unary minus in shunting yard expression parser

Question

here is my expression parser using shunting-yard algorithm it work well as expected except in one situation , when I use unary minus like in -2*3 it wont work (I think it shouldn't because I didn't find anything in algorithm to handle this ) is there a simple way that I can fix this? (this is a simple parser I only need () + - * / ^ ) Regards Pedram

#include <cctype>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cmath>
using namespace std;
int olaviat (char c) {
   /*************
    **Operator precedence 
    *************/
  switch(c) {
       case '-' : case '+' :
           return 1 ;
       case '*' : case '/' :
           return 2 ;
       case '^' :
           return 3 ;
       default :
           return 0 ;
  }
}
double eval(char *exp) {
    /*************
    **Convert to reverse polish
    *************/
    char n [50] , o[50] ;
    static int nl = 0  , ol = 0 ;

    while (*exp) {
            while(isspace(*exp)) *exp++ ;
        if(*exp == '(') {
             o[ol++]  = *exp++ ;
           }
        else if (*exp == ')'){
            while(o[--ol]!='('){
                    n[nl++] = o[ol];
                    n[nl++] = ' ';
                  }
                  *exp++;
        }
        else if (isdigit(*exp)) {
          while (isdigit(*exp)) {
            n[nl++] = *exp++ ;
          }
        n[nl++] = ' ' ;
        }
        else if (strchr("+-*/^",*exp)){
            if(olaviat(*exp) > olaviat(o[ol-1])) {
               o[ol++]  = *exp++ ;


            }
            else {
                    if(olaviat(*exp) == olaviat(o[ol-1]) && olaviat(*exp)== 3) {
                      o[ol++]  = *exp++ ;
                    }else{
                n[nl++] = o[ol-1] ;
                n[nl++] = ' ' ;
                o[--ol] = '\0' ;

                    }
            }
        }

    }

for (int k = ol-1 ; k >= 0 ; k --){
    n[nl++] = o[k];
    n[nl++] = ' ' ;
}
/*******************************/
cout << "Reverse Polish" << endl ;
for (int i = 0 ; i < nl-1 ; i++){
        cout << n[i]  ;
    }
cout << endl ;
//n[nl+1] = '\0' ;
/*******************************
**Calculate Result
*******************************/
    double temp[50];
    char *e ;
    ol = 0;
   int  nol = 0 ;
    e=n ;
    int digitcount = 0;
    while (*e) {
            while (isspace(*e)) *e++;
        if (isdigit(*e)) {
          while (isdigit(*e)) {
             o[ol++] =*e++ ;
             digitcount++ ;
          }
        temp[nol++] = atof(o) ;
        for (int i = 0 ; i < digitcount ; i++)
            o[i]='\0' ;
        ol=0;
        digitcount = 0 ;
        }
        else if (strchr("+-*/^",*e)){
          // char opr ;
           double tempAns = 0;
           switch (*e) {
              case '+' :
                  tempAns = temp[nol-2] + temp [nol-1] ;
                  break ;
              case '-' :
                  tempAns = temp [nol-2] - temp [nol-1] ;
                  break;
              case '*' :
                  tempAns = temp [nol-2] * temp [nol-1] ;
                  break;
              case '/' :
                  tempAns = temp[nol-2] / temp[nol-1];
                  break ;
              case '^' :
                  tempAns = pow(temp[nol-2],temp [nol-1]);
                  break ;
              default :
                cout << "\n Unknown error" ;
                continue;
           }
           *e++ ;
           nol--;
           temp[nol-1] = tempAns ;
           temp[nol] = NULL ;
        }
        else {
            break ;
        }
    }
    double ans = temp[0];

  return ans ;
}

int main() {

char exp[100];
char c;
start :
    cin.get (exp , 99);
    cout << "\n\tANS= " << eval(exp)  ;
    cout << endl ;
    system("PAUSE");
    return 0;
} 

Answer 1

The above option is correct, but it would get very cumbersome and buggy. Consider the case 2*-(1+2)^-(2+5*-(2+4)). As you can see you need to take in account many things. Also whenever you find *-(, for example, you know that you'll substitute that with *(0-(..., which would be coded in a cumbersome recursive function.

The best solution is much easier. When parsing the operators, take into account the cases when the operator is - and it is preceded by another operator, or preceded by a left parenthesis, or when it is the first character of the input (these cases mean that it is a unary minus rather than binary). In this case, you change it to another character, say u (this was my case), and make its precedence the same as that of ^.

Also, treating it as part of the number literal has its catch. Imagine a case such as -2^4. In Wolfram Alpha you'd get -16, not 16.

And consider using stacks. They'll make your life easier.

Let me explain what I meant. Consider you are given the input:

2 / - 7 + ( - 9 * 8 ) * 2 ^ - 9 - 5

Making the replacements I suggested, it would become like this:

2 / u 7 + ( u 9 * 8 ) * 2 ^ u 9 - 5

Now your operator precedence switch should be changed to:

switch(c)
{
       case '-' : case '+' :
           return 1 ;
       case '*' : case '/' :
           return 2 ;
       case '^' : case 'u': //note the 'u' operator we added
           return 3 ;
       default :
           return 0 ;
}

And, of course, you need to make changes to support this unary operator.

Answer 2

One option is to put a 0 in front if the first character is '-'. You have to do this also when the - is after a (.

Nicer ones are implementing either the unary minus operator or treating it as part of the number literal.