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Convert from an infix expression to postfix (C++) using Stacks

Asked 2012-10-02T03:18:41.370
9

My lecturer gave me an assignment to create a program to convert and infix expression to postfix using Stacks. I've made the stack classes and some functions to read the infix expression.

But this one function, called convertToPostfix(char * const inFix, char * const postFix) which is responsible to convert the inFix expression in the array inFix to the post fix expression in the array postFix using stacks, is not doing what it suppose to do. Can you guys help me out and tell me what I'm doing wrong?

The following is code where the functions to convert from inFix to postFix is and convertToPostfix(char * const inFix, char * const postFix) is what I need help fixing:

 void ArithmeticExpression::inputAndConvertToPostfix()
    {
       char inputChar; //declaring inputChar
       int i = 0; //inizalize i to 0

       cout << "Enter the Arithmetic Expression(No Spaces): ";

       while( ( inputChar = static_cast<char>( cin.get() ) ) != '\n' )
       {
          if (i >= MAXSIZE) break; //exits program if i is greater than or equal to 100

          if(isdigit(inputChar) || isOperator(inputChar))
          {
             inFix[i] = inputChar; //copies each char to inFix array
             cout << inFix[i] << endl;
          }
          else
             cout << "You entered an invalid Arithmetic Expression\n\n" ;

          }

          // increment i;
          i++;
          convertToPostfix(inFix, postFix);


       }




    bool ArithmeticExpression::isOperator(char currentChar)
    {

        if(currentChar == '+')
            return true;
        else if(currentChar == '-')
            return true;
        else if(currentChar == '*')
            return true;
        else if(currentChar == '/')
            return true;
        else if(currentChar == '^')
            return true;
        else if(currentChar == '%')
            return true;
        else
            return
c++ stack infix-notation postfix-notation
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1 Answer

7

This is basically a comment to the answer from Yuushi.

  • The outer while(!stack.empty()) loop is wrong. just remove it. (keep the loop body ofc). At the end of the function, check that the stack is empty, else the expression had syntax errors.
  • As Yuushi already said the precedence function looks bogus. First you should give the parameters better names: one is the operator to the left, and the other to the right. (Right now you call it precedence(rightOp, leftOp)). Then you should document what the result means - right now you return true if a rOp b lOp c == (a rOp b) lOp c (yes, the operator order doesn't match what you call - "+" and "-" are not the same in both orders for example).
  • If you find a new operator you need to loop over the old operators on the stack, for example after reading a - b * c your output is a b c and the stack is [- *]. now you read a +, and you need to pop both operators, resulting in a b c * -. I.e., the input a - b * c + d should result in a b c * - d +

Update: appended complete solution (based on Yuushi's answer): 

bool isOperator(char currentChar)
{
    switch (currentChar) {
    case '+':
    case '-':
    case '*':
    case '/':
    case '^':
    case '%':
        return true;
    default:
        return false;
    }
}

// returns whether a `lOp` b `rOp` c == (a `lOp` b) `rOp` c
bool precedence(char leftOperator, char rightOperator)
{
    if ( leftOperator == '^' ) {
        return true;
    } else if ( rightOperator == '^' ) {
        return false;
    } else if ( leftOperator == '*' || leftOperator == '/' || leftOperator == '%' ) {
        return true;
    } else if ( rightOperator == '*' || rightOperator == '/' || rightOperator == '%' ) {
        return false;
    }

    return true;
}

#include <stdexcept>
#include <cctype>
answered 2013-01-19T12:10:11.550

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