java - 如何重置这个随机字符串生成器？ [Java]

Question

我一直在寻找答案( Java: how to "restart" a static class? )，但它似乎没有回答我的问题。

我有一个类，它从单词列表中生成一个有点随机的字符串。然后，这些单词通过 read() 输出到游戏中的标牌上。我可以通过两种方式执行操作:如果buildingText()、verbText() 和nounText() 不是静态的，则每次访问该标志时我都会得到一个新字符串。如果这三个是静态的，即使游戏重新启动，我也会得到一个不会改变的字符串。

如何更改此代码以便 reset() 创建新的随机字符串？

我一直在尝试使用 boolean 值并用 if 包围 switch 语句，但这似乎不可能。

代码如下:

public class Sign {

private static final String TXT_PLACE_NAME      = buildingText();
private static final String TXT_VERBING         = verbText();
private static final String TXT_NOUN            = nounText();


protected static final String SIGNS =
        "Welcome to the  "+ "\n \n "+ TXT_PLACE_NAME+ " of the "+ TXT_VERBING + TXT_NOUN + "";



public void reset(){
    //still working on this. As it stands, will only reset if you exit the program
    //could be worse things
}

public void read(int pos) {
    GameScene.show(new WndMessage(SIGNS));


}

public static String buildingText() {

    String buildingType;

        switch (Random.Int(1, 4)) {
            case 1:
                buildingType = "Inn";
                break;
            case 2:
                buildingType = "Inn";
                break;
            case 3:
                buildingType = "Tavern";
                break;
            case 4:
                buildingType = "Pub";
                break;
            default:
                buildingType = "Inn";
                break;
        }

        return buildingType;
    }

public static String verbText()
{
    String verbName;
    switch (Random.Int(1,20))
    {
        case 1: verbName = " Dancing";
            break;
        case 2: verbName = " Prancing";
            break;
        case 3: verbName = " Eating";
            break;
        case 4: verbName = " Jigging";
            break;
        case 5: verbName = " Digging";
            break;
        case 6: verbName = " Flogging";
            break;
        case 7: verbName = " Floating";
            break;
        case 8: verbName = " Flying";
            break;
        case 9: verbName = " Laughing";
            break;
        case 10: verbName = " Smiling";
            break;
        case 11: verbName = " Drowning";
            break;
        case 12: verbName = " Golden";
            break;
        case 13: verbName = " Silver";
            break;
        case 14: verbName = " Copper";
            break;
        case 15: verbName = " Farming";
            break;
        case 16: verbName = " Running";
            break;
        case 17: verbName = " Sewing";
            break;
        case 18: verbName = " Black";
            break;
        case 19: verbName = " White";
            break;
        case 20: verbName = " Fighting";
            break;
        default: verbName = " Gesticulating";
            break;
    }
    return verbName;
}

public static String nounText()
{
    String nounName;
    switch (Random.Int(1,20))
    {
        case 1: nounName = " Pony";
            break;
        case 2: nounName = " Horse";
            break;
        case 3: nounName = " Griffin";
            break;
        case 4: nounName = " Dragon";
            break;
        case 5: nounName = " Wench";
            break;
        case 6: nounName = " Bastard";
            break;
        case 7: nounName = " Ogre";
            break;
        case 8: nounName = " Troll";
            break;
        case 9: nounName = " Ox";
            break;
        case 10: nounName = " Cow";
            break;
        case 11: nounName = " Cock";
            break;
        case 12: nounName = " Hen";
            break;
        case 13: nounName = " Ram";
            break;
        case 14: nounName = " Ewe";
            break;
        case 15: nounName = " Dog";
            break;
        case 16: nounName = " Merchant";
            break;
        case 17: nounName = " Fisherman";
            break;
        case 18: nounName = " Arborist";
            break;
        case 19: nounName = " Archer";
            break;
        case 20: nounName = " Gallbladder";
            break;
        default: nounName = " Pancreas";
            break;
    }
    return nounName;
}
}

编辑:

这是一种“最佳”解决方案。它结合了以下两个解决方案的要素并产生了所需的结果。请随意批评格式和组织!

public class Sign {


protected static String SIGNS;
protected static final String SIGNS_FORMAT = "Welcome to the %s of the %s %s";

private static final String[] BUILDING_TYPES = new String[]{
        "Inn",
        "Inn",
        "Tavern",
        "Pub",
        "Inn"
};

private static final String[] VERB_NAMES = new String[] {
        "Dancing",
        "Prancing",
        "Eating",
        "Jigging",
        "Digging",
        "Flogging",
        "Floating",
        "Flying",
        "Laughing",
        "Smiling",
        "Drowning",
        "Golden",
        "Silver",
        "Copper",
        "Farming",
        "Running",
        "Sewing",
        "Black",
        "White",
        "Fighting",
        "Gesticulating"
};

private static final String[] NOUN_NAMES = new String[] {
        "Pony",
        "Horse",
        "Griffin",
        "Dragon",
        "Wench",
        "Bastard",
        "Ogre",
        "Troll",
        "Ox",
        "Cow",
        "Cock",
        "Hen",
        "Ram",
        "Ewe",
        "Dog",
        "Merchant",
        "Fisherman",
        "Arborist",
        "Archer",
        "Gallbladder",
        "Pancreas"
};

private String currentSign = getNewSign();

public void reset(){
    buildingText();
    verbText();
    nounText();
    SIGNS = currentSign;
}

public void read(int pos) {
    //referenced by pos in order to make the sign catch fire later on
    GameScene.show(new WndMessage(SIGNS));
}

private String getNewSign(){
    return String.format(SIGNS_FORMAT, buildingText(), verbText(), nounText());
}

private String buildingText() {
    return BUILDING_TYPES[Random.Int(0, BUILDING_TYPES.length - 1)];
}

private String verbText() {
    return VERB_NAMES[Random.Int(0, VERB_NAMES.length - 1)];
}

private String nounText() {
    return NOUN_NAMES[Random.Int(0, NOUN_NAMES.length - 1)];
}

}

Answer 1

当您使用静态最终字符串时，JVM 在启动期间创建这些字符串。为了执行您想要的操作，请使用返回字符串实例的单例模式。然后reset()方法可以在您需要时重置您正在使用的实例。