我尽力在此Java代码中找到问题，但我找不到它-您能帮我吗？
我希望我提供的代码足够，但是如有必要，我会发布更多代码。
进一步致歉，我没有举一个最小的例子。
game.getGroupPlayers().list();
MoverThread[] playerThread = game.getPlayers();
System.out.println(playerThread.length);
for (int i = 0; i < playerThread.length; i++) {
try {
System.out.println(i + " -> " +playerThread[i].toString());
returnString += playerThread[i].toString() + "\n";
} catch(NullPointerException e) {
System.out.println("Problem at i = " + i);
e.printStackTrace();
}
game.getGroupPlayers().list();
}
有时会给我以下输出:
[编辑]
这是getPlayers()的来源
/*
* post returns the games players as an array
*/
public MoverThread[] getPlayers() {
synchronized(movers) {
MoverThread[] playerList = new MoverThread[players.activeCount()];
players.enumerate(playerList);
return playerList;
}
}
[编辑]
这是球员的产生方式
private ThreadGroup movers;
private ThreadGroup players;
private ThreadGroup ghosts;
private Observer observer;
/*
* constructor
*/
public Game(Maze maze, Referee referee) {
this.maze = maze;
this.referee = referee;
threadList = new ArrayList<MoverThread>();
movers = new ThreadGroup("Movers");
players = new ThreadGroup(movers, "Players");
ghosts = new ThreadGroup(movers, "Ghosts");
observer = null;
}
[编辑]
这是我如何调用产生问题的方法:
/*
* post checks if the players thread was interrupted - if not if hostfield pretends to be a goal the game gets stopped and referee is called to perform "goal-found-actions"
*/
private void checkGoal() {
if (!getThread().isInterrupted()) {
synchronized(getGame().getMovers()) {
if (!getThread().isInterrupted()) {
if (getHostField().isGoal()) {
Field goal = getHostField();
getGame().getReferee().goalFound(this, goal);
getGame().setGameOver();
}
}
}
}
}
这是整个目标的Found()
/*
* post action to be performed if a player finds a goal
* print some information
*/
public void goalFound(Player player, Field at) {
//FIXME get the Bug!!!
String returnString = "Game over - player " + player.getName() + " found a goal on (" + at.getPos()[0] + ", " + at.getPos()[1] + ")!\n";
game.getGroupPlayers().list();
MoverThread[] playerThread = game.getPlayers();
System.out.println(playerThread.length);
for (int i = 0; i < playerThread.length; i++) {
try {
System.out.println(i + " -> " +playerThread[i].toString());
returnString += playerThread[i].toString() + "\n";
} catch(NullPointerException e) {
System.out.println("Problem at i = " + i);
e.printStackTrace();
}
}
game.getGroupPlayers().list();
returnString += game.mazeString();
System.out.println(returnString);
}
最佳答案
没有一种枚举Thread的ThreadGroup的好方法。这是一个众所周知的可怕设计。
在调用ThreadGroup.activeCount和ThreadGroup.enumerate(Thread[])之间，线程可能已经启动或死亡。最好的办法是在分配数组时在activeCount中添加一个软糖因子。如果返回的值与数组长度匹配，则您可能错过了一些值，应该以更大的数组大小(可能是一个更大的因数，而不只是添加一个常数)重复进行。成功后，您将需要适本地修剪数组(或将其照这样处理)。