如果当范围破坏角$表中删除？ [英] Should angular $watch be removed when scope destroyed?
问题描述
目前工作的一个项目，我们发现了巨大的内存泄漏时，不清除广播订阅了破坏范围。下面code已经修复了这个：
VAR onFooEventBroadcast = $ rootScope上（'fooEvent'，DoSomething的）$;（范围。在$（'$破坏'，功能）{
//删除广播订阅范围时被销毁
onFooEventBroadcast（）;
}）;
如若这种做法也可以用于手表 code下面的例子：
VAR onFooChanged = $范围手表（'富'，DoSomething的）。（范围。在$（'$破坏'，功能）{
//驻足观望范围时被销毁
onFooChanged（）;
}）;
没有，你并不需要删除 $$观察家，因为它们会有效地去除得到一次的范围被破坏。
从角的源$ C $ C（v1.2.21），的范围 $摧毁方法：
$破坏：功能（）{
...
如果（家长$$ childHead ==本）父$$ childHead =这个$$ nextSibling。;
如果（家长$$ childTail ==本）父$$ childTail =这个$$prevSibling。;
如果（这$$prevSibling）这个$$prevSibling $$ nextSibling =这个$$ nextSibling。;
如果（这$$ nextSibling）。这个$$ nextSibling $$prevSibling =这个$$prevSibling。;
...
。这个$$观察家=这个$$ asyncQueue =这个$$ postDigestQueue = [];
...
因此， $$观察家数组被清空（和范围是从范围层次结构中移除）。
删除观察家从数组是所有DEREGISTER功能确实反正：
$表：功能（watchExp，监听器，objectEquality）{
...
返回功能deregisterWatch（）{
arrayRemove（数组，守望者）;
lastDirtyWatch = NULL;
};
}
因此，有在注销的 $$观察家没有一点手动。
您还是应该取消注册事件侦听器，但（为你正确地在您的文章提）！
注意：
你只需要注销的其他范围注册的侦听器。没有必要注销上被销毁范围注册的侦听器。结果
例如：
//你必须注销这些
$ rootScope $上（...）。
。$ $范围父在$（...）。//你不必注销该
$范围。在$（...）
<子>（感谢@约翰的<一个href=\"http://stackoverflow.com/questions/25113884/should-angular-watch-be-removed-when-scope-destroyed/25114028?noredirect=1#comment-49039216\">pointing吧）
此外，请确保您从活得比范围被破坏元素注销任何事件侦听器。例如。如果你有一个指令注册一个监听器父节点上或＆LT;车身方式＆gt; ，则必须注销他们太多结果
<子>同样，你也不必取消对被破坏的元素注册的侦听器。
无关类原来的问题，但现在也有在元件上的 $摧毁事件调度被破坏，所以你可以挂接到，作为以及（如果这是适合您的用例）：
链接：功能postLink（范围，ELEM）{
doStuff（）;
elem.on（'$毁灭'，清理）;
}
Currently working on a project where we found huge memory leaks when not clearing broadcast subscriptions off destroyed scopes. The following code has fixed this:
var onFooEventBroadcast = $rootScope.$on('fooEvent', doSomething);
scope.$on('$destroy', function() {
//remove the broadcast subscription when scope is destroyed
onFooEventBroadcast();
});
Should this practice also be used for watches? Code example below:
var onFooChanged = scope.$watch('foo', doSomething);
scope.$on('$destroy', function() {
//stop watching when scope is destroyed
onFooChanged();
});
No, you don't need to remove $$watchers, since they will effectively get removed once the scope is destroyed.
From Angular's source code (v1.2.21), Scope's $destroy method:
$destroy: function() {
...
if (parent.$$childHead == this) parent.$$childHead = this.$$nextSibling;
if (parent.$$childTail == this) parent.$$childTail = this.$$prevSibling;
if (this.$$prevSibling) this.$$prevSibling.$$nextSibling = this.$$nextSibling;
if (this.$$nextSibling) this.$$nextSibling.$$prevSibling = this.$$prevSibling;
...
this.$$watchers = this.$$asyncQueue = this.$$postDigestQueue = [];
...
So, the $$watchers array is emptied (and the scope is removed from the scope hierarchy).
Removing the watcher from the array is all the deregister function does anyway:
$watch: function(watchExp, listener, objectEquality) {
...
return function deregisterWatch() {
arrayRemove(array, watcher);
lastDirtyWatch = null;
};
}
So, there is no point in deregistering the $$watchers "manually".
You should still deregister event listeners though (as you correctly mention in your post) !
NOTE:
You only need to deregister listeners registered on other scopes. There is no need to deregister listeners registered on the scope that is being destroyed.
E.g.:
// You MUST deregister these
$rootScope.$on(...);
$scope.$parent.$on(...);
// You DON'T HAVE to deregister this
$scope.$on(...)
(Thx to @John for pointing it out)
Also, make sure you deregister any event listeners from elements that outlive the scope being destroyed. E.g. if you have a directive register a listener on the parent node or on <body>, then you must deregister them too.
Again, you don't have to remove a listener registered on the element being destroyed.
Kind of unrelated to the original question, but now there is also a $destroyed event dispatched on the element being destroyed, so you can hook into that as well (if it's appropriate for your usecase):
link: function postLink(scope, elem) {
doStuff();
elem.on('$destroy', cleanUp);
}
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