I've got something like ```interface MyInterface<R> { fun doIt(): R } object MyImpl : MyI...

# getting-started

[y]

I've got something like

interface MyInterface<R> {
    fun doIt(): R
}

object MyImpl : MyInterface<String> {
    override fun doIt(): String = "hello"
}

object MyOtherImpl : MyInterface<Int> {
    override fun doIt(): Int = 5
}

which works fine.
now say I've got fun <R> foo(impl: MyInterface<R>): R = impl.doIt()
is it possible to add a default value to impl?
fun <R> foo(impl: MyInterface<R> = MyImpl) = impl.doIt() doesn't compile.
the best I could come up with is to also add an overload of `foo`:
fun foo() = foo(MyImpl).

[Paul Griffith]

the overload is your only option afaik

[Loney Chou]

MyImpl emits String, while MyInterface<R> emits R, R could be something other than String, so you can't assign impl: MyInterface<R> with MyImpl.

[Loney Chou]

Maybe you're thinking of default type parameter, but that's not a thing in Kotlin.

[y]

maybe I am. I'm trying to emulate Rust's associated type.

[y]

that said, default parameters are of course not a thing over there.

[y]

anyway, the overload seems like a reasonable compromise