Why are

any

and

all

defined on

Array<out T>

instead of

Array<T>

?

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

What?

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.

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>

Array<T>

is a subtype of

Array<out T>

, so any extensions for the latter are available for the former.

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

@ilya-gh Why is it defined on

out

then if its the same either way?

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

Dammit not again. Thanks!

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

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

It can make a difference in cases when there are more than one

T

position in a function signature.