java - 递归方法上的 StackOverflowError

Question

使用递归方法时出现 StackOverflowError。我正在 Spigot API 之上进行构建。

这是我的方法:

private Location spawnPlayer(Player p) {
    int r = new Random().nextInt(ServerManager.getInstance().getServer(p).getSpawns().size());
    final Spawn s = ServerManager.getInstance().getServer(p).getSpawns().get(r);

    if (s.hasPlayer()) {
        spawnPlayer(p);
    } else {
        s.setPlayer(p);

        Bukkit.getServer().getScheduler().scheduleSyncRepeatingTask(Arsenal.p, new Runnable() {
            public void run() {
                spawnClear--;

                if (spawnClear == 0) {
                    s.setPlayer(null);
                }
            }
        }, 0L, 20L);
    }
    return s.getLocation();
}

我在 spawnPlayer(p) 处收到错误，如果生成点已经有玩家，它会调用自身。

Answer 1

看起来 s.hasPlayer() 始终为 true，因此:

if(s.hasPlayer()){
    spawnPlayer(p);
}

在 spawnPlayer(Player) 方法内部执行，这会导致该方法无限运行，从而导致 StackOverflowError .

要解决此问题，您可以在调用 spawnPlayer(p) 之前等待:

if(s.hasPlayer()){
    long timeToWait = 20L;//set the time to wait to 20 ticks (1 second)
    Bukkit.getServer().getScheduler().runTaskLater(Arsenal.p, new Runnable(){
        public void run(){
            spawnPlayer(p);
        }
    },timeToWait);
}

那么，您的代码可能如下所示:

private Location spawnPlayer(Player p) {
    int r = new Random().nextInt(ServerManager.getInstance().getServer(p).getSpawns().size());
    final Spawn s = ServerManager.getInstance().getServer(p).getSpawns().get(r);

    if(s.hasPlayer()){
        long timeToWait = 20L;//set the time to wait to 20 ticks (1 second)
        Bukkit.getServer().getScheduler().runTaskLater(Arsenal.p, new Runnable(){
            public void run(){
                spawnPlayer(p);
            }
         },timeToWait);
    }
    else{
        s.setPlayer(p);

        Bukkit.getServer().getScheduler().scheduleSyncRepeatingTask(Arsenal.p, new Runnable() {
            public void run() {
                spawnClear--;

                if (spawnClear == 0) {
                    s.setPlayer(null);
                }
            }
        }, 0L, 20L);
    }
    return s.getLocation();
}

另一个修复方法是使 s.hasPlayer() 并不总是正确，例如，通过使用 ArrayList确保玩家尚未生成:

List<String> spawned = new ArrayList<String>();

private Location spawnPlayer(Player p){
   int r = new Random().nextInt(ServerManager.getInstance().getServer(p).getSpawns().size());
   final Spawn s = ServerManager.getInstance().getServer(p).getSpawns().get(r);

   if(s.hasPlayer() && !spawned.contains(p.getName()){
      spawned.add(p.getName());
      spawnPlayer(p);
   }
   else{
     //the rest of your code...
   }
}