Benjamin Gruenbaum Benjamin Gruenbaum - 6 months ago 19
Javascript Question

Creating range in JavaScript - strange syntax

I've run into the following code in the es-discuss mailing list:

Array.apply(null, { length: 5 }).map(Number.call, Number);


This produces

[0, 1, 2, 3, 4]


Why is this the result of the code? What's happening here?

Answer

Understanding this "hack" requires understanding several things:

  1. Why we don't just do Array(5).map(...)
  2. How Function.prototype.apply handles arguments
  3. How Array handles multiple arguments
  4. How the Number function handles arguments
  5. What Function.prototype.call does

They're rather advanced topics in javascript, so this will be more-than-rather long. We'll start from the top. Buckle up!

1. Why not just Array(5).map?

What's an array, really? A regular object, containing integer keys, which map to values. It has other special features, for instance the magical length variable, but at it's core, it's a regular key => value map, just like any other object. Let's play with arrays a little, shall we?

var arr = ['a', 'b', 'c'];
arr.hasOwnProperty(0); //true
arr[0]; //'a'
Object.keys(arr); //['0', '1', '2']
arr.length; //3, implies arr[3] === undefined

//we expand the array by 1 item
arr.length = 4;
arr[3]; //undefined
arr.hasOwnProperty(3); //false
Object.keys(arr); //['0', '1', '2']

We get to the inherent difference between the number of items in the array, arr.length, and the number of key=>value mappings the array has, which can be different than arr.length.

Expanding the array via arr.length does not create any new key=>value mappings, so it's not that the array has undefined values, it does not have these keys. And what happens when you try to access a non-existent property? You get undefined.

Now we can lift our heads a little, and see why functions like arr.map don't walk over these properties. If arr[3] was merely undefined, and the key existed, all these array functions would just go over it like any other value:

//just to remind you
arr; //['a', 'b', 'c', undefined];
arr.length; //4
arr[4] = 'e';

arr; //['a', 'b', 'c', undefined, 'e'];
arr.length; //5
Object.keys(arr); //['0', '1', '2', '4']

arr.map(function (item) { return item.toUpperCase() });
//["A", "B", "C", undefined, "E"]

I intentionally used a method call to further prove the point that the key itself was never there: Calling undefined.toUpperCase would have raised an error, but it didn't. To prove that:

arr[5] = undefined;
arr; //["a", "b", "c", undefined, "e", undefined]
arr.hasOwnProperty(5); //true
arr.map(function (item) { return item.toUpperCase() });
//TypeError: Cannot call method 'toUpperCase' of undefined

And now we get to my point: How Array(N) does things. Section 15.4.2.2 describes the process. There's a bunch of mumbo jumbo we don't care about, but if you manage to read between the lines (or you can just trust me on this one, but don't), it basically boils down to this:

function Array(len) {
    var ret = [];
    ret.length = len;
    return ret;
}

(operates under the assumption (which is checked in the actual spec) that len is a valid uint32, and not just any number of value)

So now you can see why doing Array(5).map(...) wouldn't work - we don't define len items on the array, we don't create the key => value mappings, we simply alter the length property.

Now that we have that out of the way, let's look at the second magical thing:

2. How Function.prototype.apply works

What apply does is basically take an array, and unroll it as a function call's arguments. That means that the following are pretty much the same:

function foo (a, b, c) {
    return a + b + c;
}
foo(0, 1, 2); //3
foo.apply(null, [0, 1, 2]); //3

Now, we can ease the process of seeing how apply works by simply logging the arguments special variable:

function log () {
    console.log(arguments);
}

log.apply(null, ['mary', 'had', 'a', 'little', 'lamb']);
 //["mary", "had", "a", "little", "lamb"]

//arguments is a pseudo-array itself, so we can use it as well
(function () {
    log.apply(null, arguments);
})('mary', 'had', 'a', 'little', 'lamb');
 //["mary", "had", "a", "little", "lamb"]

//a NodeList, like the one returned from DOM methods, is also a pseudo-array
log.apply(null, document.getElementsByTagName('script'));
 //[script, script, script, script, script, script, script, script, script, script, script, script, script, script, script, script, script, script, script, script]

//carefully look at the following two
log.apply(null, Array(5));
//[undefined, undefined, undefined, undefined, undefined]
//note that the above are not undefined keys - but the value undefined itself!

log.apply(null, {length : 5});
//[undefined, undefined, undefined, undefined, undefined]

It's easy to prove my claim in the second-to-last example:

function ahaExclamationMark () {
    console.log(arguments.length);
    console.log(arguments.hasOwnProperty(0));
}

ahaExclamationMark.apply(null, Array(2)); //2, true

(yes, pun intended). The key => value mapping may not have existed in the array we passed over to apply, but it certainly exists in the arguments variable. It's the same reason the last example works: The keys do not exist on the object we pass, but they do exist in arguments.

Why is that? Let's look at Section 15.3.4.3, where Function.prototype.apply is defined. Mostly things we don't care about, but here's the interesting portion:

  1. Let len be the result of calling the [[Get]] internal method of argArray with argument "length".

Which basically means: argArray.length. The spec then proceeds to do a simple for loop over length items, making a list of corresponding values (list is some internal voodoo, but it's basically an array). In terms of very, very loose code:

Function.prototype.apply = function (thisArg, argArray) {
    var len = argArray.length,
        argList = [];

    for (var i = 0; i < len; i += 1) {
        argList[i] = argArray[i];
    }

    //yeah...
    superMagicalFunctionInvocation(this, thisArg, argList);
};

So all we need to mimic an argArray in this case is an object with a length property. And now we can see why the values are undefined, but the keys aren't, on arguments: We create the key=>value mappings.

Phew, so this might not have been shorter than the previous part. But there'll be cake when we finish, so be patient! However, after the following section (which'll be short, I promise) we can begin dissecting the expression. In case you forgot, the question was how does the following work:

Array.apply(null, { length: 5 }).map(Number.call, Number);

3. How Array handles multiple arguments

So! We saw what happens when you pass a length argument to Array, but in the expression, we pass several things as arguments (an array of 5 undefined, to be exact). Section 15.4.2.1 tells us what to do. The last paragraph is all that matters to us, and it's worded really oddly, but it kind of boils down to:

function Array () {
    var ret = [];
    ret.length = arguments.length;

    for (var i = 0; i < arguments.length; i += 1) {
        ret[i] = arguments[i];
    }

    return ret;
}

Array(0, 1, 2); //[0, 1, 2]
Array.apply(null, [0, 1, 2]); //[0, 1, 2]
Array.apply(null, Array(2)); //[undefined, undefined]
Array.apply(null, {length:2}); //[undefined, undefined]

Tada! We get an array of several undefined values, and we return an array of these undefined values.

The first part of the expression

Finally, we can decipher the following:

Array.apply(null, { length: 5 })

We saw that it returns an array containing 5 undefined values, with keys all in existence.

Now, to the second part of the expression:

[undefined, undefined, undefined, undefined, undefined].map(Number.call, Number)

This will be the easier, non-convoluted part, as it doesn't so much rely on obscure hacks.

4. How Number treats input

Doing Number(something) (section 15.7.1) converts something to a number, and that is all. How it does that is a bit convoluted, especially in the cases of strings, but the operation is defined in section 9.3 in case you're interested.

5. Games of Function.prototype.call

call is apply's brother, defined in section 15.3.4.4. Instead of taking an array of arguments, it just takes the arguments it received, and passes them forward.

Things get interesting when you chain more than one call together, crank the weird up to 11:

function log () {
    console.log(this, arguments);
}
log.call.call(log, {a:4}, {a:5});
//{a:4}, [{a:5}]
//^---^  ^-----^
// this   arguments

This is quite wtf worthy until you grasp what's going on. log.call is just a function, equivalent to any other function's call method, and as such, has a call method on itself as well:

log.call === log.call.call; //true
log.call === Function.call; //true

And what does call do? It accepts a thisArg and a bunch of arguments, and calls its parent function. We can define it via apply (again, very loose code, won't work):

Function.prototype.call = function (thisArg) {
    var args = arguments.slice(1); //I wish that'd work
    return this.apply(thisArg, args);
};

Let's track how this goes down:

log.call.call(log, {a:4}, {a:5});
  this = log.call
  thisArg = log
  args = [{a:4}, {a:5}]

  log.call.apply(log, [{a:4}, {a:5}])

    log.call({a:4}, {a:5})
      this = log
      thisArg = {a:4}
      args = [{a:5}]

      log.apply({a:4}, [{a:5}])

The later part, or the .map of it all

It's not over yet. Let's see what happens when you supply a function to most array methods:

function log () {
    console.log(this, arguments);
}

var arr = ['a', 'b', 'c'];
arr.forEach(log);
//window, ['a', 0, ['a', 'b', 'c']]
//window, ['b', 1, ['a', 'b', 'c']]
//window, ['c', 2, ['a', 'b', 'c']]
//^----^  ^-----------------------^
// this         arguments

If we don't provide a this argument ourselves, it defaults to window. Take note of the order in which the arguments are provided to our callback, and let's weird it up all the way to 11 again:

arr.forEach(log.call, log);
//'a', [0, ['a', 'b', 'c']]
//'b', [1, ['a', 'b', 'c']]
//'b', [2, ['a', 'b', 'c']]
// ^    ^

Whoa whoa whoa...let's back up a bit. What's going on here? We can see in section 15.4.4.18, where forEach is defined, the following pretty much happens:

var callback = log.call,
    thisArg = log;

for (var i = 0; i < arr.length; i += 1) {
    callback.call(thisArg, arr[i], i, arr);
}

So, we get this:

log.call.call(log, arr[i], i, arr);
//After one `.call`, it cascades to:
log.call(arr[i], i, arr);
//Further cascading to:
log(i, arr);

Now we can see how .map(Number.call, Number) works:

Number.call.call(Number, arr[i], i, arr);
Number.call(arr[i], i, arr);
Number(i, arr);

Which returns the transformation of i, the current index, to a number.

In conclusion,

The expression

Array.apply(null, { length: 5 }).map(Number.call, Number);

Works in two parts:

var arr = Array.apply(null, { length: 5 }); //1
arr.map(Number.call, Number); //2

The first part creates an array of 5 undefined items. The second goes over that array and takes its indices, resulting in an array of element indices:

[0, 1, 2, 3, 4]