How Function.prototype.call.bind works in this code [duplicate]

Question

I really could use an explanation from anybody.

Seriously confused about this code. Especially line 2.

Code source is https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/Array/slice

var unboundSlice = Array.prototype.slice;
var slice = Function.prototype.call.bind(unboundSlice);

function list() {
  return slice(arguments);
}

var list1 = list(1, 2, 3); // [1, 2, 3]

It hard-wires (so to speak) the .call part of Array.prototype.slice.call, turning a method of an array instance into a function that instead takes an array as an argument. — Jared Smith
– Jared Smith, Commented Jun 23, 2016 at 19:57

Cuong Ta · Accepted Answer · 2016-06-23 19:46:55Z

7

var slice = Function.prototype.call.bind(unboundSlice);

Pass unboundSlice as context (this operator) that function call will execute. So when execute: list(1,2,3) <=>

slice([1,2,3]) <=>

excecute call with context unboundSlice and parameter arguments: unboundSlice.call(arguments) <=>

execute unboundSlice function with context arguments (array [1,2,3]):[1,2,3].unboundSlice() <=>

execute: [1,2,3].slice() <=>

[1, 2, 3]

edited Jun 23, 2016 at 19:46

answered Jun 23, 2016 at 18:57

Cuong Ta

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1 Comment

W3Guy Over a year ago

Thank you so much.

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