Modelling Game Difficulty

Question

I wonder if my aproach into modelling difficulty level for my game is good.

public abstract class AbstractDifficulty {

    public AbstractDifficulty() {

    }

    public abstract int enemyWaves();
    public abstract int enemiesInWave();
    public abstract long enemyWaveIntervalMilis();
}

And then several subclasses which implement those methods, for example.

public class EasyDifficulty extends AbstractDifficulty {

    @Override
    public int enemyWaves() {
        return 1;
    }

    @Override
    public int enemiesInWave() {
        return 10;
    }

    @Override
    public long enemyWaveIntervalMilis() {
        return 500;
    }
}

I wonder if there is a simpler way to do this, but the same easy and clean to use as this.

Edit: Could someone be so kind to explain to me why this question got minus votes. Is something wrong with this code, or my explanation ? Thanks.

This looks perfectly fine by me. The only other way to do this, I think (which is less flexible, btw) is to have one class that governs your actions and setting a difficulty modifier. This modifier can then be used to calculate settings based of a base number. E.g.: enemiesInWaveBase = 10 * difficultyMod, and then setting difficultyMod to 0.7 for easy, 1.0 for normal and 1.5 for hard or something. Less code, less flexible. — RDM
– RDM, Commented Mar 18, 2015 at 12:37
It depends on what you consider difficult. This looks like an arcade style game, so other factors could include player health, the goal of the game/level (if one exists) etc. I'd personally use interfaces instead since it is easier to add more interfaces if a new feature comes up. Modifying parent classes can be a bad idea. — ryanyuyu
– ryanyuyu, Commented Mar 18, 2015 at 12:42
@ryanyuyu Could you please explain or give me some simple example, which would illustrate this situation, when you want to add new feature. Thanks. — 10101101
– 10101101, Commented Mar 18, 2015 at 14:37
@ryanyuyu OK nevermind, I think I understand now what you mean. Thanks anyway, I will consider this. — 10101101
– 10101101, Commented Mar 18, 2015 at 14:51

LastFreeNickname · Accepted Answer · 2015-03-18 12:56:10Z

5

Your design will do the job, but it seems what you want to have is a container class. So instead having a method for each value, have a (final) field with the value and a getter. They can be set e.g. in the constructor.

If your difficulty modes are known and few, consider having an enum, which are final and easier to use.

enum Difficulty {
    EASY(1, 10, 5000),
    MEDIUM(2, 15, 4000),
    HARD(4, 20, 3500);

    private final int enemyWaves;
    private final int enemiesInWave;
    private final long enemyWaveIntervalMilis;

    Difficulty(int enemyWaves, int enemiesInWave, long enemyWaveIntervalMilis ){
        this.enemyWaves = enemyWaves;
        this.enemiesInWave = enemiesInWave;
        this.enemyWaveIntervalMilis = enemyWaveIntervalMilis;
    }

    public int getEnemyWaves() {

        return enemyWaves;
    }
    public int getEnemiesInWave() {

        return enemiesInWave;
    }
    public long getEnemyWaveIntervalMilis() {

        return enemyWaveIntervalMilis;
    }

}

edited Mar 18, 2015 at 12:56

answered Mar 18, 2015 at 12:50

LastFreeNickname

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4 Comments

Giulio Franco Over a year ago

The fields have to be public final, or you'll never read them.

10101101 Over a year ago

Man that is exacty what I wanted I think ! I never used enums, but dont know why I thought about them first.

10101101 Over a year ago

Are enums polymorphic ? I dont want to use even single IF statement in this particular code, like i can have with abstract classes. Like will it work like some method will get argument Difficulty, but it could be any of the three ?

LastFreeNickname Over a year ago

Jep, you can set the current difficulty at some place (maybe some WorldConfiguration or UserSettings object) and then always access it regardless of its actual value. => worldConfiguration.getDifficulty().getEnemyWaves();

kutschkem · Accepted Answer · 2015-03-18 12:50:10Z

2

One alternative is to not have different difficulty classes, but just one difficulty class with different instances being the different difficulties (so, a class with attributes enemyWaves, enemiesInWave and enemyWaveIntervalMilis, and different instances with different settings).

answered Mar 18, 2015 at 12:50

kutschkem

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Comments

Peter · Accepted Answer · 2015-03-18 12:56:32Z

Your solution looks fine. If you want to increase readability you can just remove the constructor from the abstract class, since that is created by Java as default anyway.

I also vote for an interface instead of an abstract class unless there is a specific reason to use abstract class.

I also feel that for the specific problem you do not need different classes, but rather different instances of the same class. In that case I would use the final fields set by constructor and then the methods return the values.

public class Difficulty  {

  final private int enemyWaves;
  public Difficulty(final enemyWaves){ this.enemyWaves = enemyWaves; }

  @Override
  public int enemyWaves() {
      return this.enemyWaves;
  }
}

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