Related
I understand the below code adds a new object to an array of objects but fuzzy on a specific syntax: setProductList([array, obj])
From what I'm seeing, the setProductList function takes in an object. That object consists of an array and an object. So how does it add the object to the array? is this built into JS or React?
// array of products
const Products = [
{
item: "basketball",
price: 3,
description: "basketball desc",
},
{
item: "football",
price: 5,
description: "football desc",
},
];
// setting state
const [productList, setProductList] = useState(Products);
// handling click
const handleClick = () => {
// Don't quite understand the syntax of below block
setProductList([
...productList, // spread creates a copy of the array
{
item: "baseball",
price: 4,
description: "baseball desc",
},
]);
};
What you are seeing is Spread operator (...)
Spread syntax can be used when all elements from an object or array need to be included in a list of some kind.
By writing [ (element inside) ], you did create a new array. ...productList alone does not create a new array
You could understand ...productList as it helps you write shorter code, instead of writing like this
setProductList([
productList[0],
productList[1],
// ...
productList[productList.length - 1],
{
item: "baseball",
price: 4,
description: "baseball desc",
},
])
Another example that could help you understand the spread operator, is the use of Math.max. Basically, syntax of Math.max is Math.max(value0, value1, ... , valueN).
For example you have an array of N elements, because Math.max solely could not take an array as arguments, so to calculate max of the array, you could iterate each element and compare
const arr = [1, 3, 2]
let max = -Infinity
for (const num of arr) {
max = Math.max(max, num)
}
console.log(max)
But now with spread operator, you could achieve the same result in a shorter way
const arr = [1, 3, 2]
let max = Math.max(...arr)
console.log(max)
This feature is built into js and called Object Spreading
Example
var x = [1, 2] // an array
console.log(x)
// spreads an array, and put an extra element
var y = [...x, 6]
console.log(y)
so, the state hook returns a function here - setProductList
That just updates the state to the new array.
it's a common practise. See this detailed answer for in deth explanation
Ok, the code which you don't understand basically is setting a new copy of the current array, BUT including the new element (it'd be like push a new element in the array. BUT push would modify the original array. In this case, it's making a copy and pushing the value to that new array)
Lets go line by line.
We have:
setProductList([
...productList,
{
item: "baseball",
price: 4,
description: "baseball desc",
},
]);
We will define:
setProductList([ // Line 0
...productList, // Line 1
{ // Line 2
item: "baseball", // Part of Line 2
price: 4, // Part of Line 2
description: "baseball desc", // Part of Line 2
}, // Part of Line 2
]); // Part of Line 1(]) & 0())
With Line 0 (setProductList) we are setting a new value to productsList because react hooks and useState. (https://reactjs.org/docs/hooks-state.html)
With Line 1 we are creating a new copy of the array ([...oldArray]) will return a copy of the previous array (You can read: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Spread_syntax)
With Line 2, we are including a new value at the end of the new array.
I guess your confusion comes from not understanding the spread syntax.
Spread in array literals
A more powerful array literal
Without spread syntax, to create a new array using an existing array
as one part of it, the array literal syntax is no longer sufficient
and imperative code must be used instead using a combination of
push(), splice(), concat(), etc. With spread
syntax this becomes much more succinct:
let parts = ['shoulders', 'knees'];
let lyrics = ['head', ...parts, 'and', 'toes'];
// ["head", "shoulders", "knees", "and", "toes"]
Just like spread for argument lists, ... can be used anywhere in the
array literal, and may be used more than once.
When looking at your code:
setProductList([
...productList, // spread creates a copy of the array
{
item: "baseball",
price: 4,
description: "baseball desc",
},
]);
The statement "spread creates a copy of the array" is incorrect. The spread operator by itself does not create a copy. It takes the items in an iterable (like an array) and and places them in the context as separate arguments/elements.
[...productList]
The above would create a copy because we define a new array and place all the items of productList in this new array. Thus creating a copy.
[...productList, newItem]
Defines a new array, places all the items of productList in this new array and places newItem behind it. Since we added newItem it strictly speaking is not really a copy.
I have an object called featureSet.
Inside of featureSet there are many items, including an array called features, which contains other array, attributes.
I can add a new array inside of featureSet.features.attributes by doing the following within a for loop
featureSet.features[i].attributes.NEWITEM= [NEWITEM_ARRAY];
And when I use console.log(featureSet), I can see that the items are there.
When I use var test = JSON.stringify(featureSet), however, only the original featureSet is returned.
How can I circumvent this so that when I call JSON.stringify, the new items are there as well?
Thank you in advance.
I guess what you do is close to:
let arr = []
console.log(arr) // []
arr.push(1)
console.log(arr) // [1]
arr.abc = 2
console.log(arr.abc) // 2
arr.push(3)
console.log(arr) // [1, 3]
console.log(JSON.stringify(arr)) // '[1, 3]'
console.log(arr.abc) // 2
JSON.stringify loops through array props with the help of Symbol.iterator. Your properties do not have positive integer indexes, that's why they are ignored. There is an example on MDN as well: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/JSON/stringify
One more example to consider (continues the one above):
arr[7] = 7
console.log(arr) // [1, 3, undefined, undefined, undefined, undefined, undefined, 7]
console.log(JSON.stringify(arr)) // "[1,3,null,null,null,null,null,7]"
How can I circumvent this so that when I call JSON.stringify, the new items are there as well?
You are trying to append attributes to an array, which will not work.
You can only append attributes to an object.
/* the original feature set */
const featureSet = {
features: [
{feature: 'color',
attributes: {}}, // this is an object, not an array
{feature: 'shape',
attributes: []} // this is an array as per original scenario
]
}
/* adding attributes to an object succeeds */
featureSet.features[0].attributes.NEWITEM = ['this', 'was', 'appended'];
/* adding attributes to an array quietly fails */
featureSet.features[1].attributes.NEWITEM = ['this', 'was', 'not', 'appended'];
const featureSetAsJSONString = JSON.stringify(featureSet)
console.log(featureSetAsJSONString) // notice that feature[0] is as expected, but [1] isn't
Hope this helps.
Cheers,
Try this
featureSet.features[i].attributes.push({NEWITEM: [NEWITEM_ARRAY]})
I have some code here and I was wondering if it is the same thing or different. I am pretty sure these are both suppose to be the same but I wasnt sure if I was doing it right.
let zoneComment = updatedMap[action.comment.zone]
? [...updatedMap[action.comment.zone]] : [];
let zoneComment = updatedMap[action.comment.zone]
? Object.assign([], updatedMap[action.comment.zone]) : [];
If these are the same then which should I use or does it matter? I want to use best practice so if it is your OPINION of which is better then please state so.
In your particular case they are not the same.
The reason is that you have an array, not an object.
Doing ... on an array will spread out all the elements in the array (but not the properties)
Doing Object.assign expects an object so it will treat an array as an object and copy all enumerable own properties into it, not just the elements:
const a = [1, 2, 3];
a.test = 'example';
const one = [...a] // [1, 2, 3];
const two = Object.assign([], a); // { '0': 1, '1': 2, '2': 3, 'test': 'example' }
console.log('\none');
for (let prop in one) {
console.log(prop);
}
console.log('\ntwo');
for (let prop in two) {
console.log(prop);
}
However, if you compare the ... operator applied on an object with Object.assign, they are essentially the same:
// same result
const a = { name: 'test' }
console.log({ ...a })
console.log(Object.assign({}, a))
except ... always creates a new object but Object.assign also allows you to mutate an existing object.
// same result
const a = { name: 'test' }
const b = { ...a, name: 'change' };
console.log(a.name); // test
Object.assign(a, { name: 'change'})
console.log(a.name); // change
Keep in mind that Object.assign is already a part of the language whereas object spread is still only a proposal and would require a preprocessing step (transpilation) with a tool like babel.
To make it short, always use ... spread construction and never Object.assign on arrays.
Object.assign is intended for objects. Although arrays are objects, too, it will cause a certain effect on them which is useful virtually never.
Object.assign(obj1, obj2) gets values from all enumerable keys from obj2 and assigns them to obj1. Arrays are objects, and array indexes are object keys, in fact.
[...[1, 2, 3], ...[4, 5]] results in [1, 2, 3, 4, 5] array.
Object.assign([1, 2, 3], [4, 5]) results in [4, 5, 3] array, because values on 0 and 1 indexes in first array are overwritten with values from second array.
In the case when first array is empty, Object.assign([], arr) and [...arr] results are similar. However, the proper ES5 alternative to [...arr] is [].concat(arr) and not Object.assign([], arr).
Your question really bubbles down to:
Are [...arr] and Object.assign([], arr) providing the same result when arr is an array?
The answer is: usually, yes, but:
if arr is a sparse array that has no value for its last slot, then the length property of the result will not be the same in both cases: the spread syntax will maintain the same value for the length property, but Object.assign will produce an array with a length that corresponds to the index of the last used slot, plus one.
if arr is a sparse array (like what you get with Array(10)) then the spread syntax will create an array with undefined values at those indexes, so it will not be a sparse array. Object.assign on the other hand, will really keep those slots empty (non-existing).
if arr has custom enumerable properties, they will be copied by Object.assign, but not by the spread syntax.
Here is a demo of the first two of those differences:
var arr = ["abc"]
arr[2] = "def"; // leave slot 1 empty
arr.length = 4; // empty slot at index 3
var a = [...arr];
var b = Object.assign([], arr);
console.log(a.length, a instanceof Array); // 4, true
console.log(b.length, b instanceof Array); // 3, true
console.log('1' in arr); // false
console.log('1' in a); // true (undefined)
console.log('1' in b); // false
If however arr is a standard array (with no extra properties) and has all its slots filled, then both ways produce the same result:
Both return an array. [...arr] does this by definition, and Object.assign does this because its first argument is an array, and it is that object that it will return: mutated, but it's proto will not change. Although length is not an enumerable property, and Object.assign will not copy it, the behaviour of the first-argument array is that it will adapt its length attribute as the other properties are assigned to it.
Both take shallow copies.
Conclusion
If your array has custom properties you want to have copied, and it has no empty slots at the end: use Object.assign.
If your array has no custom properties (or you don't care about them) and does not have empty slots: use the spread syntax.
If your array has custom properties you want to have copied, and empty slots you want to maintain: neither method will do both of this. But with Object.assign it is easier to accomplish:
a = Object.assign([], arr, { length: arr.length });
This question already has answers here:
Undefined values in Array(len) initializer
(5 answers)
Closed 7 years ago.
I am confused by the results of mapping over an array created with new:
function returnsFourteen() {
return 14;
}
var a = new Array(4);
> [undefined x 4] in Chrome, [, , , ,] in Firefox
a.map(returnsFourteen);
> [undefined x 4] in Chrome, [, , , ,] in Firefox
var b = [undefined, undefined, undefined, undefined];
> [undefined, undefined, undefined, undefined]
b.map(returnsFourteen);
> [14, 14, 14, 14]
I expected a.map(returnsFourteen) to return [14, 14, 14, 14] (the same as b.map(returnsFourteen), because according to the MDN page on arrays:
If the only argument passed to the Array constructor is an integer
between 0 and 2**32-1 (inclusive), a new JavaScript array is created
with that number of elements.
I interpret that to mean that a should have 4 elements.
What am I missing here?
When you create an array like so:
var arr1 = new Array( 4 );
you get an array that has a length of 4, but that has no elements. That's why map doesn't tranform the array - the array has no elements to be transformed.
On the other hand, if you do:
var arr2 = [ undefined, undefined, undefined, undefined ];
you get and array that also has a length of 4, but that does have 4 elements.
Notice the difference between having no elements, and having elements which values are undefined. Unfortunately, the property accessor expression will evaluate to the undefined value in both cases, so:
arr1[0] // undefined
arr2[0] // undefined
However, there is a way to differentiate these two arrays:
'0' in arr1 // false
'0' in arr2 // true
var a = new Array(4);
This defines a new array object with an explicit length of 4, but no elements.
var b = [undefined, undefined, undefined, undefined];
This defines a new array object with an implicit length of 4, with 4 elements, each with the value undefined.
From the docs:
callback is invoked only for indexes of the array which have assigned
values; it is not invoked for indexes which have been deleted or which
have never been assigned values.
For array a, there are no elements that have been assigned values, so it does nothing.
For array b, there are four elements that have been assigned values (yes, undefined is a value), so it maps all four elements to the number 14.
new Array(len) creates an empty array, and does something different than filling it with undefined values: It sets its length to len. So, it translates to this code:
var newArr = [];
newArr.length = len;
Let's have some fun with newArr (assuming that len = 4):
newArr.length; //4
newArr[1] === undefined; //true
newArr.hasOwnProperty(1); //false
This is because while the is 4 items long, it does not contain any of these 4 items. Imagine an empty bullet-clip: It has space for, say, 20 bullets, but it doesn't contain any of them. They weren't even set to the value undefined, they just are...undefined (which is a bit confusing.)
Now, Array.prototype.map happily walks along your first array, chirping and whistling, and every time it sees an array item, it calls a function on it. But, as it walks along the empty bullet-clip, it sees no bullets. Sure, there are room for bullets, but that doesn't make them exist. In here, there is no value, because the key which maps to that value does not exist.
For the second array, which is filled with undefined values, the value is undefined, and so is the key. There is something inside b[1] or b[3], but that something isn't defined; but Array.prototype.map doesn't care, it'll operate on any value, as long as it has a key.
For further inspection in the spec:
new Array(len) : http://es5.github.com/#x15.4.2.2
Array.prototype.map : http://es5.github.com/#x15.4.4.19 (pay close attention to step 8.b)
One additional answer on the behavior of console.log. Plain simple, the output is sugar and technically wrong.
Lets consider this example:
var foo = new Array(4),
bar = [undefined, undefined, undefined, undefined];
console.log( Object.getOwnPropertyNames(bar) );
console.log( Object.getOwnPropertyNames(foo) );
As we can see in the result, the .getOwnPropertyNames function returns
["length"]
for the foo Array/Object, but
["length", "0", "1", "2", "3"]
for the bar Array/Object.
So, the console.log just fools you on outputting Arrays which just have a defined .length but no real property assignments.
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?
Understanding this "hack" requires understanding several things:
Why we don't just do Array(5).map(...)
How Function.prototype.apply handles arguments
How Array handles multiple arguments
How the Number function handles arguments
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:
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]
Disclaimer: This is a very formal description of the above code - this is how I know how to explain it. For a simpler answer - check Zirak's great answer above. This is a more in depth specification in your face and less "aha".
Several things are happening here. Let's break it up a bit.
var arr = Array.apply(null, { length: 5 }); // Create an array of 5 `undefined` values
arr.map(Number.call, Number); // Calculate and return a number based on the index passed
In the first line, the array constructor is called as a function with Function.prototype.apply.
The this value is null which does not matter for the Array constructor (this is the same this as in the context according to 15.3.4.3.2.a.
Then new Array is called being passed an object with a length property - that causes that object to be an array like for all it matters to .apply because of the following clause in .apply:
Let len be the result of calling the [[Get]] internal method of argArray with argument "length".
As such, .apply is passing arguments from 0 to .length , since calling [[Get]] on { length: 5 } with the values 0 to 4 yields undefined the array constructor is called with five arguments whose value is undefined (getting an undeclared property of an object).
The array constructor is called with 0, 2 or more arguments.
The length property of the newly constructed array is set to the number of arguments according to the specification and the values to the same values.
Thus var arr = Array.apply(null, { length: 5 }); creates a list of five undefined values.
Note: Notice the difference here between Array.apply(0,{length: 5}) and Array(5), the first creating five times the primitive value type undefined and the latter creating an empty array of length 5. Specifically, because of .map's behavior (8.b) and specifically [[HasProperty] .
So the code above in a compliant specification is the same as:
var arr = [undefined, undefined, undefined, undefined, undefined];
arr.map(Number.call, Number); // Calculate and return a number based on the index passed
Now off to the second part.
Array.prototype.map calls the callback function (in this case Number.call) on each element of the array and uses the specified this value (in this case setting the this value to `Number).
The second parameter of the callback in map (in this case Number.call) is the index, and the first is the this value.
This means that Number is called with this as undefined (the array value) and the index as the parameter. So it's basically the same as mapping each undefined to its array index (since calling Number performs type conversion, in this case from number to number not changing the index).
Thus, the code above takes the five undefined values and maps each to its index in the array.
Which is why we get the result to our code.
As you said, the first part:
var arr = Array.apply(null, { length: 5 });
creates an array of 5 undefined values.
The second part is calling the map function of the array which takes 2 arguments and returns a new array of the same size.
The first argument which map takes is actually a function to apply on each element in the array, it is expected to be a function which takes 3 arguments and returns a value.
For example:
function foo(a,b,c){
...
return ...
}
if we pass the function foo as the first argument it will be called for each element with
a as the value of the current iterated element
b as the index of the current iterated element
c as the whole original array
The second argument which map takes is being passed to the function which you pass as the first argument. But it would not be a, b, nor c in case of foo, it would be this.
Two examples:
function bar(a,b,c){
return this
}
var arr2 = [3,4,5]
var newArr2 = arr2.map(bar, 9);
//newArr2 is equal to [9,9,9]
function baz(a,b,c){
return b
}
var newArr3 = arr2.map(baz,9);
//newArr3 is equal to [0,1,2]
and another one just to make it clearer:
function qux(a,b,c){
return a
}
var newArr4 = arr2.map(qux,9);
//newArr4 is equal to [3,4,5]
So what about Number.call ?
Number.call is a function that takes 2 arguments, and tries to parse the second argument to a number (I'm not sure what it does with the first argument).
Since the second argument that map is passing is the index, the value that will be placed in the new array at that index is equal to the index. Just like the function baz in the example above. Number.call will try to parse the index - it will naturally return the same value.
The second argument you passed to the map function in your code doesn't actually have an effect on the result. Correct me if I'm wrong, please.
An array is simply an object comprising the 'length' field and some methods (e.g. push). So arr in var arr = { length: 5} is basically the same as an array where the fields 0..4 have the default value which is undefined (i.e. arr[0] === undefined yields true).
As for the second part, map, as the name implies, maps from one array to a new one. It does so by traversing through the original array and invoking the mapping-function on each item.
All that's left is to convince you that the result of mapping-function is the index. The trick is to use the method named 'call'(*) which invokes a function with the small exception that the first param is set to be the 'this' context, and the second becomes the first param (and so on). Coincidentally, when the mapping-function is invoked, the second param is the index.
Last but not least, the method which is invoked is the Number "Class", and as we know in JS, a "Class" is simply a function, and this one (Number) expects the first param to be the value.
(*) found in Function's prototype (and Number is a function).
MASHAL