As mentioned by @Alexander Udalov it's not possible to do directly but there's a workaround where you define the extension method on another type like so:
data class Wrapper<T>(val value: T)
val <T> T.ext: Wrapper<T> get() = Wrapper(this)
fun <T> Wrapper<T>.thing(p: T) {
println("value = $value, param = $p")
}
With the above the following compiles:
"abc".ext.thing("A")
but the next fails
"abc".ext.thing(2)
with:
Kotlin: Type inference failed: Cannot infer type parameter T in fun <T> Wrapper<T>.thing(p: T): Unit
None of the following substitutions
receiver: Wrapper<String> arguments: (String)
receiver: Wrapper<Int> arguments: (Int)
can be applied to
receiver: Wrapper<String> arguments: (Int)
As suggested by @hotkey it seems that it should be possible to avoid the need for explicit Wrapper
type with the following extension property:
val <T> T.thing: (T) -> Any? get() = { println("extension body") }
And then use it as "abc".thing("A")
but it also fails. Surprisingly the following does compile "abc".thing.invoke("A")
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