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why does `Map<K, V>.get(key)` work for `key ...
# getting-started
y
why does 
Map<K, V>.get(key)
work for 
key == null
? how does it even typecheck?
e
a
are you asking about JVM interop? https://stackoverflow.com/a/74707052/4161471
e
Copy code
operator fun <K, V> Map<out K, V>.get(key: K): V?
with 
K = T?
, 
Map<T, V>
is a 
Map<out K, V>
Adam, that's not what was asked
y
@ephemient so... since 
K
is covariant to 
K?
?
j
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?
y
oh, so it's because 
Map<K, V>
is covariant to 
Map<K?, V>
?
(probably not so apologies in advance heh)
e
Map
is not naturally covariant in 
K
but the 
get
extension is explicitly on 
out K
j
@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.
y
I believe I understand now. thank you for the explanation! as for "to" vs "in" - I was going by the wikipedia article
e
K
is a subtype of `K?`: every inhabitant of 
K
is also an inhabitant of 
K?
Map<K, V>
is a subtype of 
Map<out K?, V>
because 
K
satisfies 
out K?
because 
K
is a subtype of 
K?
❤️ 1
s
hmm, i would have assumed, it's because K isn't restricted explicitly, it's implicitly restricted to 
K: Any?
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