AngularJS 脏检查性能 [英] AngularJS dirty checking performance

问题描述

在角度作用域中，我有一个巨大的对象 scope.a，不知何故我有另一个对它的 scope.b 引用.

In the angular scope, I have a huge object scope.a, and somehow I have another scope.b reference to it.

我知道 angularJS 使用脏检查，所以我们应该减少作用域内的东西.我的问题是，因为 a 和 b 本质上是相同的 obj(reference).如果我设法去掉 b，只保留一个引用，它是否会有明显的性能提升?

I know angularJS uses dirty checking, so we should reduce the stuff inside scope. My question is, since a and b are essentially the same obj(reference). Will it have noticeable performance improvement if I manage to get rid of b, keep only one reference?

推荐答案

仅在作用域中有某些内容不会对 $digest 循环产生任何性能影响(请参阅此处与浏览器事件循环的集成:https://docs.angularjs.org/guide/scope).

Just having something in the scope does not have any performance implications on the $digest cycle (See Integration with the browser event loop here: https://docs.angularjs.org/guide/scope).

脏检查($digest cycle)调用任何注册的$watch 函数(在您的代码中手动注册或在 angular 代码中注册)，然后如果 $watch 函数返回与上次不同的任何内容，则调用侦听器函数.

The dirty checking ($digest cycle) calls any registered $watch functions (manually registered in your code or registered in the angular code), and then calls the listener function if the $watch function returned anything different from last time.

为了回答您的问题，不向范围添加内容不会有任何性能改进.通过提高任何 $watch 函数的性能最容易提高性能，因为它们总是在每个 $digest 周期至少被调用一次.

To answer your question, no there won't be any performance improvement by not adding something to the scope. Performance is most readily improved by improving performance of any $watch functions, since they are always called at least once every $digest cycle.

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