Thread: Post Fixed Math Expressions using Stack

Hello, I have been working on a silly little project for class for over a week now. The directions are below in bold

You may assume the user will enter a one line properly formatted post fixed math expression, with the tokens (operators and operands) separated by white space. You should use the API Stack class and the StringTokenizer class. Declare your Stack object to be a stack that can only store objects in the class Double.

Below is an outline of the program.


import java.util.*; // needed for the StringTokenizer class
import javax.swing.*;

// Declare a Java API Stack object that can store Double objects (Stack<Double> ds = new Stack<Double>()
// Accept the user’s input of the math String into the variable mathExpression
// Evaluate the math expression // see the code below
StringTokenizer tokens = new StringTokenizer(mathExpression, " ");
while(tokens.hasMoreTokens())
{
thisToken = tokens.nextToken();
//processing for token goes here
}
// Pop the stack and output the returned value

Here is my code, mathExpression being the user input. I used a switch statement to determine if the token is a +, -, or * sign. The default case is if the token is just a number. So basically my goal is to take user input which has spaces ex: 2 4 + (post fixed math which really means 2 + 4), and perform the math to return a result while using a stack.

import java.util.*; 
import javax.swing.*; 
 
public class Project2Option1 
{
     public static void main(String[] args) 
    {
    	String mathExpression, thisToken;
    	Double result, number, currentToken;
 
    	Stack<Double> ds = new Stack<Double>();	// Declare a Java API Stack object that can store Double objects
 
		mathExpression = JOptionPane.showInputDialog("Please enter one properly formatted post fixed math expression");
	    // Accept the user’s input of the math String into the variable mathExpression
 
		StringTokenizer tokens = new StringTokenizer(mathExpression);
 
		while(tokens.hasMoreTokens())
		{
			thisToken = tokens.nextToken();
			currentToken = Double.parseDouble(thisToken);
 
			switch(thisToken)
			{
				case "+": // Evaluate the math expression  "+"
					ds.pop(); ds.pop(); 
					result = (currentToken) + (currentToken - 1); 
					ds.push(result); break;
				case "-": //Evaluate the math expression "-"
					ds.pop(); ds.pop(); 
					result = (currentToken) - (currentToken - 1); 
					ds.push(result); break;
				case "*": //Evaluate the math expression "*"
					ds.pop(); ds.pop(); 
					result = (currentToken) * (currentToken - 1); 
					ds.push(result); break;
				default: //Converting token to a double; numbers entered by user
					number = Double.parseDouble(thisToken);
					ds.push(number); break;
			}		
		}
 
    	System.out.println("Here is the answer: " + ds.pop()); // Pop the stack and output the returned value
   }
}

My question is, how can I get three cases in my switch statement to work? (they should should examine the token, If its an math operator, pop the stack twice, do the math, and then push the result)

Code added above

Currently the only thing that works is default in my switch statement(I can output the top item in the stack) I get an error when using any of the other cases, Exception in thread "main" java.lang.NumberFormatException: For input string: "+"
at sun.misc.FloatingDecimal.readJavaFormatString(Floa tingDecimal.java:1241)
at java.lang.Double.parseDouble(Double.java:540)
at Project2Option1.main(Project2Option1.java:21)

Process completed.


I would like to be able to take input which will have spaces (ex: 2 3 -) and perform the post fixed math expr

Originally Posted by Norm

The statement on line 21 calls the parseDouble() method with the String: "+" and the method throws the exception.
Why is the code trying to convert the "+" to a double? It should wait until the after the switch statement has selected out the arithmetic operators and the default case gets control.

I do not want to convert "+" to a double, if the program encounters a + I would like it to add the two previous tokens in the string.

So essentially in case "+": When the we encounter a "+", add the two previous tokens and push the result upon the stack.

Originally Posted by Norm

Look at the code in the program. Where is line 21 with the parseDouble() where the error happens?
Why is that statement located there? It tries to convert the String to a double BEFORE the switch statement has checked the String to be an arithmetic operator like a "+".
Remove that statement to keep the error from happening.

Ok I removed that line. Now the error says the stack is empty, Exception in thread "main" java.util.EmptyStackException

 	while(tokens.hasMoreTokens())
		{
			thisToken = tokens.nextToken();
 
			switch(thisToken)
			{
				case "+": // Evaluate the math expression  "+"
					ds.pop(); ds.pop(); 
					currentToken = Double.parseDouble(thisToken); 
					result = (currentToken) + (currentToken - 1);
					ds.push(result); break;
			/*	case "-": //Evaluate the math expression "-"
					ds.pop(); ds.pop(); 
					result = (currentToken) - (currentToken - 1); 
					ds.push(result); break;
				case "*": //Evaluate the math expression "*"
					ds.pop(); ds.pop(); 
					result = (currentToken) * (currentToken - 1); 
					ds.push(result); break;
				default: //Converting token to a double; numbers entered by user
					number = Double.parseDouble(thisToken);
					ds.push(number); break; */
			}