Htydjtk

The following code output endless errors, when a user entered not a number.

Scanner scn = new Scanner(System.in);

System.out.println("Enter the number: ");   

int n = 0;  

while(true){

    try {

        n = scn.nextInt();

    } catch (Exception e) {

        e.printStackTrace();

        continue;

    }

    break;

}

I expect that the code wait for new input when user entered not a number.

From the javadoc,

When a Scanner throws an InputMismatchException, the scanner will not
pass the token that caused the exception, so that it may be retrieved
or skipped via some other method.

In your case, scn.nextInt() must have thrown InputMismatchException. But the token is not passed (or remains). So on the next iteration of the loop, scn.nextInt() reads the same token again and throws the exception again resulting in an infinite loop.

If you inspect the value of CharBuffer (HeapCharBuffer?) inside the Scanner object, it still contains the input you have typed the first time and causing this issue. Also, you cannot explicitly clear Scanner's buffer.

Also, beware of resource-leaks!

I would stay away from using nextInt and instead read input as a string and try to convert it afterwards

Scanner scn = new Scanner(System.in);

System.out.println("Enter the number: ");   

int n = 0;  

while(true) {

    String input = scn.nextLine();



    try {

        n = Integer.valueOf(input);

        break;

    } catch (Exception e) {

        System.out.println( input + " is not a valid number. Try again.");

    }

}

System.out.println("You entered " + n);

My version:

public class IsNumber {

    public static void main(String o) {

        Scanner scn = new Scanner(System.in);

        System.out.println("Enter the number: ");   

        String str = null;

        do{

            str = scn.next();

            if(isInteger(str))

                break;

        }while(true);



        System.out.println("Number Entered: " + Integer.parseInt(str));

        scn.close();

    }



    private static boolean isInteger(String s) {

        try { 

            Integer.parseInt(s); 

        } catch(Exception e) { 

            return false; 

            }

        return true;

    }

}

Quick fix

try {

    n = scn.nextInt();

} catch(Exception e) {

    e.printStackTrace();

    scn.nextLine(); // <-- add this to move to the next line

    continue;

}

Correct way to implement it. Not rely on catch exceptions.

My approach looks not very small like yours, but I avoid catching exceptions in it.

private static final IntSupplier getIntegerNumber = new IntSupplier() {

    private final Predicate<String> isIntegerString = str -> {

        str = str == null ? null : str.trim();

        str = str == null || str.isEmpty() ? null : str;



        if (str == null)

            return false;



        for (int i = 0; i < str.length(); i++) {

            if (str.charAt(i) == '-') {

                if (i != 0)

                    return false;

            } else if (str.charAt(i) < '0' || str.charAt(i) > '9')

                return false;

        }



        return true;

    };



    @Override

    public int getAsInt() {

        try (Scanner scan = new Scanner(System.in)) {

            System.out.print("Enter the number: ");



            while (true) {

                String str = scan.nextLine();



                if (isIntegerString.test(str))

                    return Integer.parseInt(str);



                System.out.print("This is not an integer number. Try again: ");

            }

        }

    }

};

Demo:

System.out.println(getIntegerNumber.getAsInt());

P.S.

According to What's the best way to check if a String represents an integer in Java? this is the fastest way to check if given string an integer or not.