什么是这个underscore.js“安全参考”代码做什么？ [英] what is this underscore.js "safe reference" code doing?

问题描述

我正在学习Backbone，它使用的是Underscore。

I'm learning Backbone, which uses Underscore.

在某些例子中，我看到初始化代码创建了一个空的子数组，如下所示：

In some examples, I see initialization code to create an empty array of children like this:

// inside a constructor function for a view object that will be extended:
this.children = _([]);

上面调用的下划线函数 _ 是在Underscore.js顶部附近定义：

The Underscore function _ above being called is defined near top of Underscore.js:

// Create a safe reference to the Underscore object for use below.
var _ = function(obj) {
    if (obj instanceof _) return obj;
    if (!(this instanceof _)) return new _(obj);
    this._wrapped = obj;
};

在调试器中单步执行显示返回新_（obj）<首先调用/ code>，因此再次调用该函数，最后执行 this._wrapped = obj 。 此似乎是指 _ 。

Stepping through in the debugger shows me the return new _(obj) is called at first, so the function is called again and finally this._wrapped = obj is executed. this appears to be referring to _.

我很困惑。为什么不首先说 this.children = [] ？

I am bewildered. Why not just say this.children = [] in the first place?

推荐答案

因为 this.children 需要是下划线的实例：包装数组的专用类，而不仅仅是常规的javascript数组文字。 _ 函数中的代码只是确保它总是包含一个常规数组的 _ 实例，即使你试图重复重写下划线实例，使用或不使用 new 关键字调用_。

Because this.children needs to be a instance of underscore: a specialized class that wraps an array, not just a regular javascript array literal. The code in the _ function just makes sure it's always one _ instance wrapping one regular array, even if you try to rewrap an underscore instance repeatedly, call _ with or without the new keyword.

//new _ instance wrapping an array. Straightforward.
var _withNew = new _([]);

//automatically calls `new` for you and returns that, resulting in same as above
var _withoutNew = _([]);

//just gives you _withoutNew back since it's already a proper _ instance
var _doubleWrapped = _(_withoutNew);

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