How I can return a function within itself? it’s li...

# announcements

[Ifvwm]

How I can return a function within itself?
it’s like
def f(x):
print(x)
return f
then I can do f(1)(2)(3)

[jw]

return ::f

[Ifvwm]

@jw then what its type is when the function return itself? it won’t produce an infinite type?

[jw]

don't know. use an expression body and find out!

[jw]

it gets rejected by the compiler

[Ifvwm]

it seems we can’t do that in any static type-check language

[streetsofboston]

Yup, the return type of f would be infinitely recursive:
(Int)->((Int)->((Int)->((Int)->... ...))))))))

[Ifvwm]

@streetsofboston but if we’ve known the times what it would be called, then we can get the result type manually,
fun f(x: Int): (Int) -> (Int) -> Any
then we can call f(3)(5)(7)

[jw]

to be fair, recursive types are allowed

[streetsofboston]

@Ifvwm But then you cannot do return ::f, unless f returns a plain Any itself,...

[streetsofboston]

@jw How do you declare such a recursive type in Kotlin?

[jw]

https://kotlinlang.org/api/latest/jvm/stdlib/kotlin/-enum/index.html

[Ifvwm]

@streetsofboston

fun test(x:String):(String)->(String)->Any{
  println(x)
  return ::test
}

test(“1”)(“2”)(“3”)

[streetsofboston]

But ::test has the type (String)->((String)->(String)->Any).

Can that be coerced/cast to (String)->(String)->Any?

If so, then (String)->((String)->(String)->Any) is a sub-type of (String)->(String)->Any...

[Ifvwm]

actually I really don’t know about that sub-type stuff, I’m from Haskell, there’s no such thing like type inheritance

[Ifvwm]

a->b->a is a->(b->a) in type level

[streetsofboston]

😀
In other words,
If so, is a (String)->((String)->(String)->Any) safely assignable to a (String)->(String)->Any...

[streetsofboston]

But in this case there's one extra (String)....
Is this true as well: a -> a -> a -> b is a -> a -> b?

[Ifvwm]

@streetsofboston yes, but when it’s concreted then no

[streetsofboston]

Then how can ::test be returned as a return-value to itself in Kotlin?

[Ifvwm]

a is a type variable, genetic in java

[Ifvwm]

a can be a->a even a->a->a but Int->Int and Int->Int->Int are definitely not the same

[Ifvwm]

@streetsofboston actually I don’t know why it can...

[streetsofboston]

You can construct the type-paramater A with a concrete type Int. If A -> A is A, shouldn't Int -> Int be Int (after constructing it with an Int)?

[Ifvwm]

@streetsofboston that’s polymorphic

[Ifvwm]

Int->Int and (Int->Int)->(Int->Int) both are a->a, I think

[streetsofboston]

Or is this possible because of type erasure in JVM?
In JVM (A)->A is ambiguous with any other lambda that looks like (A)->.... ....

[streetsofboston]

Now I'm very curious! 😀

[Ifvwm]

I don’t know type erasure in java well, but that Haskell guys tell me Java’s generics is very different with Haskell’s type variables, also some people call Java’s generics is a false polymorphic

[streetsofboston]

It's almost midnight here....I'm going to (Z)->(Z)->(Z)->(Z)->(Z) 💤

[Ifvwm]

@streetsofboston good night!

[Big Chungus]

lateinit var f: (String) -> ((String) -> Any)
f = {
    print(it)
    f
}

f("1")("2")("3")

[molikuner]

Not quite a function, but works like it.
https://pl.kotl.in/tFUO00goe

object test {
    operator fun invoke(i: Int) = also {
        println(i)
    }
}

fun main() {
    test(0)(1)(2)(3)(4)
}