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Why does the first console.log() print undefined values but second one has transformed values? I know it has to do something with function scope but not getting it
var array = [1,2,3,4,5,7];
function incrementByOne(arr) {
arr = arr.map(function(value, index, array){
arr[index] = arr[index] +1;
});
console.log(arr);
}
incrementByOne(array);
console.log(array);
// [undefined, undefined, undefined, undefined, undefined, undefined]
// [2, 3, 4, 5, 6, 8]
also i notice that the first console.log() knows how many times to iterate but what happens to the value...
js bin link
you need to return the incremented value from the function inside map. use return arr[index] +1
Also you need to return the new array formed using map and stored in arr now.
var array = [1,2,3,4,5,7];
function incrementByOne(arr) { //contains array reference
arr = arr.map(function(value, index, array){
return value +1;
});
//now arr contains a new array and doesn't refer to passed array anymore.
console.log(arr);
return arr;
}
array = incrementByOne(array);
console.log(array);
// [undefined, undefined, undefined, undefined, undefined, undefined]
// [2, 3, 4, 5, 6, 8]
if you don't want to return, you can use forEach(), as in that case arr will refer to passed array throughout. The difference is because map returns a new array.
var array = [1,2,3,4,5,7];
function incrementByOne(arr) { //contains array reference
arr.forEach(function(value, index){
arr[index] = value +1;
});
//arr still refers to the passed array.
console.log(arr);
}
incrementByOne(array);
console.log(array);
// [undefined, undefined, undefined, undefined, undefined, undefined]
// [2, 3, 4, 5, 6, 8]
Well, all the above answers are correct but they miss the most important point here. There is a concept in JavaScript called call by sharing.
Consider this code:
var num= 3;
var json = {myValue : '10'};
var json2 = {myValue : '100'};
function callBySharing(a,b,c){
a = a + 37;
b = {myValue : 'new value'};
c.myValue = 'new Value';
}
callBySharing(num,json,json2);
console.log(num);//3 *UNCHANGED*
console.log(json.myValue);//10 *UNCHANGED*
console.log(json2.myValue);//'new Value' *CHANGED*
So what you are doing is same as what is happening in json.myvalue; You are trying to update the whole object and replace it with the new value. So a very simple change in the code with do this for you:
var array = [1,2,3,4,5,7];
function incrementByOne(arr) {
arr.map(function(value, index, array){
arr[index] = arr[index] +1;
});
console.log(arr);
}
incrementByOne(array);
console.log(array);
I just replaced the arr= arr.map().... part to just arr.map().....
What this does is, it changes the function to json2.myValue example case.
So what is the difference between the 2: JS lets you update items within the object but not the whole object.By making the above said change in code, you are updating individual values of arr and not replacing the whole object with new values. I learnt this concept from SO only back when I was confused with it. So I am linking the post(Is JavaScript a pass-by-reference or pass-by-value language?)
Hope this helps!
var array = [1,2,3,4,5,7];
function incrementByOne(arr) {
arr = arr.map(v => v + 1);
console.log(arr);
}
incrementByOne(array);
console.log(array);
// [undefined, undefined, undefined, undefined, undefined, undefined]
// [2, 3, 4, 5, 6, 8]
A couple of things going on here:
The callback you pass into the .map function should return a value with the return keyword. If you don't return a value from the callback, the default return value is undefined. That's why the 'arr' array you define in incrementByOne only has undefined values inside of it when you log it with your first console.log. I think what you're really trying to do is simply return value + 1 inside of the callback.
You're making a classic passed-by-value passed-by-reference error here; it's a mistake everyone makes when they're first learning JS. I'd recommend checking out posts like this one. In short, inside of your .map callback you are mutating the array that you passed into incrementByOne, which is the same array you log out with the second console.log; that's why the values appear to have been incremented correctly.
I've been trying to understand this function for quite a while, but it just doesn't make sense to me. The goal of the function is to remove any numbers within the array of arguments that match the other argument numbers.
Why is it necessary to slice the array for the function to work?
Is args.splice(0,1) redundant? I removed it and nothing changed.
It seems like the filter function does the bulk of the work, but I don't see how it actually filters for the numbers...
function destroyer(arr) {
var args = Array.prototype.slice.call(arguments);
args.splice(0, 1);
return arr.filter(function(element) {
return args.indexOf(element) === -1;
});
}
destroyer([1, 2, 3, 1, 2, 3], 2, 3);
Let's go through it line by line:
var args = Array.prototype.slice.call(arguments);
JavaScript's arguments variable is similar to an array but it's not an array. You can try this yourself: arguments instanceof Array will give false. So applying the slice method from the Array prototype will simply convert arguments to a real array.
args.splice(0, 1);
This is to remove the first argument, which is arr in your case.
return arr.filter(function(element) {
return args.indexOf(element) === -1;
});
This will go through all the numbers in arr and will check each one of them if it exists in the arguments. When indexOf() returns -1 it means the element was not found in the array.
Slice does not alter. It returns a shallow copy of elements from the original array. Elements of the original array are copied into the returned array.
Take this example
var object = {
'0': 'zero',
'1': 'one',
'2': 'two',
'3': 'three',
'4': 'four',
length: 5
};
var sliced = Array.prototype.slice.call( object, 3 );
['three','four']; //output
passe Arguments [Array[6], 2, 3]
Arguments after splicing or remove first element of argument [2, 3]
so closure function filters element which is present in first element of array with other two element. returning just [1, 1]
To understand what's going on, we need to understand the Function.prototype.call method.
It invokes the Array.prototype.slice method on the first argument you pass to it, which in this case is the magical JS arguments object, and then passes in whatever arguments follow.
Thus Array.prototype.splice is unnecessary, and you can just write:
function destroyer(arr) {
var rest = Array.prototype.slice.call(arguments, 1);
return arr.filter(function(element) {
return rest.indexOf(element) === -1;
});
}
in fact, this has been implemented in ES2015+ with the spread operator, so you could write:
function destroyer(arr, ...rest) {
return arr.filter(function(element) {
return rest.indexOf(element) === -1;
});
}
I've added comments, please see if it helps you to understand the function.
function destroyer(arr) {
// arr just holds [1, 2, 3, 1, 2, 3]
var args = Array.prototype.slice.call(arguments);
// args contains nested array with all input params [[1, 2, 3, 1, 2, 3], 2, 3]
args.splice(0, 1);
//args is spliced and we have [2,3] in args
//Filter arr=[1, 2, 3, 1, 2, 3] elements, condition it must not be in args i.e [2,3]
return arr.filter(function(element) {
return args.indexOf(element) === -1;
});
}
destroyer([1, 2, 3, 1, 2, 3], 2, 3);
Please refer to the below documentation to read about arguments object used in this function:
https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Functions/arguments
In arr we will have [1,2,3,1,2,3] and
in args we will have [[1,2,3,1,2,3],2,3]
The filter function loop over arr
and args.indexOf(element) will return -1 if element is not in args.
So, for the first time in loop the element value is 1 and inside loop
args.indexOf(1) returns -1 because 1 is not present in args because at 0 index we have array and at 1st index we have 2 and at 2nd index we have 3. So the condition === -1 is true and returns 1 to the array that is going to be printed to the console.
for next element, i.e., 2 in arr, the statement args.indexOf(2) returns the first index at which 2 is present i.e., 1 in args array. for likewise entire loop will be executed for arr
I'm working through Eloquent Javascript and I'm having trouble understanding something. Perhaps I've missed something along the way. This is the solution given for chapter 5 (higher-order functions), exercise 1, which takes the elements in the different arrays and puts them all in a single array:
var arrays = [[1, 2, 3], [4, 5], [6]];
console.log(arrays.reduce(function(flat, current) {
return flat.concat(current);
}, []));
My problem is: I have absolutely no clue why the arguments "flat" and "current" work in this situation. The entire chapter reads through assuming the reader understands what's going on here but I have absolutely no idea why this works. It doesn't appear that "flat" and "current" are defined anywhere. Another short example is this one where the author explains how the reduce method works (problem area in bold):
function reduce(array, combine, start) {
var current = start;
for (var i = 0; i < array.length; i++)
current = combine(current, array[i]);
return current;
}
**console.log(reduce([1, 2, 3, 4], function(a, b) {
return a + b;
}, 0));**
Where in the world did "a" and "b" come from and why does this piece of code work? Any help would be much appreciated, thank you.
flat and current don't need to be declared anywhere, they are parameters to the anonymous function that is passed to Array.reduce.
One way to illustrate this is to modify your second example to use Array.reduce directly using an anonymous function with parameters a and b. Look at this:
[1, 2, 3, 4].reduce(function(a, b) {
console.log("a: " + a + " b: " + b);
return a + b;
});
The console will now show:
a: 1 b: 2
a: 3 b: 3
a: 6 b: 4
10
What's happening is that the anonymous function(a, b) {...} is called with (1, 2), which returns 3, which is passed in again (3, 3) which returns 6, which is passed in as the first argument (6, 4), which returns the final answer 10.
Another illustration is to use a second argument to Array.reduce, say 10, to see what's going on. That 10 is used as the initialValue. So:
[1, 2, 3, 4].reduce(function(a, b) {
console.log("a: " + a + " b: " + b);
return a + b;
}, 10);
The trace is:
a: 10 b: 1
11 b: 2
13 b: 3
16 b: 4
20
You can work out how that happened.
Yes, reduce can be a little confusing in the beginning. It is a native function that takes two parameters. One is a callback function and the other one is any value.
The idea is that in the callback function you can use the values of the array one at a time to process a result. To do that it iterates over the array values and passes them to the callback function you defined one at a time and for every loop it takes the value of the last loop and passes it as well.
Let's say you want to sum all numbers in an array:
//your array
var numbers = [4,7,3];
//your initial value
var initialValue = 0;
//your function
function sum(iteratingValue, arrayValue) {
return iteratingValue + arrayValue;
}
var result = numbers.reduce(sum, initialValue);
Now, you can name your callback function parameters whatever you like, a and b, start and finish, fish and duck. It won't matter, reduce will pass the values in the same order.
Here is the definition by MDN:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/Reduce
reduce executes the callback function once for each element present in the array, excluding holes in the array, receiving four arguments: the initial value (or value from the previous callback call), the value of the current element, the current index, and the array over which iteration is occurring.
First, let us touch the core of the question. Let's suppose you have a function, like this:
function func1(myfunc, arr) {
//Do something, which will result in having variables called param1 and param2
myfunc(param1, param2);
}
Let's see what happens. func1 takes myfunc and arr as parameters. myfunc will be passed when func1 is called. arr is an array.
Now, let's suppose you call it this way:
func1(function(a, b) {
//do something with a and b
}, [1, 2, 3, 4]);
You are calling func1, by passing a function and an array. The array is obvious, so let's see the function. The function expects two parameters and will do something with them. You do not have to define a or b when you call func1, since it is func1's internal job to create and initialize them. So, func1 will do its internal things and call your function passed as a parameter. Now, let's see your example:
var arrays = [[1, 2, 3], [4, 5], [6]];
console.log(arrays.reduce(function(flat, current) {
return flat.concat(current);
}, []));
Here, you call arrays.reduce (which is very similar to func1 in the general description). You pass a function and an array. Again, the array is obvious, but the question is, how flat and current are defined. The answer is that it is arrays.reduce's internal job to create and initialize them. As about the reduce prototype function, you can read more about it here.
I found an implementation using using prototypes. This is a simplification to show the structure:
function Thingie(){
this.content = [];
}
Thingie.prototype = {
push: function(arg) {
this.content.push(arg);
},
pop: function() {
return this.content.pop();
}
};
var t = new Thingie();
forEach([10, 3, 4, 8, 2, 9, 7, 1, 2, 6, 5],
method(t, "push"));
What is "method" in the example on the last line? I've never seen this construct. I use t.push like everyone else.
I tried find how "method()" is defined online, but it is impossible to search for a function called "method" using any possible set of search terms. All you get are how functions and methods are defined and used. There also seems to be no information when I look at forEach documentation.
Does this make sense to anybody?
method(t, "push")
would be defined as:
function method(obj, name) {
return obj[name].bind(obj);
}
That forEach looks like UnderscoreJS's function _.each
_.each(list, iteratee, [context]) Alias: forEach
example:
_.each([1,2,3], function(item) { console.log(item); });
// console output:
// 1
// 2
// 3
That method probably looks like this (Beware: wild guess!) , giving you the function to use as iteratee paramater
function method(obj, name){
if(typeof(obj[name]) != "function")
throw new Error("Not a function");
return obj[name];
}
A function like that is lodash.bindKey, which does exactly what you want. Also, the forEach method could be lodash.forEach, or simply Array.prototype.forEach, which is built-in.
[1,2,3].forEach(_.bindKey(t, 'push'));
However, this works because Thinghie#push expects only one argument. If the same call would be made on an array, the result would not be as expected, since forEach methods take 3 arguments: value, index, array, and [].push can handle multiple arguments. So, the code
var array = [];
[1,2].forEach(_.bindKey(array, 'push'));
console.log(array); // outputs [1, 0, [1, 2], 2, 1, [1, 2]]
In this case (and in any case when we would like the function returned by method applied on only one argument), I guess the solution is to write
function method(obj, name) {
return function(arg) { return obj[name](arg); }
}
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