JavaScript "new Array(n)" and "Array.prototype.map" weirdness

I had a task that I only knew the length of the array and needed to transform the items. I wanted to do something like this:

let arr = new Array(10).map((val,idx) => idx);

To quickly create an array like this:

[0,1,2,3,4,5,6,7,8,9]

But it didn't work because: (see Jonathan Lonowski's answer a few answers above)

The solution could be to fill up the array items with any value (even with undefined) using Array.prototype.fill()

let arr = new Array(10).fill(undefined).map((val,idx) => idx);

console.log(new Array(10).fill(undefined).map((val, idx) => idx));

Update

Another solution could be:

let arr = Array.apply(null, Array(10)).map((val, idx) => idx);

console.log(Array.apply(null, Array(10)).map((val, idx) => idx));

It appears that the first example

x = new Array(3);

Creates an array with a length of 3 but without any elements, so the indices [0], [1] and [2] is not created.

And the second creates an array with the 3 undefined objects, in this case the indices/properties them self are created but the objects they refer to are undefined.

y = [undefined, undefined, undefined]
// The following is not equivalent to the above, it's the same as new Array(3)
y = [,,,];

As map runs on the list of indices/properties, not on the set length, so if no indices/properties is created, it will not run.

With ES6, you can do [...Array(10)].map((a, b) => a) , quick and easy!

ES6 solution:

[...Array(10)]

Doesn't work on typescript (2.3), though