Is it possible to omit a type parameter in Kotlin?

Question

I'm translating a very old and big Java server to Kotlin, and I'm trying to make as few logic changes as possible during the conversion. I'm looking for a way to translate this snippet of Java code without fundamentally changing it.

Consider the following toy code:

interface Pet

class Dog : Pet
class Cat : Pet

class PetSitter<T : Pet> {
  fun walk(pet: T) {}
}

fun main() {
  val sitters: Array<PetSitter<*>> =
      arrayOf(PetSitter<Cat>(), PetSitter<Dog>())

    sitters[0].walk(Cat()) // Type mismatch.
                           // Required: Nothing
                           // Found: Cat
}

This fails to compile with the message Type mismatch: inferred type is Cat but Nothing was expected.

In Java the following works fine:

PetSitter[] sitters = ... // Note: PetSitter's generic is omitted
sitters[0].walk(new Cat());

Is it possible to similarly define a Array<PetSitter> (with no pet generic) in Kotlin? Or some other way to make the parameter type for PetSitter<*>.walk to be Pet instead of Nothing?

Answer 1

In Java, this is called a raw type. It is only supported by necessity for backward compatibility, and it is discouraged to ever use, because it defeats the purpose of using generics. For the same reason, Kotlin forbids it outright because it doesn't have a pre-generics version of the language it must support.

The equivalent of using a raw type is to cast the types to be permissive. This requirement is to force you to consider if the cast is appropriate.

(sitters[0] as PetSitter<Cat>).walk(Cat())