广播框架

Question

当您使用生产者/消费者关系设计两个类时，如何避免循环依赖？这里ListenerImpl需要对Broadcaster的引用才能注册/取消注册，而Broadcaster需要一个引用回侦听器才能发送消息。此示例使用Java，但它可以应用于任何OO语言。

public interface Listener {
  void callBack(Object arg);
}
public class ListenerImpl implements Listener {
  public ListenerImpl(Broadcaster b) { b.register(this); }
  public void callBack(Object arg) { ... }
  public void shutDown() { b.unregister(this); }
}
public class Broadcaster {
  private final List listeners = new ArrayList();
  public void register(Listener lis) { listeners.add(lis); }
  public void unregister(Listener lis) {listeners.remove(lis); }
  public void broadcast(Object arg) { for (Listener lis : listeners) { lis.callBack(arg); } }
}

Answer 1

我不认为这是一种循环依赖。

听众不依赖。

ListenerImpl依赖于Listener和Broadcaster

广播公司取决于听众。

        Listener
       ^        ^
      /          \
     /            \
Broadcaster <--  ListenerImpl

所有箭头在Listener处结束。没有周期。所以，我觉得你没事。

Answer 2

任何OOP语言？好。这是CLOS的十分钟版本。

广播框架

(defclass broadcaster ()
  ((listeners :accessor listeners
              :initform '())))

(defgeneric add-listener (broadcaster listener)
  (:documentation "Add a listener (a function taking one argument)
  to a broadcast's list of interested parties"))

(defgeneric remove-listener (broadcaster listener)
  (:documentation "Reverse of add-listener"))

(defgeneric broadcast (broadcaster object)
  (:documentation "Broadcast an object to all registered listeners"))

(defmethod add-listener (broadcaster listener)
  (pushnew listener (listeners broadcaster)))

(defmethod remove-listener (broadcaster listener)
  (let ((listeners (listeners broadcaster)))
    (setf listeners (remove listener listeners))))

(defmethod broadcast (broadcaster object)
  (dolist (listener (listeners broadcaster))
    (funcall listener object)))

示例子类

(defclass direct-broadcaster (broadcaster)
  ((latest-broadcast :accessor latest-broadcast)
   (latest-broadcast-p :initform nil))
  (:documentation "I broadcast the latest broadcasted object when a new listener is added"))

(defmethod add-listener :after ((broadcaster direct-broadcaster) listener)
  (when (slot-value broadcaster 'latest-broadcast-p)
    (funcall listener (latest-broadcast broadcaster))))

(defmethod broadcast :after ((broadcaster direct-broadcaster) object)
  (setf (slot-value broadcaster 'latest-broadcast-p) t)
  (setf (latest-broadcast broadcaster) object))

示例代码

Lisp> (let ((broadcaster (make-instance 'broadcaster)))
        (add-listener broadcaster 
                      #'(lambda (obj) (format t "I got myself a ~A object!~%" obj)))
        (add-listener broadcaster 
                      #'(lambda (obj) (format t "I has object: ~A~%" obj)))
        (broadcast broadcaster 'cheezburger))

I has object: CHEEZBURGER
I got myself a CHEEZBURGER object!

Lisp> (defparameter *direct-broadcaster* (make-instance 'direct-broadcaster))
      (add-listener *direct-broadcaster*
                  #'(lambda (obj) (format t "I got myself a ~A object!~%" obj)))
      (broadcast *direct-broadcaster* 'kitty)

I got myself a KITTY object!

Lisp> (add-listener *direct-broadcaster*
                    #'(lambda (obj) (format t "I has object: ~A~%" obj)))

I has object: KITTY

不幸的是，Lisp通过消除对它们的需求来解决大多数设计模式问题（例如你的问题）。

Answer 3

与Herms的回答相反，我做看到一个循环。它不是依赖循环，它是一个引用循环：LI持有B对象，B对象持有（一个数组）LI对象。他们不容易释放，需要注意确保他们在可能的情况下自由。

一种解决方法就是让LI对象为广播公司提供WeakReference。从理论上讲，如果广播公司已经离开，那么无论如何都没有注销，所以你的注销将只是检查是否有广播公司注销，如果存在则这样做。

Answer 4

我不是java开发者，但是像这样：

public class ListenerImpl implements Listener {
  public Foo() {}
  public void registerWithBroadcaster(Broadcaster b){ b.register(this); isRegistered = true;}
  public void callBack(Object arg) { if (!isRegistered) throw ... else ... }
  public void shutDown() { isRegistered = false; }
}

public class Broadcaster {
  private final List listeners = new ArrayList();
  public void register(Listener lis) { listeners.add(lis); }
  public void unregister(Listener lis) {listeners.remove(lis); }
  public void broadcast(Object arg) { for (Listener lis : listeners) { if (lis.isRegistered) lis.callBack(arg) else unregister(lis); } }
}

Answer 5

使用弱引用来打破循环。

请参阅this answer。

Answer 6

这是Lua中的一个示例（我在这里使用我自己的Oop lib，请参阅代码中对'Object'的引用。）

与Mikael Jansson的CLOS示例一样，您可以直接使用函数，无需定义侦听器（请注意使用'...'，它是Lua的varargs）：

Broadcaster = Object:subclass()

function Broadcaster:initialize()
    self._listeners = {}
end

function Broadcaster:register(listener)
    self._listeners[listener] = true
end

function Broadcaster:unregister(listener)
    self._listeners[listener] = nil
end
function Broadcaster:broadcast(...)
    for listener in pairs(self._listeners) do
        listener(...)
    end
end

坚持你的实现，这是一个可以用任何动态语言编写的例子：

--# Listener
Listener = Object:subclass()
function Listener:callback(arg)
    self:subclassResponsibility()
end

--# ListenerImpl
function ListenerImpl:initialize(broadcaster)
    self._broadcaster = broadcaster
    broadcaster:register(this)
end
function ListenerImpl:callback(arg)
    --# ...
end
function ListenerImpl:shutdown()
    self._broadcaster:unregister(self)
end

--# Broadcaster
function Broadcaster:initialize()
    self._listeners = {}
end
function Broadcaster:register(listener)
    self._listeners[listener] = true
end
function Broadcaster:unregister(listener)
    self._listeners[listener] = nil
end
function Broadcaster:broadcast(arg)
    for listener in pairs(self._listeners) do
        listener:callback(arg)
    end
end

使用回调时如何避免循环依赖？

6 个答案:

广播框架

示例子类

示例代码