How to correctly throw an exception

Question

I'm new to java and I'm trying to account for possible errors that might come up in my program. I'm creating a calculator and converting the inputs from infix to postfix to calculate the result. I want to try accounting for mismatched parenthesis in the input, but I'm having trouble. For example, when converting from infix to postfix, when a ) is reached, it will pop numbers from the stack and put them onto the new postfix list. In one case where there may not be a matching left parenthesis present (the while loop reaches the end of the stack without coming across a (, it should throw an exception. I've implemented the following code, but it doesn't seem to be working:

    else if(tok.equals(")")){
      while(stack.peek().equals("(") == false){
        try{
          Operator o = stack.pop();
          nlist.add(o);                
        } catch(EmptyStackException e){
          System.out.println(e.getMessage());
        }
        stack.pop();
      }    

Then in the other file which constructs the GUI and processes the inputs, I entered:

try{
  java.util.List<Token> postfix = ExpressionManager.infixToPostfix(infix);
  // build the expression
  Expression exp = ExpressionManager.buildExpression(postfix);
  // display the results
}catch(ArithmeticException e){
     entryField.setText(e.getMessage())
}

Any suggestions?

Answer 1

stack.peek() throws EmptyStackException before you enter inside try-catch block (I assume you expect that pop will throw the exception).

Second block doesn't show how the ArithmeticException is thrown, so, not sure what you expecting here.

Answer 2

If you want to re-throw an exception you can do

catch(EmptyStackException e){
    throw new ArithmeticException("empty stack, bad mathematical expression.", e);
}