SS - INDEX OF EXPERIMENTS

SS 1.A - LEX - RECOGNIZE VALID ARITHMETIC EXPRESSION

1 -A) Write a LEX program to recognize valid arithmetic expression. Identifiers in the expression could be only integers and operators could be + and * . Count the identifiers & operators present and print them separately.

lab1.l - PROGRAM

%{
#include<stdio.h>
int v=0,op=0,id=0,flag=0;
%}

%%
[a-zA-Z]+[0-9A-Za-z]* {id++;}
[0-9]+ {id++;}
[\+\-\*/\=] {op++;}
"(" {v++;}
")" {v--;}
";" {flag=1;}
.|\n {return 0;}
%%

int main()
{
        printf("Enter the expression:");
        yylex();
        if((op+1)==id && v==0 && flag==0)
        {
                printf("\n Expression is Valid\n");
                printf("No of identifier = %d \n No of Operators = %d \n",id,op);
        }
        else
                printf("\n Expression is Invalid\n");
return 0;

}

OUTPUT :( click on image to zoom )

SS 1.B - YACC - EVALUATE ARITHMETIC EXPRESSION

1 - B) Write YACC program to evaluate arithmetic expression involving operators: +, -, *, and /

lab1b.l - PROGRAM

%{
#include "y.tab.h"
extern yylval;
%}

%%
[0-9]+ {yylval=atoi(yytext);return num;}
[\+\-\*\/] {return yytext[0];}
[)] {return yytext[0];}
[(] {return yytext[0];}
. {;}
\n {return 0;}
%%

lab1b.y - PROGRAM

%{
#include<stdio.h>
#include<stdlib.h>
%}
%token num
%left '+' '-'
%left '*' '/'

%%
input:exp{printf("%d\n",$$);exit(0);}
exp:exp'+'exp{$$=$1+$3;}
|exp'-'exp{$$=$1-$3;}
|exp'*'exp{$$=$1*$3;}
|exp'/'exp{ if($3==0){printf("Divide by Zero error\n");exit(0);}
else  $$=$1/$3;}
|'('exp')'{$$=$2;}
|num{$$=$1;};
%%

int yyerror()
{
printf("error");
exit(0);
}

int main()
{
printf("Enter an expression:\n");
yyparse();
}

STEPS & OUTPUT :( click on image to zoom )

SS 2 - YACC - RECOGNIZE ALL STRINGS

2 - Develop, Implement and Execute a program using YACC tool to recognize all strings ending with b preceded by n a’s using the grammar an b (note: input n value)

lab2.l - PROGRAM

%{
#include "y.tab.h"
%}
%%
a {return A;}
b {return B;}
[\n] return '\n';
%%

lab2.y - PROGRAM

%{
#include<stdio.h>
#include<stdlib.h>
%}
%token A B
%%
input:s'\n' {printf("Successful Grammar\n");exit(0);}
s: A s1 B| B s1: ; | A s1
%%

main()
{
printf("Enter A String\n"); yyparse();
}

int yyerror()
{
printf("Error \n"); exit(0);
}

OUTPUT :( click on image to zoom )

SS 3 - C - PREDICTIVE/LL(1) PARSING TABLE

3 - Design, develop and implement YACC/C program to construct Predictive / LL(1) Parsing Table for the grammar rules: A ®aBa , B ®bB | e. Use this table to parse the sentence: abba$

lab3.c - PROGRAM

/*GRAMMER RULES ---- A ->aBa , B ->bB | @*/

#include<stdio.h>

#include<stdlib.h>

#include<string.h>

char prod [3][10]={"A->aBa","B->bB","B->@"};

char first[3][10]={"a","b","@"};

char follow[3][10]={"$","a","a"};

char table[3][4][10];

char input[10];

int top=-1;

char stack[25];

char curp[20];

void push(char item)

{

  stack[++top]=item;

}

void pop()

{

  top=top-1;

}

void display()

{

  int i;

  for(i=top;i>=0;i--)

  printf("%c",stack[i]);

}

int numr(char c)

{

  switch(c)

  {

    case'A':return 1;

    case'B':return 2;

    case'a':return 1;

    case'b':return 2;

    case'@':return 3;

  }

  return 1;

}

int main()

{

  char c;

  int i,j,k,n;

  for(i=0;i<3;i++){

    for(j=0;j<4;j++){

      strcpy(table[i][j],"EMPTY");

    }

  }

  printf("\nGrammar\n");

  for(i=0;i<3;i++)

  printf("%s\n",prod[i]);

  printf("\nfirst={%s,%s,%s}",first[0],first[1],first[2]);

  printf("\nfollow={%s,%s}\n",follow[0],follow[1]);

  printf("\nPredictive parsing table for the given grammar :\n");

  strcpy(table[0][0],"");

  strcpy(table[0][1],"a");

  strcpy(table[0][2],"b");

  strcpy(table[0][3],"$");

  strcpy(table[1][0],"A");

  strcpy(table[2][0],"B");

  for(i=0;i<3;i++)

  {

    if(first[i][0]!='@')

    strcpy(table[numr(prod[i][0])][numr(first[i][0])],prod[i]);

    else

    strcpy(table[numr(prod[i][0])][numr(follow[i][0])],prod[i]);

  }

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

  for(i=0;i<3;i++){

    for(j=0;j<4;j++)

    {

      printf("%-30s",table[i][j]);

      if(j==3) printf("\n-------------------------------------------------------------------\n");

    }

  }

  printf("Enter the input string terminated with $ to parse:-");

  scanf("%s",input);

  for(i=0;input[i]!='\0';i++){

    if((input[i]!='a')&&(input[i]!='b')&&(input[i]!='$'))

    {

      printf("Invalid String");

      exit(0);

    }

  }

if(input[i-1]!='$')

  {

    printf("\n\nInput String Entered Without End Marker $");

    exit(0);

  }

push('$');

  push('A');

  i=0;

printf("\n\n");

  printf("Stack\t Input\tAction");

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

while(input[i]!='$'&&stack[top]!='$')

  {

    display();

    printf("\t\t%s\t",(input+i));

    if(stack[top]==input[i])

    {

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

      pop();

      i++;

    }

    else

    {

      if(stack[top]>=65&&stack[top]<92)

      {

        strcpy(curp,table[numr(stack[top])][numr(input[i])]);

        if(!(strcmp(curp,"e")))

        {

          printf("\nInvalid String - Rejected\n");

          exit(0);

        }

        else

        {

          printf("\tApply production %s\n",curp);

          if(curp[3]=='@')

          pop();

          else

          {

            pop();

            n=strlen(curp);

            for(j=n-1;j>=3;j--)

            push(curp[j]);

          }

        }

      }

    }

  }

  display();

  printf("\t\t%s\t",(input+i));

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

  if(stack[top]=='$'&&input[i]=='$')

  {

    printf("\nValid String - Accepted\n");

  }

  else

  {

    printf("Invalid String - Rejected\n");

  }

}

OUTPUT :( click on image to zoom )

SS 4 - C - SHIFT REDUCE PARSING TECHNIQUE

4 - Design, develop and implement YACC/C program to demonstrate Shift Reduce Parsing technique for the grammar rules: E ®E+T | T, T ®T*F | F, F ®(E) | id and parse the sentence: id + id * id.

lab4.c - PROGRAM

#include<stdio.h>
#include<string.h>

int k=0,z=0,i=0,j=0,c=0;
char a[16],ac[20],stk[15],act[10];
void check();

int main()
{
    puts("GRAMMAR is E->E+E \n E->E*E \n E->(E) \n E->id");
    puts("\nEnter input string :");
    gets(a);
    c=strlen(a);
    strcpy(act,"SHIFT->");
    puts("stack \t input \t action");
    for(k=0,i=0; j<c; k++,i++,j++)
    {
        if(a[j]=='i' && a[j+1]=='d')
        {
            stk[i]=a[j];
            stk[i+1]=a[j+1];
            stk[i+2]='\0';
            a[j]=' ';
            a[j+1]=' ';
            printf("\n$%s\t%s$\t%sid",stk,a,act);
            check();
        }
        else
        {
            stk[i]=a[j];
            stk[i+1]='\0';
            a[j]=' ';
            printf("\n$%s\t%s$\t%ssymbols",stk,a,act);check();
        }
    }
}

void check()
{
    strcpy(ac,"REDUCE TO E");
    for(z=0; z<c; z++)
        if(stk[z]=='i' && stk[z+1]=='d')
        {
            stk[z]='E';
            stk[z+1]='\0';
            printf("\n$%s\t%s$\t%s",stk,a,ac);
            j++;
        }
    for(z=0; z<c; z++)
        if(stk[z]=='E' && stk[z+1]=='+' && stk[z+2]=='E')
        {
            stk[z]='E';
            stk[z+1]='\0';
            stk[z+2]='\0';
            printf("\n$%s\t%s$\t%s",stk,a,ac);
            i=i-2;
        }
    for(z=0; z<c; z++)
        if(stk[z]=='E' && stk[z+1]=='*' && stk[z+2]=='E')
        {
            stk[z]='E';
            stk[z+1]='\0';
            stk[z+2]='\0';
            printf("\n$%s\t%s$\t%s",stk,a,ac);
            i=i-2;
        }
    for(z=0; z<c; z++)
        if(stk[z]=='(' && stk[z+1]=='E' && stk[z+2]==')')
        {
            stk[z]='E';
            stk[z+1]='\0';
            stk[z+1]='\0';
            printf("\n$%s\t%s$\t%s",stk,a,ac);
            i=i-2;
        }

}

OUTPUT :( click on image to zoom )

SS 5 - C - GENERATE MACHINE CODE USING TRIPLES

5 - Design, develop and implement a C/Java program to generate the machine code using
Triples for the statement A = -B * (C +D) whose intermediate code in three-address form:

T1 = -B
T2 = C + D
T3 = T1 + T2
A = T3

lab5.c - PROGRAM

#include<stdio.h>
#include<stdlib.h>
#include<ctype.h>

char op[2],arg1[5],arg2[5],result[5];
int main()
{
    FILE *fp1,*fp2;
    fp1=fopen("input.txt","r");
    fp2=fopen("output.txt","w");
    while(!feof(fp1))
    {
        fscanf(fp1,"%s%s%s%s",result,arg1,op,arg2);
        if(strcmp(op,"+")==0)
        {
            fprintf(fp2,"\nMOV R0,%s",arg1);
            fprintf(fp2,"\nADD R0,%s",arg2);
            fprintf(fp2,"\nMOV %s,R0",result);
        }

        if(strcmp(op,"*")==0)
        {
            fprintf(fp2,"\nMOV R0,%s",arg1);
            fprintf(fp2,"\nMUL R0,%s",arg2);
            fprintf(fp2,"\nMOV %s,R0",result);
        }

        if(strcmp(op,"-")==0)
        {
            fprintf(fp2,"\nMOV R0,%s",arg1);
            fprintf(fp2,"\nSUB R0,%s",arg2);
            fprintf(fp2,"\nMOV %s,R0",result);
        }

        if(strcmp(op,"/")==0)
        {
            fprintf(fp2,"\nMOV R0,%s",arg1);
            fprintf(fp2,"\nDIV R0,%s",arg2);
            fprintf(fp2,"\nMOV %s,R0",result);
        }

        if(strcmp(op,"=")==0)
        {
            fprintf(fp2,"\nMOV R0,%s",arg1);
            fprintf(fp2,"\nMOV %s,R0",result);
        }
    }
    fclose(fp1);
    fclose(fp2);
}

input.txt

T1 -B = ?

T2 C + D

T3 T1 * T2

A T3 = ?

OUTPUT :( click on image to zoom )

SS 6.A - LEX - ELIMINATE COMMENT LINES IN A C PROGRAM

6 - a) Write a LEX program to eliminate comment lines in a C program and copy the resulting program into a separate file.

lab6.l - PROGRAM

%{
#include<stdio.h>
int sl=0;
int ml=0;
%}
%%
"/*"[a-zA-Z0-9' '\t\n]+"*/" ml++;
"//".* sl++;
%%

main()
{
yyin=fopen("f1.c","r");
yyout=fopen("f2.c","w");
yylex();
fclose(yyin);
fclose(yyout);
printf("\n Number of single line comments are = %d\n",sl);
 printf("\nNumber of multiline comments are =%d\n",ml);
}

f1.c FILE

#include<stido.h>

int main()
{
// this is a comment
printf("hello");
/* this is another comment */
}


OUTPUT :( click on image to zoom )

SS 6.B - YACC - RECOGNIZE VALID IDENTIFIER, OPERATORS AND KEYWORDS

6 - b) Write YACC program to recognize valid identifier, operators and keywords in the given text (C program) file.

lab6.l PROGRAM

%{
#include <stdio.h>
#include "y.tab.h"
extern yylval;
%}

%%
[ \t];
[+|-|*|/|=|<|>] {printf("operator is %s\n",yytext);return OP;}
[0-9]+ {yylval = atoi(yytext); printf("numbers is %d\n",yylval); return DIGIT;}
int|char|bool|float|void|for|do|while|if|else|return|void {printf("keyword is %s\n",yytext);return KEY;}
[a-zA-Z0-9]+ {printf("identifier is %s\n",yytext);return ID;}
. ;
%%

lab6.y PROGRAM

%{
#include <stdio.h>
#include <stdlib.h>
int id=0, dig=0, key=0, op=0;
%}
%token DIGIT ID KEY OP

%%
input:
DIGIT input { dig++; }
| ID input { id++; }
| KEY input { key++; }
| OP input {op++;}
| DIGIT { dig++; }
| ID { id++; }
| KEY { key++; }
| OP { op++;}
;
%%

#include <stdio.h>
extern int yylex();
extern int yyparse();
extern FILE *yyin;
main()
{
FILE *myfile = fopen("inputfile.c", "r");
if (!myfile)
{
printf("I can't open inputfile.c!");
return -1;
}
yyin = myfile;
do{
yyparse();
}while (!feof(yyin));
printf("numbers = %d\nKeywords = %d\nIdentifiers = %d\noperators = %d\n",dig, key,id, op);
}

void yyerror()
{
printf("EEK, parse error! Message: ");
exit(-1);
}

inputfile.txt FILE

#include<stdio.h>

int main()
{
int a ;
int b ;
a = 1 ;
b = 2 ;
a = a+b;
return 0 ;
}


OUTPUT :( click on image to zoom )

SS 7 - C - ROUND ROBIN (RR) SCHEDULING ALGORITHM

7 - Design, develop and implement a C/C++/Java program to simulate the working of Shortest remaining time and Round Robin (RR) scheduling algorithms. Experiment with different quantum sizes for RR algorithm.

lab7.c PROGRAM

// CPU Scheduling -Round Robin
#include<stdio.h>
struct process
{
char name;
int at,bt,wt,tt,rt;
int completed;
}p[10];

int n;
int q[10];  //queue
int front=-1,rear=-1;
void enqueue(int i)
{
    if(rear==10)
        printf("overflow");
    rear++;
    q[rear]=i;
    if(front==-1)
        front=0;
}

int dequeue()
{
    if(front==-1)
        printf("underflow");
    int temp=q[front];
    if(front==rear)
        front=rear=-1;
    else
        front++;
    return temp;
}

int isInQueue(int i)
{
    int k;
    for(k=front;k<=rear;k++)
    {
        if(q[k]==i)
        return 1;
    }
    return 0;
}

void sortByArrival()
{
    struct process temp;
    int i,j;
    for(i=0;i<n-1;i++)
    {
        for(j=i+1;j<n;j++)
        {
            if(p[i].at>p[j].at)
            {
                temp=p[i];
                p[i]=p[j];
                p[j]=temp;
            }
        }
    }
}

int main()
{
    int i,j,time=0,sum_bt=0,tq;
    char c;
    float avgwt=0;
     printf("Enter Number of Processes:\n");
     scanf("%d",&n);
     for(i=0,c='A';i<n;i++,c++)
     {
         p[i].name=c;
         printf("\nProcess %c\n",c);
         printf("\tArrival Time :");
         scanf("%d",&p[i].at);
         printf("\tBurst Time :");
         scanf("%d",&p[i].bt);
         p[i].rt=p[i].bt;
         p[i].completed=0;
         sum_bt+=p[i].bt;
    }
    printf("\nEnter the time quantum:");
    scanf("%d",&tq);
    sortByArrival();
    enqueue(0);
    printf("\nProcess execution order: ");
    for(time=p[0].at;time<sum_bt;)
    {
        i=dequeue();
        if(p[i].rt<=tq)
        {
            time+=p[i].rt;
            p[i].rt=0;
            p[i].completed=1;
            printf(" %c ",p[i].name);
            p[i].wt=time-p[i].at-p[i].bt;
            p[i].tt=time-p[i].at;
            for(j=0;j<n;j++)
            {
                if(p[j].at<=time && p[j].completed!=1&& isInQueue(j)!=1)
                {
                    enqueue(j);
                }
            }
        }
        else
        {
            time+=tq;
            p[i].rt-=tq;
            printf(" %c ",p[i].name);
            for(j=0;j<n;j++)
            {
                if(p[j].at<=time && p[j].completed!=1&&i!=j&& isInQueue(j)!=1)
                {
                    enqueue(j);
                }
            }
            enqueue(i);
        }
    }
    printf("\n\nName\tArrival Time\tBurst Time\tResponse Time\tTurnAround Time");
    for(i=0;i<n;i++)
    {
        avgwt+=p[i].wt;
        printf("\n%c\t\t%d\t\t%d\t\t%d\t\t%d",p[i].name,p[i].at,p[i].bt,p[i].wt,p[i].tt);
    }
    printf("\n\nAverage waiting time:%f",avgwt/n);
}

OUTPUT :( click on image to zoom )

SS 8 - C - BANKER’S ALGORITHM

8 - Design, develop and implement a C/C++/Java program to implement Banker’s algorithm. Assume suitable input required to demonstrate the results.

lab8.c PROGRAM

#include <stdio.h>
#include <stdlib.h>

int main()
{
int Max[10][10], need[10][10], alloc[10][10], avail[10], completed[10], safeSequence[10];
int p, r, i, j, process, count = 0;
printf("Enter the no of processes : ");
scanf("%d", &p);
for(i = 0; i< p; i++)
completed[i] = 0;

printf("\n\nEnter the no of resources : ");
scanf("%d", &r);
printf("\n\nEnter the Max Matrix for each process : ");
for(i = 0; i < p; i++)
{
printf("\nFor process %d : ", i + 1);
for(j = 0; j < r; j++)
scanf("%d", &Max[i][j]);
}

printf("\n\nEnter the allocation for each process : ");
for(i = 0; i < p; i++)
{
printf("\nFor process %d : ",i + 1);
for(j = 0; j < r; j++)
scanf("%d", &alloc[i][j]);
}

printf("\n\nEnter the Available Resources : ");
for(i = 0; i < r; i++)
scanf("%d", &avail[i]);
for(i = 0; i < p; i++)
for(j = 0; j < r; j++)
need[i][j] = Max[i][j] - alloc[i][j];

do
{
printf("\n Max matrix:\tAllocation matrix:\n");
for(i = 0; i < p; i++)
{
for( j = 0; j < r; j++)
printf("%d ", Max[i][j]);
printf("\t\t");
for( j = 0; j < r; j++)
printf("%d ", alloc[i][j]);
printf("\n");
}
process = -1;
for(i = 0; i < p; i++)
{
if(completed[i] == 0)//if not completed
{
process = i ;

for(j = 0; j < r; j++)
{
if(avail[j] < need[i][j])
{
process = -1;
break;
}
}
}
if(process != -1)
break;
}
if(process != -1)
{
printf("\nProcess %d runs to completion!", process + 1);
safeSequence[count] = process + 1;
count++;
for(j = 0; j < r; j++)
{
avail[j] += alloc[process][j];
alloc[process][j] = 0;
Max[process][j] = 0;
completed[process] = 1;
}
}
} while(count != p && process != -1);

if(count == p)
{
printf("\nThe system is in a safe state!!\n"); printf("Safe Sequence : < "); for( i = 0; i < p; i++)
printf("%d ", safeSequence[i]);
printf(">\n");
}
else
printf("\nThe system is in an unsafe state!!");
}


OUTPUT :( click on image to zoom )

SS 9 - C - PAGE REPLACEMENT ALGORITHMS LRU AND FIFO

9 - Design, develop and implement a C/C++/Java program to implement page replacement algorithms LRU and FIFO. Assume suitable input required to demonstrate the results.

lab9.c PROGRAM

#include<stdio.h>
#include<stdlib.h>

void FIFO(char [ ],char [ ],int,int);
void lru(char [ ],char [ ],int,int);
void opt(char [ ],char [ ],int,int);

int main()
{
    int ch,YN=1,i,l,f;
    char F[10],s[25];
    printf("\nEnter the no of empty frames: ");
    scanf("%d",&f);
    printf("\nEnter the length of the string: ");
    scanf("%d",&l);
    printf("\nEnter the string: ");
    scanf("%s",s);
    for(i=0;i<f;i++)
        F[i]=-1;

    do
    {
        printf("\n*********** MENU ***********");
        printf("\n1:FIFO\n2:LRU \n3:EXIT");
        printf("\nEnter your choice: ");
        scanf("%d",&ch);

        switch(ch)
        {
            case 1: for(i=0;i<f;i++)
                        F[i]=-1;
                    FIFO(s,F,l,f);
                    break;

            case 2: for(i=0;i<f;i++)
                        F[i]=-1;
                    lru(s,F,l,f);
                    break;

            case 3: exit(0);
        }
        printf("\n\nDo u want to continue IF YES PRESS 1\nIF NO PRESS 0 : ");
        scanf("%d",&YN);
    } while(YN==1);
    return(0);
}


void FIFO(char s[],char F[],int l,int f)
{
    int i,j=0,k,flag=0,cnt=0;
    printf("\n\tPAGE\t FRAMES\t\t\t FAULTS");
    for(i=0;i<l;i++)
    {
        for(k=0;k<f;k++)
        {
            if(F[k]==s[i])
                flag=1;
        }

        if(flag==0)
        {
            printf("\n\t%c\t",s[i]);
            F[j]=s[i];
            j++;
            for(k=0;k<f;k++)
                printf(" %c",F[k]);
            printf("\tPage-fault%d",cnt);
            cnt++;
        }

        else
        {
            flag=0;
            printf("\n\t%c\t",s[i]);
            for(k=0;k<f;k++)
                printf(" %c",F[k]);
            printf("\tNo page-fault");
        }
        if(j==f)
            j=0;
    }
}


void lru(char s[],char F[],int l,int f)
{
    int i,j=0,k,m,flag=0,cnt=0,top=0;
    printf("\n\tPAGE\t FRAMES\t\t\t FAULTS");
    for(i=0;i<l;i++)
    {
        for(k=0;k<f;k++)
        {
            if(F[k]==s[i])
            {
                flag=1;
                break;
            }
        }
        printf("\n\t%c\t",s[i]);
        if(j!=f && flag!=1)
        {
            F[top]=s[i];
            j++;
            if(j!=f)
                top++;
        }

        else
        {
            if(flag!=1)
            {
                for(k=0;k<top;k++)
                    F[k]=F[k+1];
                F[top]=s[i];
            }

            if(flag==1)
            {
                for(m=k;m<top;m++)
                    F[m]=F[m+1];
                F[top]=s[i];
            }
        }

        for(k=0;k<f;k++)
            printf(" %c",F[k]);

        if(flag==0)
        {
            printf("\tPage-fault%d",cnt);
            cnt++;
        }
        else
            printf("\tNo page fault");
        flag=0;
    }
}

OUTPUT :( click on image to zoom )

SS 10.A - C - A NUMERICAL CALCULATOR

10 - a). Design, develop and implement a C/C++/Java program to simulate a numerical calculator.

lab10a.c PROGRAM

#include <stdio.h>

int main()
{
char operator;
float num1, num2, result;
printf("Simulation of a Simple Calculator\n\n");
printf("Enter two numbers \n");
scanf("%f %f", &num1,&num2);

fflush(stdin);
printf("\nEnter the operator [+,-,*,/] \n");
scanf("%s", &operator);

switch(operator)
{
case '+': result = num1 + num2; break;
case '-': result = num1 - num2; break;
case '*': result = num1 * num2; break;
case '/': result = num1 / num2; break;
default : printf("Error in operation");     break;
}
printf("\n%.2f %c %.2f = %.2f\n", num1, operator, num2, result);
}

OUTPUT :( click on image to zoom )

SS 10.B - C - PAGE REPLACEMENT TECHNIQUE

10 - b). Design, develop and implement a C/C++/Java program to simulate page replacement technique.

lab10b.c PROGRAM

#include<stdio.h>

int main()
{
      int reference_string[10], page_faults = 0, m, n, s, pages, frames;
      printf("\nEnter Total Number of Pages:\t");
      scanf("%d", &pages);
      printf("\nEnter values of Reference String:\n");
      for(m = 0; m < pages; m++)
      {
            printf("Value No. [%d]:\t", m + 1);
            scanf("%d", &reference_string[m]);
      }
      printf("\nEnter Total Number of Frames:\t");
      scanf("%d", &frames);
      int temp[frames];
      for(m = 0; m < frames; m++)
            temp[m] = -1;
      for(m = 0; m < pages; m++)
      {
            s = 0;
            for(n = 0; n < frames; n++)
            {
                  if(reference_string[m] == temp[n])
                  {
                        s++;
                        page_faults--;
                  }
            }
            page_faults++;
            if((page_faults <= frames) && (s == 0))
                  temp[m] = reference_string[m];
            else if(s == 0)
                  temp[(page_faults - 1) % frames] = reference_string[m];
            printf("\n");
            for(n = 0; n < frames; n++)
                  printf("%d\t", temp[n]);
      }
      printf("\nTotal Page Faults:\t%d\n", page_faults);
      return 0;
}

OUTPUT :( click on image to zoom )

SS 11 - LEX - SENTENCE IS SIMPLE/COMPOUND

11. Program to recognize whether a given sentence is simple or compound.

vi lab11.l

%{

int flag=0;

%}

%%

" and " |

" or " |

" but " |

" because " |

" than "|

" nevertheless " {flag=1;}

%%

int main()

{

printf("Enter the sentence:\n");

yylex();

if(flag==1)

printf("compound statement");

else

printf("simple statements\n");

}

OUTPUT :( click on image to zoom )

SS 12 - C++ EMULATE UNIX ln COMMAND

12. Write a C++ program to emulate the Unix ln command

lab12.cpp PROGRAM

#include<iostream>

#inculde<unistd.h>

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

{

using namespace std;

if(argc!=3)

{

cout<<"Usage ./a.out sourcefile destination file\n";

return 0;

}

if(link(argv[1],argv[2])==-1)

{

cout<<"cant link\n";

return 1;

}

else

{

cout<<"Files have been Linked\n";

}

return 0;

}

OUTPUT :( click on image to zoom )