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Wednesday, 12 October 2016

Predictive Parsing Of the grammar

/*This Program implements the Predictive Parsing Of the grammar
E->E+T/T
F->F*T/F
F->id(Identifier)*/
#include<string.h>
#include<stdio.h>
char a[10];
int top=-1,i;
void error(){
printf("Syntax Error");
}
void push(char k[]) //Pushes The Set Of Characters on to the Stack
{
  for(i=0;k[i]!='\0';i++)
  {
    if(top<9)
    a[++top]=k[i];
  }
}
char TOS()        //Returns TOP of the Stack
{
  return a[top];
}
void pop()       //Pops 1 element from the Stack
{
  if(top>=0)
    a[top--]='\0';
}
void display()  //Displays Elements Of Stack
{
  for(i=0;i<=top;i++)
    printf("%c",a[i]);
}
void display1(char p[],int m) //Displays The Present Input String
{
  int l;
  printf("\t");
  for(l=m;p[l]!='\0';l++)
    printf("%c",p[l]);
}
char* stack(){
return a;
}
int main()
{
 char ip[20],r[20],st,an;
 int ir,ic,j=0,k;
 char t[5][6][10]=  {"$","$","TH","$","TH","$",
      "+TH","$","e","e","$","e",
      "$","$","FU","$","FU","$",
      "e","*FU","e","e","$","e",
      "$","$","(E)","$","i","$"};
 printf("\nEnter any String(Append with $)");
 gets(ip);
 printf("Stack\tInput\tOutput\n");
 push("$E");
 display();
 printf("\t%s\n",ip);
 for(j=0;ip[j]!='\0';)
 {
  if(TOS()==an)
  {
   pop();
   display();
   display1(ip,j+1);
   printf("\tPOP\n");
   j++;
  }
  an=ip[j];
  st=TOS();
  if(st=='E')ir=0;
  else if(st=='H')ir=1;
  else if(st=='T')ir=2;
  else if(st=='U')ir=3;
  else if(st=='F')ir=4;
  else 
  {
   error();
   break;
  }
  if(an=='+')ic=0;
  else if(an=='*')ic=1;
  else if(an=='(')ic=2;
  else if(an==')')ic=3;
  else if((an>='a'&&an<='z')||(an>='A'&&an<='Z')){ic=4;an='i';}
  else if(an=='$')ic=5;
  strcpy(r,strrev(t[ir][ic]));
  strrev(t[ir][ic]);
  pop();
  push(r);
  if(TOS()=='e')
  {
   pop();
   display();
   display1(ip,j);
   printf("\t%c->%c\n",st,238);
  }
  else{
   display();
   display1(ip,j);
   printf("\t%c->%s\n",st,t[ir][ic]);
  }
  if(TOS()=='$'&& an=='$')
   break;
  if(TOS()=='$')
  {
   error();
   break;
  }
 }
 k=strcmp(stack(),"$");
 if(k==0 && j==strlen(ip)-1)
  printf("\nGiven String is accepted");
 else
  printf("\nGiven String is not accepted");
 return 0;
}

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