The output is sorted due to the fact that for each number, a timer is started that prints out the number after waiting a number of milliseconds equal to said number.

Therefore, 1 is printed first after delaying for 1 millisecond, 5 is printed second after 5 milliseconds etc.

[–] ptu@sopuli.xyz 4 points 16 hours ago (1 children)

So all items in the array are launched simultaneuously and ran in parallel instead of sequentially?

[–] scrion@lemmy.world 3 points 14 hours ago (1 children)

They are launched sequentially, but run simultaneously, yes - at least some of them. And they run concurrently but not in parallel - using a single execution context, there is only a single thread, so no parallelism exist.

[–] ptu@sopuli.xyz 1 points 14 hours ago

I see, I was only aware of sleep but that makes sense. Thanks for your insight.

[–] JargonWagon@lemmy.world 13 points 1 day ago

Perfectly explained, thank you!

[–] theunknownmuncher@lemmy.world 23 points 1 day ago* (last edited 1 day ago) (1 children)

The program goes through the collection of numbers and prints each one after a delay of milliseconds equal to that number: "Print the number 20 after a 20 millisecond delay. Print the number 5 after a 5 millisecond delay. Print the number 100 after a 100 millisecond delay... etc..." effectively sorting the collection because the numbers will be printed in order from smallest to largest.

This is a clever (but impractical) way to sort a collection, because it does not require comparing any of the elements of the collection.

[–] ulterno@programming.dev 0 points 9 hours ago

because it does not require comparing any of the elements of the collection

Well, if you are comparing x + a to y and x + b to y and then both to y', then y'' and so on, then are you really not comparing a to b?