函数可以是默认参数值吗？

Question

Kotlin docs states that "functions are first-class". I'm trying to use a function as a default value of a function extension. However the compiler isn't having any of it:

fun <T> identity(x: T): T = x
fun <T, P> Channel<T>.dedupe(by: (T) -> P = ::identity): ReceiveChannel<T>
{
    ...
}

The error is Function invocation 'identity(...)' expected which kinda indicates Kotlin isn't really understanding what I want to do at all.

Is there a way?

Answer 1

I don't know why you get this error message, but the problem is type mismatch: the default value must make sense for any type parameters (subject to bounds). Ie you need a (T) -> P , but ::identity can give you (T) -> T or (P) -> P .

Proof: if you change to

fun <T, P> identity(x: T): P = throw Exception()
fun <T, P> List<T>.dedupe(by: (T) -> P = ::identity): Unit {}

it compiles.

Answer (which came out in comments below):

If P is changed to Any? , we should be able to use ::identity because (T) -> T is a subtype of (T) -> Any? . Unfortunately, it doesn't work, but using a lambda instead of a function reference does:

fun <T> identity(x: T): T = x
fun <T> Channel<T>.dedupe(by: (T) -> Any? = { it }): ReceiveChannel<T>
{
    ...
}