When implementing an interface in Kotlin with an inline
function:
interface Foo {
fun qux(fn: () -> Unit)
}
open class Bar : Foo {
final
I know this is pretty late, but here's why. You got this right:
It is impossible to inline a virtual method.
But you should consider that, while Bar.foo
is not virtual in the sense that it could be overridden (it can't be), it is virtual in the sense that a decision can be made to run it at runtime. Consider the following example which builds on your own:
interface Foo {
fun qux(fn: () -> Unit)
}
open class Bar : Foo {
final override inline fun qux(fn: () -> Unit){TODO()}
}
class Baz : Foo {
override fun qux(fn: () -> Unit) = TODO()
}
fun main() {
var foo: Foo = Bar()
foo.qux { } // calls Bar.qux
foo = Baz()
foo.qux { } // calls Foo.qux
}
Here, Bar.qux
is invoked initially, but Baz.qux
is invoked the second time. So, not every call can be inlined. So why is this a warning and not a compiler error, like when we declare an open fun
with the inline
modifier? Consider the following:
val bar: Bar = Bar()
bar.qux { }
In this case, the compiler can safely inline all calls to qux
on the variable bar
because it's declared as a Bar
. Even if the class is open as in your example, since the method itself is final
, any derived type will always use exactly that qux
. So this statement,
In the example above, when I invoke bar.qux(), the compiler can ensure that only this particular implementation will be used, and can safely inlined.
is true only when the compiler statically knows the type of bar
to be an actual Bar
. It's an error when it's open
because none of them can be inlined which definitely isn't the desired behavior, it's a warning when it's an override because only some of them can be inlined, which might not be the desired behavior.
This example is available at play.kotlinlang.org