Skip to content

RPC的暴露服务的代码不会创建无限多线程吗? #2

Description

@ohMyJason
 /**
     * 暴漏服务
     * @param serviceImpl 服务的实现
     * @param port 服务所处的端口号
     */
    public static void export(Object serviceImpl, int port) {
        try {
            try (ServerSocket server = new ServerSocket(port)) {
                while (!Thread.currentThread().isInterrupted()) {
                    Socket socket = server.accept();
                    new Thread(() -> {
                        try (ObjectInputStream in = new ObjectInputStream(socket.getInputStream())) {
                            Method method = serviceImpl.getClass().getMethod(in.readUTF(), (Class<?>[]) in.readObject());
                            Object result = method.invoke(serviceImpl, (Object[]) in.readObject());
                            try (ObjectOutputStream out = new ObjectOutputStream(socket.getOutputStream())) {
                                out.writeObject(result);
                            }
                            System.out.println("Invoke method [" + method.getName() + "()] success, form " + socket.getRemoteSocketAddress());
                        } catch (Exception e) {
                            throw new RuntimeException("Export service fail .", e);
                        }
                    }).start();
                }
            }
        } catch (Exception e) {
            throw new RuntimeException("Export service fail .", e);
        }
        System.out.println("Export service " + serviceImpl.getClass().getSimpleName() + " success on port " + port);
    }

socket.accept的时候返回为null,下面就直接创建线程执行了。

Activity

Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment

Metadata

Metadata

Assignees

No one assigned

    Labels

    No labels
    No labels

    Projects

    No projects

      Milestone

      No milestone

      Relationships

      None yet

      Development

      No branches or pull requests

      Issue actions