CODEWITHSUNDEEP
javascript
georg
georg

Reputation: 214949

Is explicit ".prototype" really needed?

I often see something like this in other people's scripts:

bar = Array.prototype.slice.call(whatever, 1)

However, the following shorter notation works fine as well:

bar = [].slice.call(whatever, 1)

Are these two constructs fully equivalent? Are there engines (browsers) that treat them differently?

Upvotes: 5

Views: 58

Answers (2)

c-smile
c-smile

Reputation: 27460

They are not equivalent strictly speaking. This construction:

[].slice.call(whatever, 1)

allocates new instance of array on the heap just for the purpose of getting property from it. So it has side effect - leaves garbage in the heap.

Upvotes: 1

jAndy
jAndy

Reputation: 235992

Yes, fully equivalent.

It happens to be that the access via .prototype is slightly faster because no new object instance needs to get created. However, thats what we call microoptimization.

A nice way to get entirely rid of deep chaining, is to invoke Function.prototype.bind.

Example

(function( slice ) {
    slice( whatever, 1 );
}( Function.prototype.call.bind( Array.prototype.slice )));

Upvotes: 5

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