This Program implements the Predictive Parsing Of the grammar
E->E+T/T
F->F*T/F
F->id(Identifier)
To Implement Predictive Parsing in C
//To Implement Predictive Parsing
#include<string.h>
#include<conio.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","$"};
clrscr();
printf("\nEnter any String(Append with $)");
gets(ip);
printf("Stack\tInput\tOutput\n\n");
push("$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 && i==strlen(ip))
printf("\n Given String is accepted");
else
printf("\n Given String is not accepted");
return 0;
}
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