Why are `any` and `all` defined on `Array<out T...

# announcements

[Ruckus]

Why are any and all defined on Array<out T> instead of Array<T>?

[karelpeeters]

I've been trying for about 10 minutes now but I can't find any combination where it makes any difference.

[Ruckus]

What?

[karelpeeters]

I've tried to define my own version of any that isn't defined on Array<out T> but on Array<T>, but couldn't find a case where that didn't work exactly as the real one.

[Ruckus]

I expect it to work the same. My issue is that I have an Array<T> and can't use any because it's defined on Array<out T>

[ilya.gorbunov]

Array<T> is a subtype of Array<out T>, so any extensions for the latter are available for the former.

[Ruckus]

Hmm, that makes sense. I must have been doing something else wrong. I'll have to check when I get back to a computer.

[karelpeeters]

@ilya-gh Why is it defined on out then if its the same either way?

[Ruckus]

@karelpeeters Wrong Ilya. You meant @ilya.gorbunov.

[karelpeeters]

Dammit not again. Thanks!

[ilya.gorbunov]

We use out variance to emphasize that an out-projected array is enough for the particular operation.

[karelpeeters]

And are there cases where it actually does something? Ie. does it allow more programs to compile?

[ilya.gorbunov]

It can make a difference in cases when there are more than one T position in a function signature.