#SIDU 5/5

So how does this work during actual execution? How do Iterator methods "know" to do shortcuts, and how is this comunicated across method calls?

#SIDU 4/?

In the instant case, `neighbors()` only returns four elements. But what if it was 100K elements? Does the Iterator have to loop 100K times before `choose()` gets called?

Now, "obviously" that is not what is happening, because that would be monstrously inefficient. But in order for that to *not* happen, it seems to me that `choose()` needs to somehow "reach back" to the previous method and tell the Iterator, "Look, you don't need to loop over everything; just give me element N."

#SIDU 3/?

neighbours (origin, size).into_iter().choose (&mut self.rng)

The way my C-brain interprets this is that the current method can't do its work until the previous method has fully completed. This means that `choose()` can't pick something until the Iterator has looped over all array elements returned from `neighbors()`.

#SIDU 2/?

Consider the following fragment, which I hunted down after watching a YooToob video (I really wish Mastodon supported Markdown):

https://github.com/olafurw/poke-fighting-rust/blob/e74ec08821c6bb8097431fe3d2cec89673f7fd3a/src/battle.rs#L117

neighbours (origin, size).into_iter().choose (&mut self.rng)

`neighbors()` return an array, which is converted into an Iterator, which is then fed into the `choose()` method which, in this case, picks a random element from the array.