Why is scanner reading no input and exiting function?

Question

I've created a method that asks a user for their favorite number. When I call this function more than once in my main, all method calls thereafter will jump to closing and declare Your favorite number is: 0 without asking for input. I'd like to know why a fresh scanner is reading and accepting no input.

    public static void favNumber(){
    Scanner scanner = new Scanner(System.in);
    System.out.println("Enter you favorite number: ");
    int favNumber = 0;
    boolean hasNextInt = scanner.hasNextInt();

    if(hasNextInt){
        favNumber = scanner.nextInt();
        scanner.nextLine(); //reads enter character
    }

    System.out.println("Your favorite number is: " + favNumber);
    scanner.close();
    }

You are closing `stdin` - with `scanner.close();` - don't do that. — Scary Wombat, Jun 01 '21 at 05:49
@ScaryWombat Please forgive my noobness, but why is 'scanner.close();' a bad way to close 'stdin'? And why would that be preventing a new scanner from asking for input? — croakPedlar, Jun 01 '21 at 05:55
You do not need to close `stdin` and if you do you can not re-open it. — Scary Wombat, Jun 01 '21 at 05:55
However closing input streams is good practice so you shouldn't _not_ close the scanner. — Roddy of the Frozen Peas, Jun 01 '21 at 05:56

Roddy of the Frozen Peas · Accepted Answer · 2021-06-01T07:05:56.567

Your Scanner instance wraps System.in (stdin). When you call scanner.close() at the end of your method, you're actually closing System.in in addition to your scanner. Once the input stream is closed, it can't be opened again.

The solution is, of course, to share a single instance of the Scanner. I note that you're using static methods, so I will similarly use a static instance in the following example. Note how I create the scanner, then I call favNumber(), and I only close the scanner after I finish calling favNumber(). This allows me to call favNumber() multiple times and share the same scanner instance.

private static Scanner scanner; // note that this is a class-level static variable.
public static void main(String[] args) throws Exception {
    try {
        scanner = new Scanner(System.in); // create the scanner
        favNumber(); // call the method
    } finally {
        scanner.close(); // only close the scanner when we're completely done
    }
}
public static void favNumber(){
    System.out.println("Enter you favorite number: ");
    int favNumber = 0;
    boolean hasNextInt = scanner.hasNextInt();

    if(hasNextInt){
        favNumber = scanner.nextInt();
        scanner.nextLine(); //reads enter character
    }

    System.out.println("Your favorite number is: " + favNumber);
}

An alternative solution would be to pass the scanner instance as a method parameter like this:

public static void main(String[] args) throws Exception {
    Scanner scanner = new Scanner(System.in); // create the scanner
    favNumber(scanner); // call the method, pass the scanner
    scanner.close(); // only close the scanner when we're completely done
}

public static void favNumber(Scanner scanner){
    System.out.println("Enter you favorite number: ");
    int favNumber = 0;
    boolean hasNextInt = scanner.hasNextInt();

    if(hasNextInt){
        favNumber = scanner.nextInt();
        scanner.nextLine(); //reads enter character
    }

    System.out.println("Your favorite number is: " + favNumber);
}

You can read more about closing scanners wrapping System.in in this Q&A here: Close a Scanner linked to System.in

thank you so much for your patience and explanation. This is the answer and I will mark it as soon as SO allows me to. — croakPedlar, Jun 01 '21 at 05:58

Why is scanner reading no input and exiting function?

1 Answers1