Exit this Do-While w/out an int?

[Brian Pianczk]

Hi all, I am trying to figure out a way to exit this do-while. I am wondering how it can be done w/out using a number.
I originally did it with a while num > 0 and used an int num variable, but a neg number can be even or odd.
My new approach gives me an error (unknown source) when it reads a blank entry.

import java.util.Scanner;
 
public class Test {
    
    public static void main (String [] args) { 
		Test even = new Test();
		even.getNum();
	}
	
	public void getNum(){
	
		Scanner sc = new Scanner(System.in);
		String num;
		do {
		System.out.print("Enter an integer: ");
		num = sc.nextLine();
		calculate(Integer.parseInt(num));	
		} while (!num.isEmpty());
	}	
	
	private void calculate(int calcNum) {
		boolean isEven = (calcNum % 2 == 0);
		String even;
			if (isEven){
			even = "Yes";
			} else {
			even = "No";
			}
		System.out.println("Is " + calcNum + " even? " + even);
	}
	
}
		    

Error when I just use a space or enter.
c:\Program Files\Java>java Test
Enter an integer:
Exception in thread "main" java.lang.NumberFormatException: For input string: ""

at java.lang.NumberFormatException.forInputString(Unknown Source)
at java.lang.Integer.parseInt(Unknown Source)
at java.lang.Integer.parseInt(Unknown Source)
at Test.getNum(Test.java:17)
at Test.main(Test.java:7)

c:\Program Files\Java>

[Balu Sadhasivam]

Why dont you add the same check num.isEmpty() inside the loop and return; the method.

  

       num = sc.nextLine();   
 // check here and return.
        calculate(Integer.parseInt(num));      
        } while (!num.isEmpty());  // can be simply while(true)

   

Also check if the String is number otherwise you get numberFormat Exception

[Jeanne Boyarsky]

Brian,
How about using a boolean?

[amitabh mehra]

or use a try-catch block:

do {
		System.out.print("Enter an integer: ");
		num = sc.nextLine();
		try {
			calculate(Integer.parseInt(num));
		} catch(NumberFormatException e) {
			System.out.println("Not a number");
		}
		} while (!num.isEmpty());

[Brian Pianczk]

amitabh mehra wrote:or use a try-catch block:

do {
		System.out.print("Enter an integer: ");
		num = sc.nextLine();
		try {
			calculate(Integer.parseInt(num));
		} catch(NumberFormatException e) {
			System.out.println("Not a number");
		}
		} while (!num.isEmpty());

Thanks for the quick responses!

I like this idea, but I am guessing it would be bad programming to use an exception to exit a program?

Like if I were to say "Enter an integer, or just return to exit: " then use the exception to exit? and display "goodbye" in the exception.


Edit: never mind this idea, doesn't solve the issue if someone enters a non integer.

[Brian Pianczk]

Jeanne Boyarsky wrote:Brian,
How about using a boolean?

Still thinking about how I can apply this. Are you saying make the whole method a boolean?

[amitabh mehra]

Brian Pianczk wrote:

amitabh mehra wrote:or use a try-catch block:

do {
		System.out.print("Enter an integer: ");
		num = sc.nextLine();
		try {
			calculate(Integer.parseInt(num));
		} catch(NumberFormatException e) {
			System.out.println("Not a number");
		}
		} while (!num.isEmpty());

Thanks for the quick responses!

I like this idea, but I am guessing it would be bad programming to use an exception to exit a program?

Like if I were to say "Enter an integer, or just return to exit: " then use the exception to exit? and display "goodbye" in the exception.

the try-catch I suggested was never meant as an "exit" route. It was for that "dirty" exception stack trace which you would not want the user to see.
For exit purpose, perhaps you can prompt the user to enter something like 'quit' and change your code a bit in the way like:

    public void getNum(){

		Scanner sc = new Scanner(System.in);
		String num;
		System.out.print("Enter an integer: ");
		num = sc.nextLine();
		while (!num.equals("quit")) {
			try {
				calculate(Integer.parseInt(num));
			} catch(NumberFormatException e) {
				System.out.println("Not a number");
			}
			System.out.println("Enter another integer or type \'quit\' to exit");
		    num = sc.nextLine();
		};
	}

Now this will still give "Not a number" message to user in case non-number is entered and allow the user to continue. else 'quit' the application.

[Brian Pianczk]

amitabh mehra wrote:
the try-catch I suggested was never meant as an "exit" route. It was for that "dirty" exception stack trace which you would not want the user to see.
For exit purpose, perhaps you can prompt the user to enter something like 'quit' and change your code a bit in the way like:

    public void getNum(){

		Scanner sc = new Scanner(System.in);
		String num;
		System.out.print("Enter an integer: ");
		num = sc.nextLine();
		while (!num.equals("quit")) {
			try {
				calculate(Integer.parseInt(num));
			} catch(NumberFormatException e) {
				System.out.println("Not a number");
			}
			System.out.println("Enter another integer or type \'quit\' to exit");
		    num = sc.nextLine();
		};
	}

Now this will still give "Not a number" message to user in case non-number is entered and allow the user to continue. else 'quit' the application.

Thanks, I am trying to learn java, and all these suggestions help. It is amazing how many solutions there are to one problem, and each one I learn something new.