处理列表到一个线程
基本上我有一个整数数组，我想将其移交给线程，但我无法获得正确的语法。
// Create list
List <Integer> list = new ArrayList<Integer>();
// Create thread
TPServer server = new TPServer(port, <Integer> list);
new Thread(server).start();
// Below is the TPServer class
// TPServer Class
public class TPServer implements Runnable {
private List <Integer> list = null;
private int port = 0;
private boolean isStopped = false;
public TPServer(int port, List <Integer> list) {
this.list = list;
this.port = port;
}
}
Basically I have an integer array which i want to hand over to a thread, but i can't get the syntax right.
// Create list
List <Integer> list = new ArrayList<Integer>();
// Create thread
TPServer server = new TPServer(port, <Integer> list);
new Thread(server).start();
// Below is the TPServer class
// TPServer Class
public class TPServer implements Runnable {
private List <Integer> list = null;
private int port = 0;
private boolean isStopped = false;
public TPServer(int port, List <Integer> list) {
this.list = list;
this.port = port;
}
}
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评论（3）
您还没有实现
Runnable，它也应该是new TPServer(port, list);You haven't implemented
Runnableand also it should benew TPServer(port, list);您可以像传递任何其他参数一样传递通用参数。
You pass a generic argument the same as any other.
您可能想要使用使用信号量的列表来防止并发访问，或者使用线程安全的列表类型（如果您的列表也在线程外使用）。 （如果它仅在线程内部使用，显然在内部创建它:=)）
You might wanna use a list that uses a semaphore to prevent concurent access Or use a list type that is thread safe, if your list is being used also out of the thread. (and if it is used only inside the thread, create it inside obviously :=) )