Parser first and Follow LEX C program for Syntax analysis

 Below is the source code to:-

Develop a program to implement

a. Predictive parser

b. Operator precedence parser

#include<stdio.h> 

#include<conio.h> 

#include<string.h> 

char prol[7][10]={"S","A","A","B","B","C","C"}; 

char pror[7][10]={"A","Bb","Cd","aB","@","Cc","@"}; 

char prod[7][10]={"S->A","A->Bb","A->Cd","B->aB","B->@","C->Cc","C->@"}; char first[7][10]={"abcd","ab","cd","a@","@","c@","@"}; char 

follow[7][10]={"$","$","$","a$","b$","c$","d$"}; 

char table[5][6][10]; 

int numr(char c) 

{ 

switch(c){ 

case 'S': return 0; 

case 'A': return 1; 

case 'B': return 2; 

case 'C': return 3; 

case 'a': return 0; 

case 'b': return 1; 

case 'c': return 2; 

case 'd': return 3; 

case '$': return 4; 

} 

return(2); 

} 

void main() 

{ 

int i,j,k; 

clrscr(); 

for(i=0;i<5;i++) 

for(j=0;j<6;j++) 

strcpy(table[i][j]," "); 

printf("\nThe following is the predictive parsing table for the following grammar:\n"); for(i=0;i<7;i++) 

printf("%s\n",prod[i]); 

printf("\nPredictive parsing table is\n"); 

fflush(stdin); 

for(i=0;i<7;i++){ 

k=strlen(first[i]); 

for(j=0;j<10;j++)

if(first[i][j]!='@') 

strcpy(table[numr(prol[i][0])+1][numr(first[i][j])+1],prod[i]); } 

for(i=0;i<7;i++){ 

if(strlen(pror[i])==1) 

{ 

if(pror[i][0]=='@') 

{ 

k=strlen(follow[i]); 

for(j=0;j<k;j++) 

strcpy(table[numr(prol[i][0])+1][numr(follow[i][j])+1],prod[i]); } 

} 

} 

strcpy(table[0][0]," "); 

strcpy(table[0][1],"a"); 

strcpy(table[0][2],"b"); 

strcpy(table[0][3],"c"); 

strcpy(table[0][4],"d"); 

strcpy(table[0][5],"$"); 

strcpy(table[1][0],"S"); 

strcpy(table[2][0],"A"); 

strcpy(table[3][0],"B"); 

strcpy(table[4][0],"C"); 

printf("\n--------------------------------------------------------\n"); for(i=0;i<5;i++) 

for(j=0;j<6;j++){ 

printf("%-10s",table[i][j]); 

if(j==5) 

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

getch(); 

}