CODEWITHSUNDEEP
javascriptunderscore.js
Zack Andrews
Zack Andrews

Reputation: 11

Why does _.extend make a shallow copy?

I've looked through the documentation and source code, and it appears that underscore's extend method _.extend makes a shallow copy. One of the collaborators also said that the extend method does a shallow copy of the object "intentionally" here.

Now my question is : Why was it made that way ? Could you please provide me with some scenarios or use cases where this shallow copy would be better?

I tried to look for some answers and I couldn't find much about this except that "it's a good thing".

Upvotes: 0

Views: 269

Answers (1)

LordTribual
LordTribual

Reputation: 4249

Deep copies or object clones are very computational expensive. _.extend is essentially just an alias for _.assignIn(object, [sources]) which does pretty much the same as _.assign but iterates over own and inherited source properties. The native ES6 Object.assign does not iterate over inherited source properties.

Taking the example from lodash's documentation on assignIn, Object.assign would give you the following output:

function Foo() {
  this.a = 1;
}

function Bar() {
  this.c = 3;
}

Foo.prototype.b = 2;
Bar.prototype.d = 4;

console.log(Object.assign({ 'a': 0 }, new Foo, new Bar));
// { a: 1, c: 3 }

Whereas _assignIn or _.extend gives you

// { 'a': 1, 'b': 2, 'c': 3, 'd': 4 }

As you can see, also inherited source properties (set via the prototype), that is b and d, were also iterated.

If you want to deeply clone an object, you can use _.cloneDeep(value). But keep in mind that such deep clone is very computational expensive compared to a shallow copy.

Upvotes: 2

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