Infix to postfix printing out incorrectly sometimes..any ideas?

Question

alright, my program this time is to make an infix to postfix converter.
I have it coded, and it works fine except when I use parenthesis. I'm supposed to test it with these expressions:
a + b
a * b + c
a * ( b + c )
a + b * c - d
( a + b ) * ( c - d )
a - ( b - ( v - ( d - ( e - f ))))
     // initialize two stacks: operator and operand
     private Stack operatorStack = new Stack();
     private Stack operandStack = new Stack();
     // method converts infix expression to postfix notation
     public String toPostfix(String infix)
          StringTokenizer s = new StringTokenizer(infix);
          // divides the input into tokens for input
          String symbol, postfix = "";
          while (s.hasMoreTokens())
          // while there is input to be read
               symbol = s.nextToken();
               // if it's a number, add it to the string
               if (Character.isDigit(symbol.charAt(0)))
                    postfix = postfix + " " + (Integer.parseInt(symbol));
               else if (symbol.equals("("))
               // push "("
                    Character operator = new Character('(');
                    operatorStack.push(operator);
               else if (symbol.equals(")"))
               // push everything back to "("
                    /** ERROR OCCURS HERE !!!! **/
                             while (((Character)operatorStack.peek()).charValue() != '(')
                         postfix = postfix + " " + operatorStack.pop();
                    operatorStack.pop();
               else
               // print operatorStack occurring before it that have greater precedence
                    while (!operatorStack.isEmpty() && !(operatorStack.peek()).equals("(") && prec(symbol.charAt(0)) <= prec(((Character)operatorStack.peek()).charValue()))
                         postfix = postfix + " " + operatorStack.pop();
                    Character operator = new Character(symbol.charAt(0));
                    operatorStack.push(operator);
          while (!operatorStack.isEmpty())
               postfix = postfix + " " + operatorStack.pop();
          return postfix;
// method compares operators to establish precedence
     public int prec(char x)
          if (x == '+' || x == '-')
               return 1;
          if (x == '*' || x == '/' || x == '%')
               return 2;
          return 3;
/** MY STACK **/
import java.util.LinkedList;
public class StackL {
  private LinkedList list = new LinkedList();
  public void push(Object v) {
    list.addFirst(v);
  public Object peek() {
    return list.getFirst();
  public Object pop() {
    return list.removeFirst();
  public boolean isEmpty()
      return (list.size() == 0);
}weird, it erased my question/errors I put in...
When I use any of the expressions with parenthesis (except the 2nd to last one) I get an emptystackexception pointing to the area noted in the code.
When I test the 2nd to last one (the one I would expect it to give me the most trouble) it prints out an answer, but it is incorrect. It prints out this : "Expression in postfix: a ( b - ( c - ( d - ( e - f ) -" which is incorrect, as it hsould have no parenthesis
Edited by: Taco_John on Apr 6, 2008 12:46 PM
Edited by: Taco_John on Apr 6, 2008 12:47 PM
Edited by: Taco_John on Apr 6, 2008 12:49 PM

Answer 1

the algorithm we were told to use is here:
While not stack error and not the end of infix expression
a.) Extract the next input token from the infix expression (can be constant value, variable, arithmetic operator, left or right parenthesis)
b.) If token is
- left parenthesis : push it onto the stack
- right parenthesis : pop and display stack elements until the left parenthesis is popped (don't sisplay right parenthesis). It's an error if stack becomes empty with no matching right parenthesis found.
- Operator : if the stack is empty or token is higher priority than the top element, push it onto the stack. Otherwise, pop and display the top stack element and repeat comparison of token with new top element
Note : left parenthesis in the stack is assumed to have a lower priority than any operator.
- operand : display it
When the end of the infix expression is reached, pop and display remaining stack elements until it is empty.
it works fine on anything without a parenthesis, and it prints an answer (however it is incorrect) for the a - (b-(c-(d-(e-f)))) expression.
still looking for an idea if I can get one
Ok, I just noticed this. If i do a * ( b + c ) I get the error. But if I type " a * ( ( b + c ))" with spaces between the left parenthesis and adding an extra parenthesis as well, and NOT spacing the right parenthesis, I get a result that works just like that 2nd to last was doing. So it's something about the spaces...The answer is incorrect when I use the parenthesis as well. So it's not ignoring white space correctly for some reason and it's printing incorrect answers when I use parenthesis.