Javascript: argument unpacking in function.prototype.bind()? [duplicate]

Question

The closest I've seen is this, but it doesn't really help me since I need to bind some parameters for later use with setInterval.

More specifically:

[in] var d = function(l, m) {
       console.log(l);
       console.log(m);
     }
[in] d.apply(window, [1,2])
[out] 1
[out] 2
[out-return-value] undefined
[in] d.bind(window, [1,2])()
[out] [1, 2]
[out] undefined
[out-return-value] undefined

As can be seen, arrays are unpacked with .apply(), but not with .bind(). Is there any way to unpack the arguments with .bind()?

Seems I answered such a question before :P

Felix Kling
– Felix Kling

2016-02-06 03:14:03 +00:00
Commented Feb 6, 2016 at 3:14 — Felix Kling
– Felix Kling, Commented Feb 6, 2016 at 3:14

Felix Kling · Accepted Answer · 2016-02-06 03:05:07Z

2

.bind is just another function. Call it with .apply if you want to call it with an array of arguments:

var bound = d.bind.apply(d, [window, 1, 2]);
// If the `this` value is known, but the arguments is an array, concat:
// var bound = d.bind.apply(d, [window].concat(args))

answered Feb 6, 2016 at 3:05

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Felix Kling

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Comments

Quarter2Twelve · Accepted Answer · 2016-02-06 03:14:34Z

2

Try this.

Function.prototype.bindArray(ctx, array) {
  if (array && array.length && array.pop && array.length > 0) {
    var v = array.pop();
    return this.bindArray(ctx, array).bind(ctx, v);
  }
  else return this;
};

It will iteratively bind every value in array.

Use it like:

var d = function(l, m) {
  console.log(l);
  console.log(m);
};
var c = d.bindArray(null, [1,2]);
c(); // 1 2

Also see below @felix's answer. That's even cool.

edited Feb 6, 2016 at 3:14

answered Feb 6, 2016 at 2:47

Quarter2Twelve

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4 Comments

Matthew Herbst Over a year ago

I really like this. Very elegant - nice!

13steinj Over a year ago

Question: since I think I found a solution without iteration, would this work as well, or if not, why?: ` Function.prototype.bindArray(thisArg, array) {` ` array.unshift(thisArg);` ` return this.bind.apply(this, array);` ` }`

Felix Kling Over a year ago

This seems overly complex (for the task) and creates unnecessary intermediate function objects.

Quarter2Twelve Over a year ago

@13steinj yeah it works. This answer is actually cool.

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