In the following code: ```class A<Any>(val a: Any) fun <T: Comparable<T>> A(a: T): A<T> = A(a)...

# getting-started

[Norbi]

In the following code:

class A<Any>(val a: Any)

fun <T: Comparable<T>> A(a: T): A<T> = 
    A(a) // <--- how to call the constructor of class A instead of calling function A() recursively?

Is this the most correct solution?

fun <T: Comparable<T>> A(a: T): A<T> =
    A<Any>(a) as A<T>

Thanks.

EDIT: it is a theoretical question, just to know the edge cases of the language ;)

[Adam S]

Thanks, I hate it!

Not ideal - but you could use a typealias

data class A<Any>(val a: Any)

typealias AConstructor<T> = A<T>

fun <T: Comparable<T>> A(a: T): A<T> =
    AConstructor(a)

fun main() {
    println(A("x"))
    println(AConstructor(1))
}

[Klitos Kyriacou]

By the way, do you realise that in the declaration class A<Any>(val a: Any), the word Any doesn't refer to the well-known type Any? It is just a generic parameter. To avoid confusion, name it something like T instead.

[Norbi]

Yes, it should have been

class A<T: Any>(val a: T)

Sorry for the confusion...