Description In this assignment you need to implement a recur

Description: In this assignment, you need to implement a recursive descent parser in C++ for the following CFG:

1. exps --> exp | exp NEWLINE exps

2. exp --> term {addop term}

3. addop --> + | -

4. term --> factor {mulop factor}

5. mulop --> * | /

6. factor --> ( exp ) | INT

The 1st production defines exps as an individual expression, or a sequence expressions separated by NEWLINE token.

The 2nd production describes an expression as a term followed by addop term zero, one, or more times, i.e. exp can be: term, or term addop term, or term addop term addop term, or term addop term addop term addop term, …

The 4th production describes the definition of term, which is pretty much similar to 2nd production.

The 6th production defines a factor either as an expression enclosed by a pair of parentheses or an integer.

In recursive descent parsing, a function is defined for each non-terminal, i.e. exps, exp, term, and factor in our case. The non-terminals addop and mulop are too simple so that we will process them directly in functions of exp and term respectively.

Solution

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

#include<ctype.h>

char input[100];
int i,error;
void E();
void T();
void F();
void EXPS();

main()
{
               i=0;                   // Global Variable
               error=0;               //Global Variable
printf(\"Enter an arithmetic expression : \"); // Eg: a+a*a
gets(input);
EXPS();                   //Start Evaluation of CFG
if(strlen(input)==i&&error==0)
printf(\"\ Accepted\ \");
else printf(\"\ Rejected\ \");
}
                      
void EXPS()
{
E();               //call expression

     
}
void E()
{
T();               //Term
   if(input[i]==\'\ \')       //after term it check if next char is newline token which means sequence of expressions
   {
       i++;
       E();
   }
}
void T()
{  
   F();                   //calls factor
   if(input[i]==\'*\' || input[i]==\'/\')            //checks for next char
   {
       i++;
       F();
   }
   else if(input[i]==\'+\' || input[i]==\'-\')           //checks for addop if it was present after one occurence
   {
       i++;
       T();
   }
}

void F()
{
if(isalnum(input[i]))
           i++;
else if(input[i]==\'(\')
           {
               i++;
               E();
               if(input[i]==\')\')
                   i++;

               else error=1;
}
      
       else if(input[i]==\'*\' || input[i]==\'/\')
   {
       i++;
       F();
   }

else error=1;
}