how could I make this a singleton? it seems like `object` can’t do generics? ```class Mapper<T : ...

# getting-started

[poohbar]

how could I make this a singleton? it seems like object can’t do generics?

class Mapper<T : Account> : RecordMapper<Record, T> {
    override fun map(record: Record?): T? {
        TODO("Not yet implemented")
    }
}

Even better, I really just need a function, where the input is Record and the output is T: Account how do I declare that?

[poohbar]

Ideally I need something like this:

private val MAPPER: (Record) -> Account = { record: Record ->
    TODO("Not yet implemented")
}

But the output should be T: Account, not Account

[Dominaezzz]

Check out the implementation of emptyList<T>() .

[poohbar]

I think I got it, I can just do this:

fun <T : Account> mapToAccount(record: Record): T {
    TODO("Not yet implemented")
}

[poohbar]

and then pass it as ::mapToAccount

[poohbar]

well now i dont understand this

fun <T : Account> mapToAccount(record: Record): T {
    val x: Account = Account(...)

    return x // error here
}

[poohbar]

why doesn’t Account fit into T? 😮

[Dominaezzz]

I think you need in or out .

[poohbar]

maybe i am putting it in the wrong place but it says:

Variance annotations are only allowed for type parameters of classes and interfaces

[poohbar]

trivial example:

open class Person

fun <T: Person> foo(): T {
    return Person()
}

[diesieben07]

Function references cannot be generic in this way. As soon as you do ::mapToAccount the T gets bound to a concrete value.
For example:
val foo: (Record) -> SomeAccountSubclass = ::mapToAccount binds T as SomeAccountSubclass.
(Record) -> T is not a valid type.

[diesieben07]

So when you want to store it in a reference, you have to choose a concrete type for T.

[diesieben07]

So really you do not want a singleton, because a singleton by definition means that T is always the same and as such redundant...

[ephemient]

well, if you have your variance right, it would be fine to write a singleton

object Mapper : RecordMapper<Record, Nothing>

but there isn't too much you could do with that

[diesieben07]

Yes, it could never return anything

[jbnizet]

@poohbar your function’s return type is T. T can be any class that extends Person: the caller of the function decides which class it is. So it could be a Child, an Adult, a Doctor... anything that extends Person. For example: val doctor: Doctor = foo<Doctor>().` So returning an instance of Person is incorrect: a Doctor is a Person, but a Person is not a Doctor.

[ephemient]

hence why Nothing works: it is a subtype of every other type. of course, there are no actual instances of Nothing, so returning is not possible (only abnormal termination or non-termination), but it's the only type that works

[poohbar]

maybe that’s why the emptyList() is implemented as:

internal object EmptyList : List<Nothing>

[poohbar]

strange

[ephemient]

not strange at all, List<Nothing> is a List<T> because Nothing is a T and List<out T> is covariant

[ephemient]

it may help you to think: you can get(): T, for any type T, out of a `List<Nothing>`… with type safety. of course, at runtime, you can't "get" anything from an empty list, but that doesn't matter to the types