why does

Map<K, V>.get(key)

work for

key == null

? how does it even typecheck?

operator fun <K, V> Map<out K, V>.get(key: K): V?

with

K = T?

,

Map<T, V>

is a

Map<out K, V>

@ephemient so... since

K

is covariant to

K?

?

The variance is not a property of

K

, but rather of the type using

K

. But what you can say is that

K

is a subtype of

K?

oh, so it's because

Map<K, V>

is covariant to

Map<K?, V>

?

Map

is not naturally covariant in

K

but the

get

extension is explicitly on

out K

@y I think you're using "covariant" wrong. A generic type can be covariant in one of its type parameters, but a type being "covariant to" another type is not a thing. The declaration of the

get

operator allows you to use

Map<K,V>.get<K?, V>(nullableK)

because

Map<out K, V>

is covariant in its first type parameter

K

, which in turn means that

Map<K,V>

is a subtype of

Map<K?,V>

in this context.

I believe I understand now. thank you for the explanation! as for "to" vs "in" - I was going by the wikipedia article

K

is a subtype of `K?`: every inhabitant of

K

is also an inhabitant of

K?

hmm, i would have assumed, it's because K isn't restricted explicitly, it's implicitly restricted to

K: Any?