Exiting from while loop not working in java

Question

I am new to java programming.I want to calculate the sum and want to exit the program if user enters "N" and again loop if user enters "Y".But,it is not getting me out of loop even I enter "N".

public class Program {



    public static void main(String[] args) {
        boolean a=true;
        while (a) {
        System.out.println("enter a number");
        Scanner c=new Scanner(System.in);
        int d=c.nextInt();

        System.out.println("enter a number2");
        Scanner ce=new Scanner(System.in);
        int df=ce.nextInt();

        int kk=d+df;
        System.out.println("total sum is"+kk);

        System.out.println("do you want to continue(y/n)?");
        Scanner zz=new Scanner(System.in);
        boolean kkw=zz.hasNext();
        if(kkw) {
           a=true;
        }
        else {
            a=false;
            System.exit(0);
        }
        }
}

I didnt know where I made the mistake? Is there any other way?

You are asking the user to enter y/n, but then ignoring what he says — jr593
– jr593, Commented Mar 16, 2020 at 15:55
ericlippert.com/2014/03/05/how-to-debug-small-programs -- and step through your running code with a debugger. — slim
– slim, Commented Mar 16, 2020 at 16:10

Eldar B. · Accepted Answer · 2020-03-16 16:39:48Z

4

First of all, your a variable is true if scanner.hasNext() is true, leading to a being true with every input, including "N" which means, your while loop will keep on going until there are no more inputs.

Second of all, you could optimize your code the next way:

I suggest getting rid of a and kkw to make your code cleaner and shorter.
Use only one Scanner and define it outside of the loop. You don't need more than one Scanner for the same input. Also, initializing a Scanner with every loop is resource-consuming.
Use meaningful variable names. Programming should not only be efficient, but also easy to read. In this tiny code it's a minor issue but imagine having an entire program and, instead of adding features and bug-fixing, you had to search for the meaning of every variable.

Here's an optimized and working version of your code:

Scanner scanner = new Scanner(System.in);
while (true) {
    System.out.println("Enter a number");
    int input1 = scanner.nextInt();
    scanner.nextLine(); // nextInt() doesn't move to the next line

    System.out.println("Enter a second number:");
    int input2 = scanner.nextInt();
    scanner.nextLine();

    System.out.println("Total sum is " + (input1 + input2)); /* Important to 
    surround the sum with brackets in order to tell the compiler that 
    input1 + input2 is a calculation and not an appending of
    "Total sum is "*/

    System.out.println("Do you want to continue? (Y/N)");
    if (scanner.hasNext() && scanner.nextLine().equalsIgnoreCase("n"))
        break;
}
scanner.close();

edited Mar 16, 2020 at 16:39

answered Mar 16, 2020 at 16:04

Eldar B.

1,38912 silver badges24 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Eldar B. Over a year ago

Thanks for the reminder, I haven't used Scanner for ages. Also, I didn't say it's an error, it's solely a suggestion.

Eldar B. Over a year ago

@user85421 could you please edit my answer regarding the hasNext() issue?

Ryan · Accepted Answer · 2020-03-16 16:08:22Z

0

try (Scanner in = new Scanner(System.in)) {
    boolean done = false;
    while (!done) {
        System.out.println("enter first number");
        int d = in.nextInt();
        System.out.println("enter second number");
        int df = in.nextInt();
        int kk = d + df;
        System.out.println(String.format("total sum is %d", kk));

        System.out.println("do you want to continue(y/n)?");
        String cont = in.next();
        done = cont.equalsIgnoreCase("n");

    }
} catch(Exception e) {
    e.printStackTrace();
}

answered Mar 16, 2020 at 16:08

Ryan

1,7588 silver badges11 bronze badges

3 Comments

Ryan Over a year ago

You should absolutely close the Scanner. Using System.in is the failing, but it doesn't matter since this is throw away code. Ideally, the program would open a dedicated stream.

Ryan Over a year ago

Never leave resources open. It's programming 101. It invites memory leaks. Even if it's something silly like a ByteArrayOutputStream, you should always be in the habit of closing it.

David Conrad Over a year ago

You should absolutely not close the Scanner. You should always close every resource you open, but you did not open System.in, so you should not close it.

Collectives™ on Stack Overflow

Exiting from while loop not working in java

2 Answers 2

2 Comments

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related