myfifo”;
mkfifo(myfifo, 0666);
printf(“Run Reader process to read the FIFO File\n”);
fd = open(myfifo, O_WRONLY);
write(fd,”Hi”, sizeof(“Hi”));
/* write “Hi” to the FIFO */
close(fd);
unlink(myfifo);
return 0;
}
/* remove the FIFO */
Reader process
#include <fcntl.h>
#include <sys/stat.h>
#include <sys/types.h>
#include <unistd.h>
#include <stdio.h>
#define MAX_BUF 1024
int main()
{
int fd;
/* A temp FIFO file is not created in reader */
char *myfifo = “/tmp/myfifo”;
char buf[MAX_BUF];
/* open, read, and display the message from the FIFO */
fd = open(myfifo, O_RDONLY);
read(fd, buf, MAX_BUF);
printf(“Writer: %s\n”, buf);
close(fd);
return 0;
}
9. Write a C/C++ program to illustrate the race condition.
#include<stdio.h>
#include<sys/types.h>
#include<unistd.h>
static void charatatime(char *);
int main(void)
{
pid_t pid;
if ((pid = fork()) < 0)
{
printf(“fork error”);
}
else if (pid == 0)
{
charatatime(“output from child\n”);
}
else
{
charatatime(“output from parent\n”);
}
return 0;
}
static void charatatime(char *str)
{
char *ptr; int c;
setbuf(stdout, NULL);
/* set unbuffered */
for (ptr = str; (c = *ptr++) != 0; )
putc(c, stdout);
}
10. Write a C/C++ POSIX compliant program to check the following limits:
(i) No. of clock ticks (ii) Max. no. of child processes (iii) Max. path length (iv) Max. no. of Characters in a file name (v) Max. no. of open files/ process
#define _POSIX_SOURCE
#define _POSIX_C_SOURCE 199309L
#include<stdio.h>
#include<unistd.h> int main()
{
int res;
if((res = sysconf(_SC_OPEN_MAX)) == -1)
perror(“sysconf”);
else printf(“OPEN_MAX:%d\n”,res);
if((res = pathconf(“/”, _PC_NAME_MAX)) == -1)
perror(“pathconf”);
else
printf(“max_path name:%d\n”,res);
if((res = sysconf(_SC_CLK_TCK))== -1)
perror(“sysconf”);
else
printf(“clock ticks:%d\n”,res);
if((res = sysconf (_SC_CHILD_MAX)) == -1)
perror(“sysconf”);
else printf(“max_childs:%d\n”,(res));
if((res = pathconf(“/”,_PC_PATH_MAX)) == -1)
perror(“pathconf”);
else
printf(“max path name length:%d\n”,res);
return 0;
}
11 Write a C/C++ program to implement the system function.
#include<sys/wait.h>
#include<errno.h>
#include<unistd.h>
#include<stdio.h>
#include<stdlib.h>
int system(const char *cmdstring)
{
pid_t pid; int status;
if (cmdstring == NULL)
return(1);
if ((pid = fork()) < 0)
{
status = -1;
}
else if (pid == 0)
{
/* child */
execl(“/bin/sh”, “sh”, “-c”, cmdstring, (char *)0);
_exit(127); /* execl error */
}
else
/* parent */
while (waitpid(pid, &status, 0) < 0)
{
if (errno != EINTR)
status = -1; /* error other than EINTR from waitpid() */ break;
}
return(status);
}
int main()
{
int status;
if ((status = system(“date”)) < 0) printf(“system() error”);
if ((status = system(“who”)) < 0) printf(“system() error”); exit(0);
}
12. Write a C/C++ program to set up a real-time clock interval timer using the alarm API.
#include<stdio.h>
#include<stdlib.h>
#include<unistd.h>
#include<signal.h>
#define INTERVAL 5
void callme(int sig_no)
{
alarm(INTERVAL); printf(“Hello!!\n”);
}
int main()
{
struct sigaction action;
action.sa_handler=(void(*)(int))callme;
sigaction(SIGALRM,&action,0);
alarm(2);
sleep(5); return 0;
}
13.a) Write and Execute Non-recursive shell script that accepts any number of arguments and prints them in the Reverse order, (For example, if the script is named rargs,then executing rargs A B C should produce C B A on the standard output).
if [ $# -eq 0 ]
then
echo “No argument” exit
fi
echo “Num of arguments:$#”
echo “Arguments in reverse order” for x in $*
do
y=”$x $y”
done
echo $y
13b) Write and Execute C program that creates a child process to read commands from the standard input and execute them (a minimal implementation of a shell – like program). You can assume that no arguments will be passed to the commands to be executed.
#include<stdio.h>
#include<sys/types.h>
#include<stdlib.h>
main()
{
pid_t p; char cmd[20]; p=fork(); if(p<0)
{
printf(“\n fork error”); exit(0);
}
else if(p>0)
{
printf(“\n enter the command\n”);
scanf(“%s”,cmd);
system(cmd);
}
exit(0);
}
14 a) Write a C/C++ program to avoid zombie process by forking twice.
#include<stdio.h>
#include <sys/wait.h>
#include<errno.h>
#include<stdlib.h>
int main()
{
pid_t pid;
if ((pid = fork()) < 0)
{
printf(“fork error”);
}
else if (pid == 0)
{ /* first child */
if ((pid = fork()) < 0)
printf(“fork error”);
else if
(pid > 0)
exit(0);
sleep(2);
printf(“second child, parent pid = %d\n”, getppid());
exit(0);
}
if (waitpid(pid, NULL, 0) != pid) /* wait for first child */
printf(“waitpid error”); exit(0);
}
14.b) Write and Execute C program to do the following: Using fork( ) create a child process. The child process prints its own process-id and id of its parent and then exits. The parent process waits for its child to finish (by executing the wait( )) and prints its own process-id and the id of its child process and then exits.
#include<stdio.h>
#include<stdlib.h>
#include<unistd.h>
int main()
{
int pid,id; pid=fork(); if(pid==0)
{
printf(“\n child process:report”);
printf(“\n my pid is:%d”,getpid());
printf(“\n my parent pid is:%d”,getpid());
}
else
{
wait();
printf(“\n parent process:report”);
printf(“\n my pid is:%d”,getpid());
printf(“\n my child pid is:%d”,pid);
}
exit(0);
}
