Java - Bug in while loop using newLine Scanner

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Java - Bug in while loop using newLine Scanner

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#1 by (2 votes)

3

I am creating a text adventure, and I have a while loop (game loop) in which an action is requested from the user, of which are: Explore, Move, Take, Inventory and Position. And user information is received using the Scanner class.

To make it easier to ask for input, I have this class:

public class Input {

    private static Input input = null;
    public Scanner scannerInput = new Scanner(System.in);

    protected Input() {
        // o:
    }

    public static Input getInput() {
        if (input == null)
            input = new Input();
        return input;
    }

    public String getStringInput() {
        return scannerInput.nextLine();
    }

    public char getCharacterInput() {
        return scannerInput.next(".").charAt(0);
    }

}

And this is the loop:

String action;
while (true) {
    System.out.println(currentRoom.getActionsString());
    action = input.getStringInput();

    if (action.split(" ").length > 1)
        player.processAction(action.split(" "));
    else
        player.processAction(action);
}

Method to process actions within the Player class:

public void processAction(String... actionDecision) {
    // Better-Better for sure!
    // Now make take and move much more
    // user friendly, so that you can
    // directly introduce the place you want to
    // move to or the name of the item you want to
    // take, and if not, it is alright too!

    switch(actionDecision[0].toLowerCase()) {
        case "move":
            this.move();
            break;
        case "take":
            this.take();
            break;
        case "explore":
            this.explore();
            break;
        case "inventory":
            this.getFormattedInventory();
            break;
        case "position":
            this.getFormattedPosition();
            break;
        default:
            System.out.println("Sorry ):");
    }
}

The problem occurs when the player chooses "move" as his action. The movement system is by cardinal points, and according to the adjacent rooms, I can or can not move to such points. But when you choose the move action, the action is normally processed, and the loop starts again, but automatically calls processAction and throws the default value, "Sorry": "and then it starts again normally.

I already noticed that the problem comes from this method in the Input class:

public String getStringInput() {
    return scannerInput.nextLine();
}

If I change it to this:

public String getStringInput() {
    return scannerInput.next();
}

I do not have that bug, but using next instead nextLine is not convenient for me, because next does not seem to read more than one word, and I'll be using varargs a lot, so I need the full string.

Why is this happening?

java input

asked by ManuAlvarado22 20.03.2018 в 12:09

source

1 answer

Convert date in normal format mouse event in C #

score 2 · Accepted Answer

Consider the following entry:

move
N
move
S

What really exists in that entry, includes line breaks:

move\n
N\n
move\n
S\n

When reading the action with scannerInput.nextLine() you get move without the line break, but the scanner still advances the complete line and is located just after, remaining to be processed:

N\n
move\n
S\n

After reading the parameter for the action move with scannerInput.next(".").charAt(0) , you consume only one character of the entry and you get N , but what remains to be consumed for the scanner is:

\n
move\n
S\n

If then, you try to read another action, with scannerInput.nextLine() , you will get an empty string "" and the scanner will continue advancing, remaining:

move\n
S\n

That's why you have to be careful when mixing next() and nextLine() . next() does not consume the line breaks, nextLine() if.

For your case, it is enough to consume the rest of the line in your method:

public char getCharacterInput() {
    char c = scannerInput.next(".").charAt(0);
    scannerInput.nextLine(); //resto de la linea, avanzar a la siguiente
    return c;
}