If Else error in java code

Question

I'm getting an error at the line

else break;

I think it might have to do with the location of the "{}"

public class Palindrome {

    private static String number;

    public static void main(String[] args) {
        display();
    }

    private static int getUserInput() {
        int inputNumber = 0;
        String answer = JOptionPane.showInputDialog(null, "Please enter a five digit number","Enter a number");
        inputNumber = Integer.parseInt(answer);

        if(number.length() !=5 ){
            JOptionPane.showMessageDialog(null,"Please enter a 5 digit number!","Try again",
            JOptionPane.PLAIN_MESSAGE);

        else break;
        }

        return inputNumber;
    }

    private static boolean check(){
        int inputNumber = getUserInput();
        int number = inputNumber;
        int[] myArray = new int[5];
        for (int i = 0; i < myArray.length; i++) {
            myArray[i] = (int) (number /(Math.pow(10,i)) % 10);
        }
        if(myArray[0] == myArray[4] && myArray[1] == myArray[3])
            return true;
        else
            return false;
    }

    public static boolean display(){
        if (check() == true) {
        JOptionPane.showMessageDialog(null, "This number is a Palindrome",
        "Excellent!",JOptionPane.INFORMATION_MESSAGE);
        } else
            JOptionPane.showMessageDialog(null,
                "Number is not a Palindrome!",
                "Sorry",
                JOptionPane.ERROR_MESSAGE);
            return false;
        }
    }
}

Thanks

You'll find it a lot easier to fix these errors if you indent your code - you should find your IDE is willing to do this for you automatically. — Jon Skeet
– Jon Skeet, Commented Mar 9, 2011 at 18:02

Mahesh · Accepted Answer · 2011-03-09 18:01:18Z

4

else cannot be part of if. So, change this -

if(number.length() !=5 )
{
    JOptionPane.showMessageDialog(null,"Please enter a 5 digit number!","Try again",
    JOptionPane.PLAIN_MESSAGE);
}
else break;

Also, I don't understand why you use break here ?

answered Mar 9, 2011 at 18:01

Mahesh

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2 Comments

Erik Over a year ago

There's no loop to break out of.

Mahesh Over a year ago

@Erik - That is what my doubt is. Why @Mike is using break here when there is no loop to come out of !!

Jigar Joshi · Accepted Answer · 2011-03-09 18:02:23Z

1

if(/*some condition*/){
}else{
     break;//invalid use of break it won't compile see below
}

you are writting else without matching if

moreover you can't break without loop or switch case

answered Mar 9, 2011 at 18:02

Jigar Joshi

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2 Comments

Paŭlo Ebermann Over a year ago

Actually, you can break out of a named statement, too, but there is no such statement here, too. (And the break would need the name of the statement then.)

Jigar Joshi Over a year ago

@Paŭlo You mean break with label ?

julx · Accepted Answer · 2011-03-09 18:03:20Z

1

Also a note: code like "if (condition) return true; else return false;" can be substituted for "return condition;". Much cleaner.

answered Mar 9, 2011 at 18:03

julx

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Comments

Kris Babic · Accepted Answer · 2011-03-09 18:06:33Z

1

There are two problems:

else statement is not associated with an if statement. You need to move it outside of the {} for the else
You are using a break outside of a while or for loop (or named statement).

If you are trying to exit the application, then you can use System.exit method. However a better approach is to return a value that states that the user did not select a value, e.g., -1 or null (change return type of method to Integer).

answered Mar 9, 2011 at 18:06

Kris Babic

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Peter Lawrey · Accepted Answer · 2011-03-09 18:22:35Z

1

You cannot break; from a method, instead you must return from a method. However in your case it would appear you want a retry loop.

private static int getUserInput() {
  while(true) {
    String answer = JOptionPane.showInputDialog(null, "Please enter a five digit number","Enter a number");
    if(answer.length() == 5 ) 
        return Integer.parseInt(answer);
    JOptionPane.showMessageDialog(null,"Please enter a 5 digit number!","Try again",
        JOptionPane.PLAIN_MESSAGE);
 }
}

edited Mar 9, 2011 at 18:22

answered Mar 9, 2011 at 18:05

Peter Lawrey

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2 Comments

Mike Over a year ago

I'm getting an error at the bottom line "return inputNumber;" It says unreachable code.

Mike Over a year ago

When I plug in a number that is not equal to five it brings up the original box to enter the number, it doesn’t give me the error box. Now when I go and plug a number again that is not equal to 5 then it brings up the error box. So it’s only after the second wrong attempt is when it works correctly.

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