Underscore.js core

Question

I'm going through the code of the underscore library just to see how it's been built:

I understand most of it, it contains comments, however there are parts which I do not understand yet:

// 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;
};

I've changed it a little just to learn how it works:

function _(obj) {

    if (obj instanceof _) {
        return obj;
    }

    if (!(this instanceof _)) {
        return new _(obj);
    }

    this._wrapped = obj;
}

// just an example, if statement to check if array 
// is really an array should be added

_.sum = function (array) {
    var sum = 0;
    for (var i = 0, ilen = array.length; i < ilen; i += 1) {
        sum += array[i];
    }
    return sum;
}

_.sum([1, 2, 3, 4, 5]); // 15

Ok, so now I'll create a new instance with the _ constructor. The new keyword will be used even if I don't provide it, everything works fine.

var _a = _([1, 2, 3, 4, 5]); // { _wrapped: [1, 2, 3, 4, 5] }

What's the purpose of the following line?

if (obj instanceof _) return obj;

I can remove it and the code still works correctly. Is it used only to handle the following case?

var _b = _(_a);

Benjamin Gruenbaum · Accepted Answer · 2014-01-27 13:18:46Z

2

Yes, it is used to handle the 'already wrapped' case.

If you pass an underscore instance to _ it'll return that instance instead of wrapping it. That's the purpose of :

if (obj instanceof _) return obj;

As proof - here is the bug adding this in solved.

edited Jan 27, 2014 at 13:18

answered Jan 27, 2014 at 13:13

Benjamin Gruenbaum

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1 Comment

Benjamin Gruenbaum Over a year ago

Here is the commit where it was changed github.com/jashkenas/underscore/commit/… it contains a unit test added to enforce this behavior.

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