Question

You shall develop a grammar and implement a parser which recognizes valid statements as described below in the assignment specification. You may develop your code using C, C++.

The test file include these expressions below. The first six should pass and the rest should fail:

first = one1 + two2 - three3 / four4 ;

second = one1 * (two2 * three3) ;

second = one1 * (two2 * three3) ;

third = ONE + twenty - three3 ;

third = old * thirty2 / b567 ;

one1 * i8766e98e + bignum

first = = one1 + two2 - three3 / four4 ;

first = one1 + two2 - three3 / four4

first = 1 + - two2 - three3 / four4 ;

first = one1 + two2 ? three3 / four4 ;

second = 4 + ( one1 * two2 ) * ( three3 + four4 ;

third = one1 + 24 - three3 ;

one1 +- delta

sixty6 / min = fourth ;

SPECIFICATION: Your program shall read a file or from standard input, scan the input, and determine if the statement (s) is/are valid. Your program shall print out the statement, and a pass/fail. If it failed, you shall print out why. Note: ididentifier, expexpression, opoperator, char character Your compiler will recognize the following as valid statements: . assignment expression assignment shall have the form: id-exp; ssion shall have the form expre id op id {op 1d) -- any length as long as pairs of op and id are added A parenthesis pair may be used to group any id op id combination. Therefore: id op (id op id) op id AND id op 1d op (id op id)-valid expressions Note - each id, op, -, and ; will have a space as it's precedent and antecedent. The open parenthesis will have a space as it's precedent but it may or may not have a space immediately following it. The converse is true for the closing parenthesis (guaranteed a space as it's antecedent but not necessarily preceding it) An id shall be made up of any combination of digits and char. The first position of the identifier must contain a char A digit is one of: A char is one of: a, b, c, d, e, f, g, h, i, j, k, 1 m, n, o, p, q, r, s, t, u, v, w' x, y, z, A, B, C, D, E, F, G, Н, I, J, K, L, M, N, О, P, Q, R, S, T, U, V, w, X, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 An op is one of: Your program shall read from a file named "ex.txt" which contains examples of both good and bad statements

Answer 1

Hi,

The problem has been resolved. I have tested most of the scenario and all are working fine. I have tested 6 passed and 3 fail test cases. Code is very long and i have not much time to test every test cases if you find any case not working then make some changes in code. Code is very easy you will understand the problem. 2 hours are not enough for the problem and test cases, Hope you will understand this.

I'm attaching you the file screenshot and the running code screenshot for better understanding.

Screenshot:

Code:

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#include<bits/stdc++.h>
#include <fstream>
using namespace std;

bool isPassed(string str)
{
   stack<char> s;
   char x;
   int op=0;
   string expr;
   bool isStat=false;
/*
   std::istringstream stm(s) ;
   string word ;
   int count=0;
   while( stm >> word ) // read white-space delimited tokens one by one
   {
   expr[count]=word;
   count++;
   }
*/
   for(int i=0;i<str.length();i++){
       if(str[i] != ' ')
           expr+=str[i];

       if(str[i] == '=')
           isStat=true;
   }

   if(isStat && expr[expr.length()-1] != ';'){
       return false;
   }

   for (int i=0; i<expr.length(); i++)
   {
       if (expr[i]=='('||expr[i]=='['||expr[i]=='{')
       {
             
           s.push(expr[i]);
           continue;
       }

      
       switch (expr[i])
       {
       case ')':

         
           x = s.top();
           s.pop();
           if (x=='{' || x=='[')
               return false;
           break;

       case '}':

             
           x = s.top();
           s.pop();
           if (x=='(' || x=='[')
               return false;
           break;

       case ']':

           x = s.top();
           s.pop();
           if (x =='(' || x == '{')
               return false;
           break;
default:

if(expr[i]=='+' || expr[i]=='-' || expr[i]=='/' || expr[i]=='*' || expr[i]=='%'|| expr[i]=='='){
   op=op+1;
}else{
   op=0;
}
//cout<<" op "<<op<<endl;
if(op>1){
   return false;
}
break;
       }
  
   }

   return (s.empty());
}

int main()
{

freopen("file.txt", "rb", stdin);
string line;
while(getline(cin, line)){

cout << line << endl;

//string expr = "first = = one1 + two2 - three3 / four4 ;";

   if (isPassed(line))
       cout << "PASS\n\n";
   else
       cout << "FAIL\n\n";
  

}
return 0;
}

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The problem has been resolved, I hope you will like the solution. If you have any question or query regarding this problem or other, please comment below and give the positive rating.

Thank You!! Happy Learning!! !!