How to get around Type mismatch Required: Foo<Type>, Found: Foo<Type?> - kotlin

Question

Given the following Kotlin code:
class Foo<T>(val t : T?)
fun <T : Any, R : Any> Foo<T?>.transform(transformer : (T) -> R) : Foo<R?> {
return when (t) {
null -> Foo(null)
else -> Foo(transformer(t))
}
}
fun main(args : Array<String>) {
val foo = Foo(args.firstOrNull())
val bar = foo.transform<String, Int> { t -> t.length }
val baz = bar.transform<Int, IntRange> { t -> t..(t + 1) }
}
Why do I get the following error:
Type mismatch. Required: Foo<String?> Found: Foo<String>
If I remove the ? from the extension function to be Foo<T>.transform I instead get the following error:
Type mismatch. Required: Foo<Int> Found: Foo<Int?>
I can understand the second error, because you cannot assign Int? to Int, but the first doesn't make any sense, as you can assign String to String?
EDIT:
I have modified the class Foo<T> to be class Foo<out T> and this works for me as the value t will only ever be read after the initial assignment. With this option I do not need to define the type parameters at the call site of transform.
Another option I have found that I think is a bit messy (and not sure why it makes a difference) is adding a third type parameter to the extension function as follows:
fun <T : Any, U : T?, R : Any> Foo<U>.transform(transformer : (T) -> R) : Foo<R?>
The call site of this on the other hand I find a bit odd. Looking at the above code the call of foo.transform MUST NOT include the type parameters, but the call of bar.transform<Int, Int?, IntRange> MUST include the type parameters in order to work.
This option allows setting the value t at some later point if it were a var instead of val. But it also removes the smart casting on t in the transform function. Although that can be gotten around with a !! if you are not worried about race conditions or (with some additional effort) ?: or ?. if you are worried about race conditions.

Answer 1

You can change your Foo<T> class to be not invariant (see https://kotlinlang.org/docs/reference/generics.html):
class Foo<out T>(val t : T?)
fun <T : Any, R : Any> Foo<T?>.transform(transformer : (T) -> R) : Foo<R?> {
return when (t) {
null -> Foo(null)
else -> Foo(transformer(t))
}
}
fun main(args : Array<String>) {
val foo = Foo(args.firstOrNull())
val bar = foo.transform<String, Int> { t -> t.length }
val baz = bar.transform<Int, IntRange> { t -> t..(t + 1) }
}
The out T specifies precisely the behavior you want.

Answer 2

Since you specify the property t in the constructor as T? you don't need to specify Foo<T?> as receiver and Foo<R?> as return type. Instead use Foo<T> and Foo<R> and it will work.
class Foo<T>(val t : T?)
fun <T: Any, R: Any> Foo<T>.transform(transformer : (T) -> R) : Foo<R> {
return when (t) {
null -> Foo(null)
else -> Foo(transformer(t))
}
}
fun main(args : Array<String>) {
val foo = Foo(args.firstOrNull())
val bar = foo.transform { t -> t.length }
val baz = bar.transform { t -> t..(t + 1) }
}
Note: You don't need to specify the generic types for transform because they can be inferred (at least in this example).