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

问题描述

在角度范围中，我有一个很大的对象范围.a，并且以某种方式，我还有另一个范围.b对它的引用.

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，因此我们应该减少范围内的内容.我的问题是，由于a和b本质上是相同的obj(reference).如果我设法摆脱b，只保留一个参考，它会在性能上有明显的改善吗?

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

仅在范围内进行操作就不会对$ digest周期产生任何性能影响(请参见与浏览器事件循环集成 : 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循环)调用任何已注册的 $观看函数(手动在您的代码中注册或在角度代码中注册)，然后在$watch函数返回与上次不同的内容时调用侦听器函数.

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.

要回答您的问题，不向示波器添加任何内容不会改善任何性能.通过改善任何$watch函数的性能，最容易提高性能，因为它们总是每个$ digest循环至少调用一次.

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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