Hi, I've found this limitation just recently and w...

# getting-started

[Joshua Hansen]

Hi, I've found this limitation just recently and was wondering if this is a known issue, or if it is expected to not work.

It seems that vararg constructor parameters cannot be accessed in override methods. Here is a simple example:

fun main() {
    val foo = Foo(1, "Hello", "World")
    foo.doIt()
}

interface DoSomething {
    fun doIt()
}

class Foo(val bar1: Int, vararg bar2: String) : DoSomething {
    override fun doIt() {
        println(bar1)
        println(bar2) // Unresolved Reference: bar2
    }
}

[Casey Brooks]

You still need to mark a vararg parameter as var or val. Otherwise, it’s simply a value passed into the constructor, but not set to an actual class property. Try class Foo(val bar1: Int, vararg val bar2: String)

[Joshua Hansen]

Ahh yeah that worked. I didn't know you could mark vararg parameters as val. Thanks

[ephemient]

vararg is Array but List is a better data structure in most cases. unless there are other reasons to keep the Array around, I would write

class Foo(val bar1: Int, vararg bar2: String) {
    val bar2: List<String> = bar2.toList()

[Klitos Kyriacou]

Perhaps asList() instead of toList() since there's no point in copying a vararg parameter.

[ephemient]

well, it depends. if the caller is Java, they can hold onto and mutate the array after making the call

[ephemient]

e.g. in Java this mutates the incoming parameter

String[] bar2 = new String[] { "foo" };
new Foo(0, bar2);
bar2[0] = "bar";

the equivalent in Kotlin always copies the varargs array, though

[Joshua Hansen]

In my case, the vararg param is nullable. So I opted to not mark it with val, and instead declare a property in the class body like
val bar2NotNull = bar2.filterNotNull()