Java: Terminating the while loop after pressing "Q" key

Question

I have this following java program which is working fine without while loop, but I want to run the execution until user pressed Q key from the keyboard.

So What condition should be put in while loop which breaks the loop?

import java.awt.event.KeyEvent;
import java.util.Scanner;
import static javafx.scene.input.KeyCode.Q;

public class BinaryToDecimal {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);        
        while(kbhit() != Q){
            System.out.print("Input first binary number: ");
            try{          
                String n = in.nextLine();
                System.out.println(Integer.parseInt(n,2));
            }
            catch(Exception e){
                System.out.println("Not a binary number");
            }
        }
    }
}

Any help would be great. Thank You.

So what exactly is your question? Do you want to know, how you can read a keyboard input? — dunni, Sep 24 '17 at 07:31
what condition should be put in while loop which breaks the loop? — learning new things, Sep 24 '17 at 07:34

OneCricketeer · Answer 1 · 2017-09-24T07:53:31.543

3

I don't think you can use KeyEvent within a console application as there's no keyboard listener defined.

Try a do-while loop to watch for an input of the letter q. And you should compare strings using equals method

    Scanner in = new Scanner(System.in);        
    String n;
    System.out.print("Input first binary number: ");
    do {
        try{          
            n = in.nextLine();
            // break early 
            if (n.equalsIgnoreCase("q")) break;
            System.out.println(Integer.parseInt(n,2));
        }
        catch(Exception e){
            System.out.println("Not a binary number");
        }
        // Prompt again 
        System.out.print("Input binary number: ");
    } while(!n.equalsIgnoreCase("q"));

edited Sep 24 '17 at 07:53

answered Sep 24 '17 at 07:33

OneCricketeer

126,858
14
92
185

The output is looking same whether we put " if " statement in the try block. Let me know if it reduces the unnecessary overhead or not? – learning new things Sep 24 '17 at 08:21
Not sure what you mean. Obviously you'd like to prevent parsing letter q as an integer – OneCricketeer Sep 24 '17 at 08:25
In try block, "if (n.equalsIgnoreCase("q")) break;" Will it reduces the execution time or performance time? In case if user's input is "q" in the beginning. – learning new things Sep 24 '17 at 08:30
Of course it adds execution time, it's another statement. But it's negligible. Why do you care about performance of such simple code? – OneCricketeer Sep 24 '17 at 08:34

P3trur0 · Answer 2 · 2017-09-24T07:54:12.920

0

What about this?

public class BinaryToDecimal {
    public static void main(String[] args) {
        System.out.print("Input first binary number: ");
        Scanner in = new Scanner(System.in);
        String line = in.nextLine();
        while(!"q".equalsIgnoreCase(line.trim())){
            try{
                System.out.println(Integer.parseInt(line,2));
                System.out.print("Input next binary number: ");          
                line = in.nextLine();
            }
            catch(Exception e){
                System.out.println("Not a binary number");
            }
        }
    }
}

edited Sep 24 '17 at 07:54

answered Sep 24 '17 at 07:37

P3trur0

2,887
1
11
26

Well, you are prompting for input before "Input first binary number", which might not be obvious – OneCricketeer Sep 24 '17 at 07:51
thanks for pointing it out! fixed. – P3trur0 Sep 24 '17 at 07:54

Java: Terminating the while loop after pressing "Q" key

2 Answers2