Why does adding [0] after the delete count in splice() return just a number?

Question

First of all sorry for the long title, I really didn't know how to word it better.

In the middle of solving _.shuffle in underscore, I encountered the use of splice. Here is my original code :

shuffle = function(array) {
  var shuffledArray = [];
  var total = array.length;
  var copiedArray = array.slice();
  while (total){
    var randomNum = Math.floor(Math.random() * total);
    shuffledArray.push(copiedArray.splice(randomNum,1));
    total--;
  }
  return shuffledArray;
};
var originArray = [1, 2, 3, 4];
console.log(shuffle(originArray));

However after testing just I realized it will return each value inside a [ ], instead of just the value.

i.e.

[ [ 4 ], [ 3 ], [ 1 ], [ 2 ] ]
//instead of [ 4, 3, 1, 2]

When I changed this line (added '[0]' after the deleteCount in splice)

shuffledArray.push(copiedArray.splice(randomNum,1));

into this

//edited
shuffledArray.push(copiedArray.splice(randomNum,1)[0]);

the return array that I get is what I wanted, which is

[ 3, 1, 2, 4 ] //values are not in [ ]

Can someone explain how adding [0] after splice() makes the value not return in [ ] or why not having [0] does?

`1[0]` return _undefined_ so you call same as `copiedArray.splice(randomNum)` — Grundy, Sep 23 '15 at 08:18
@Grundy: `1[0]` is indeed `undefined`, but `copiedArray.splice(randomNum)` and `copiedArray.splice(randomNum, undefined)` are **not** the same thing. — T.J. Crowder, Sep 23 '15 at 08:21
I'm just trying to understand why writing 1[0] will return the value as is and not inside [ ] — Tinah, Sep 23 '15 at 08:21
Are you sure you're not using `copiedArray.splice(randomNum,1)[0]`? — ralh, Sep 23 '15 at 08:22
@Tinah: As far as I can tell, `copiedArray.splice(randomNum, undefined)` returns a blank array. Your question is unclear about what you're trying to do, why you would ever think of doing `1[0]`, your ultimate goal... — T.J. Crowder, Sep 23 '15 at 08:22
@T.J.Crowder, yep, you right :-) with _undefined_ returned empty array — Grundy, Sep 23 '15 at 08:22
@Luaan: You mean, like, [the specification](http://www.ecma-international.org/ecma-262/6.0/index.html#sec-array.prototype.splice)? Note the opening words relate to the number of arguments, not what their values are. Or you can open a console in a browser and try it. — T.J. Crowder, Sep 23 '15 at 08:24
@Tinah, i think you want `[0]` after `splice` like `array.splice(0, 1)[0]` — Grundy, Sep 23 '15 at 08:24
@T.J.Crowder Browser are known to differ in details like this, but yeah, the spec link is exactly what I wanted :) And just to note, when I try Tinah's code in Chrome, it works *exactly* as with `undefined` (because, well, `1[0]` *is* `undefined`). — Luaan, Sep 23 '15 at 08:25
@T.J.Crowder @ralh yes you're right, I was using `copiedArray.splice(randomNum,1)[0]` not, `copiedArray.splice(randomNum,1[0])` smh — Tinah, Sep 23 '15 at 08:25
so does adding [0] just mean the element I removed is now being added to the zero index of the new array? — Tinah, Sep 23 '15 at 08:26
@Tinah, are you sure that you need `splice`? do you see [doc](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/splice)? — Grundy, Sep 23 '15 at 08:27
No :) You're simply taking the first element of the (single element) array returned by `splice`, and pushing that. — Luaan, Sep 23 '15 at 08:29
@Luaan yea, now I see it. I am just pushing the zero index of the array that splice() extracted. Thank you. — Tinah, Sep 23 '15 at 08:42

score 1 · Answer 1 · answered Sep 23 '15 at 08:27

1

splice returns an array. When you push an array into an array, you get an array of arrays. By adding the [0], you push the first element of the array instead. This is no Javascript weirdness - it's perfectly reasonable.

answered Sep 23 '15 at 08:27

Luaan

62,244
7
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@Tinah It's a very weird misunderstanding, I must say. Frighteningly, I can imagine this working the way you (wrongly) expect in some unsafe language like C under some specific conditions - and failing miserably in real world use once in a while. – Luaan Sep 23 '15 at 08:31
I think I was more confused when I thought my code was `copiedArray.splice(randomNum,1[0]` now that I realized that mistake. It makes sense. – Tinah Sep 23 '15 at 08:35

score 0 · Answer 2 · answered Sep 23 '15 at 08:27

As I suspected and was confirmed in the comments, you were using copiedArray.splice(randomNum,1)[0], not copiedArray.splice(randomNum,1[0]).

EDIT: That was the original question, and it was since edited.

And since splice returns an array (in your case, an array with one number), taking the first element of the array achieves exactly what you wanted.

score 0 · Answer 3 · answered Sep 23 '15 at 08:29

0

Splice returns an array of the deleted elements. By adding [0] you are effectively selecting the first item in the returned array.

answered Sep 23 '15 at 08:29

Armando Canals

290
1
7

score -2 · Answer 4 · answered Sep 23 '15 at 08:25

Try not to reinvent the wheel ! http://php.net/manual/fr/function.shuffle.php

If you really want to, try this in your while:

  while (total){
    var randomNum = Math.floor(Math.random() * total);
    ShuffledArray[] = array[randomNum]; //add an item to next pos
    unset(array[randomNum]); //remove the item added
    array = array_values(array); //reindex array
    total--;
  }

Best of luck !

Why does adding [0] after the delete count in splice() return just a number?

4 Answers4