为什么Java的NIO.2中的异步通道需要这些暂停？

Question

我已经写了最少的例子，但它仍然太长了，所以让我知道我应该发布指向 Pastebin 的链接。

服务器：

import java.io.IOException;
import java.net.InetSocketAddress;
import java.net.StandardSocketOptions;
import java.nio.ByteBuffer;
import java.nio.channels.AsynchronousServerSocketChannel;
import java.nio.channels.AsynchronousSocketChannel;
import java.nio.channels.CompletionHandler;
public class Srv {
    static final int    PORT = 9001;
    static final String HOST = "127.0.0.1";
    public static void runInstance() {
        try (AsynchronousServerSocketChannel asynchronousServerSocketChannel =
                     AsynchronousServerSocketChannel.open()) {
            if (asynchronousServerSocketChannel.isOpen()) {
                asynchronousServerSocketChannel.setOption(StandardSocketOptions.SO_RCVBUF, 1024);
                asynchronousServerSocketChannel.setOption(StandardSocketOptions.SO_REUSEADDR, true);
                asynchronousServerSocketChannel.bind(new InetSocketAddress(HOST, PORT));
                System.out.println(String.format("Launched master on %s:%d", HOST, PORT));
                asynchronousServerSocketChannel.accept(null, new CompletionHandler<AsynchronousSocketChannel, Void>() {
                    @Override
                    public void completed(AsynchronousSocketChannel result, Void attachment) {
                        asynchronousServerSocketChannel.accept(null, this);
                        try {
                            result.read(ByteBuffer.allocate(1024));
                            System.out.println("Conn from:" + result.getRemoteAddress());
                        } catch (Exception exn) {
                            exn.printStackTrace();
                        } finally {
                            try {
                                result.close();
                            } catch (IOException exn) {
                                exn.printStackTrace();
                            }
                        }
                    }
                    @Override
                    public void failed(Throwable exn, Void attachment) {
                        asynchronousServerSocketChannel.accept(null, this);
                        throw new UnsupportedOperationException("can't accept");
                    }
                });
                System.in.read();
            } else {
                System.out.println("The asynchronous server-socket channel cannot be opened");
            }
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}

客户：

import java.io.IOException;
import java.net.InetSocketAddress;
import java.net.StandardSocketOptions;
import java.nio.ByteBuffer;
import java.nio.channels.AsynchronousSocketChannel;
import java.nio.channels.CompletionHandler;
import static java.lang.Thread.sleep;
public class Wrk {
    static final int    PORT = 9001;
    static final String HOST = "127.0.0.1";
    public static void runInstance() throws InterruptedException {
        sleep(1000); //HERE
        try(AsynchronousSocketChannel asynchronousSocketChannel = AsynchronousSocketChannel.open()) {
            if (asynchronousSocketChannel.isOpen()) {
                asynchronousSocketChannel.setOption(StandardSocketOptions.SO_RCVBUF, 1024);
                asynchronousSocketChannel.setOption(StandardSocketOptions.SO_SNDBUF, 1024);
                asynchronousSocketChannel.setOption(StandardSocketOptions.SO_KEEPALIVE, true);
                asynchronousSocketChannel.connect(new InetSocketAddress(HOST, PORT), null,
                    new CompletionHandler<Void, Void>() {
                        @Override
                        public void completed(Void result, Void attachment) {
                            try {
                                System.out.println("Connected to: " + HOST + PORT);
                                asynchronousSocketChannel.read(ByteBuffer.allocate(1024)).get();
                            } catch (Exception exn) {
                                exn.printStackTrace();
                            } finally {
                                try {
                                    asynchronousSocketChannel.close();
                                } catch (IOException exn) {
                                    exn.printStackTrace();
                                }
                            }
                        }
                        @Override
                        public void failed(Throwable throwable, Void o) {
                            System.out.println("Connection cannot be established");
                        }
                    });
                sleep(1000); //AND HERE
            } else {
                System.out.println("The asynchronous socket channel cannot be opened");
            }
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}

要运行的样板：

public class Main {
    public static void main(String[] args) throws InterruptedException {
        Thread srv = new Thread(new Runnable() {
            @Override
            public void run() {
                Srv.runInstance();
            }
        });
        srv.start();
        Thread wrk1 = new Thread(new Runnable() {
            @Override
            public void run() {
                try {
                    Wrk.runInstance();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }
        });
        wrk1.start();
    }
}

因此，如果我有这样的代码，它会给我以下输出：

Launched master on 127.0.0.1:9001
Connected to: 127.0.0.1 9001
Conn from:/127.0.0.1:50415

但是如果我在 Srv.runInstance() 中删除这些sleep()，我会得到：

Connection cannot be established
Launched master on 127.0.0.1:9001
Conn from:/127.0.0.1:50438

那么，这是否意味着客户端连接到服务器，但服务器拒绝？我不清楚这里发生了什么，文档相当差，所以我不知道在哪里寻找解决方案。

Answer 1

我在发布的代码中看到的唯一睡眠是在 Wrk.runInstance 中。 但是，输出的顺序非常清楚发生了什么。 您的工作正在尝试在服务器完全初始化之前连接到服务器。 正如您在输出中看到的，"连接消息"位于"启动主节点"之前。

服务器需要一点时间才能启动，因此在没有睡眠的情况下，您的客户端正在尝试连接到尚不存在的东西。