Asks user to enter values until an integer is entered

Question

I am a noob in programming. I wanted to write code for a prog which asks user to enter value until an integer is entered.

public class JavaApplication34 {

    /**
    * @param args the command line arguments
    */
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int flag = 0;
        while(flag == 0) {
            int x = 0;
            System.out.println("Enter an integer");

            try {
                x = sc.nextInt();
                flag = 1;
            } catch(Exception e) {
                System.out.println("error");        
            }

            System.out.println("Value "+ x);
        }
    }
}

I think the code is correct and it should ask me to enter the value again if i have entered anything other than an integer. But when i run it , and say i enter xyz it iterates infinite time without asking me to enter the value.

test run :
Enter an integer
xyz
error
Value 0
Enter an integer
error
Value 0
Enter an integer
error
Value 0
Enter an integer
error
Value 0
Enter an integer
error
Value 0
Enter an integer
error
Value 0
Enter an integer
error
Value 0
Enter an integer
error
Value 0
Enter an integer
error
Value 0

In a catch block such as your catch(Exception e){, the e holds valuable information about what happened. Use e.printStackTrace() to find out more. — f1sh
– f1sh, Commented May 9, 2017 at 13:59

Ousmane D. · Accepted Answer · 2017-05-09 14:00:27Z

4

When a scanner throws an InputMismatchException, the scanner will not pass the token that caused the exception.

Hence sc.nextInt() reads the same token again and throws the same exception again.

...
...
...
catch(Exception e){
     System.out.println("error");
     sc.next(); // <---- insert this to consume the invalid token
}

answered May 9, 2017 at 14:00

Ousmane D.

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Comments

Nishesh Pratap Singh · Accepted Answer · 2017-05-09 14:05:24Z

You can change your logic as shown below :

    int flag = 0;
    int x = 0;
    String str="";
    while (flag == 0) {         
        System.out.println("Enter an integer");
        try {
            str = sc.next();
            x = Integer.parseInt(str);
            flag = 1;
        } catch (Exception e) {
            System.out.println("Value " + str);
        }           
    }

Here we have first read the input from Scanner and then we are trying to parse it as int, if the input is not an integer value then it will throw exception. In case of exception we are printing what user has enter. When user enters an integer then it will parsed successfully and value of flag will update to 1 and it will cause loop to exit.

T.J. Crowder · Accepted Answer · 2017-05-09 14:09:40Z

1

In the error case, you need to clear out the string you've entered (for instance, via nextLine). Since it couldn't be returned by nextInt, it's still pending in the scanner. You also want to move your line outputting the value into the try, since you don't want to do it when you have an error.

Something along these lines:

public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);

    int flag = 0;
    while(flag == 0)
    {
        int x = 0;
        System.out.println("Enter an integer");
        try
        {
            x = sc.nextInt();
            flag = 1;
            System.out.println("Value "+ x);
        }
        catch (Exception e){
            System.out.println("error");
            if (sc.hasNextLine()) { // Probably unnecessary
                sc.nextLine();
            }
        }
    }
}

Side note: Java has boolean, there's no need to use int for flags. So:

boolean flag = false;

and

while (!flag) {

and

flag = true; // When you get a value

edited May 9, 2017 at 14:09

answered May 9, 2017 at 14:00

T.J. Crowder

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4 Comments

Mridul Mittal Over a year ago

can you help ??

T.J. Crowder Over a year ago

@MridulMittal: That is help.

Sanket Makani Over a year ago

If you will not try to solve them on your own even after knowing the error, You will remain a NOOB.

T.J. Crowder Over a year ago

@MridulMittal: The best way, by far, to become a non-n00b is to work things through. But I've added an example of what you might do above.

Community · Accepted Answer · 2017-05-23 12:18:11Z

0

The answers to this question might help you

It makes use of Scanners .hasNextInt() function!

edited May 23, 2017 at 12:18

CommunityBot

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answered May 9, 2017 at 14:37

Rachel Solomon

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