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Question: I'm creating a poker client, how can I make the exception handling more clean?


I'm checking if the values are valid. The if parts looks still messy for me, checking a lot of || operator, and there is multiple InvalidArgumentException, but I always check for that.

How can this be more clean ?

This is part of my script :

public Card(String cardCode) throws IllegalArgumentException {

    this.cardCode = cardCode;

    String cardColor = this.cardCode.substring(0, 1).toUpperCase();
    String cardValue = cardCode.substring(1).toUpperCase();
    Integer intCardValue = Integer.parseInt(cardValue);

    if (!colors.contains(cardColor))
        throw new IllegalArgumentException("card color isn't valid: " + cardColor);

    if (alphabeticCardValue.get(cardValue) == null || intCardValue > 10 || intCardValue < 2 ) {
        throw new IllegalArgumentException("card number isn't valid: " + intCardValue);

Thank you

Question author Dmbdnr | Source



What you actually trying to do is validating the input. Using IllegalArgumentException is not appropriate, imo, because the purpose is of this exception is defined in JavaDoc as follows:

  • Thrown to indicate that a method has been passed an illegal or inappropriate argument.

What I would do is as follows:

  1. Define an enum for possible colors:

    public enum Color { BLACK, RED, ... }
  2. Define an enum for possible card values:

    public enum CardValues {
        THREE(3); // ...
        private int value;
        private CardValues(final int v) {
            value = v;
        public getValue() { return value;}
  3. Change the constructor as follows:

    public Card(Color color, CardValues cardValues) {
        if (color == null || cardValues == null) {
            throw new IllegalArgumentException("....");
        // doSomething else

    Note: IllegalArgumentException is an unchecked exception. So you don't need to specify it in the throws clause.

Answer author Ujulu

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