Kotlin vararg array of array of any

Question

I have a method accepting a vararg of the form

fun arrayOfArrays(vararg aoa: Array<Any>) {
}

Now, I have trouble understanding how to call this method, e.g.

fun callArrayOfArrays() {
    arrayOfArrays(arrayOf(1), arrayOf(1))      // 0) works
    val a = arrayOf(1)
    arrayOfArrays(a, a)                        // 1) type mismatch: inferred type Array, but Array was expected
    val aoa = arrayOf(a)
    arrayOfArrays(aoa)                         // 2) type mismatch: inferred type Array<array>, but Array was expected
    arrayOfArrays(*aoa)                        // 3) type mismatch: inferred type Array<array>, but Array<out array> was expected
    arrayOfArrays(aoa.toList().toTypedArray()) // 4) works
}

UPDATE: After getting a heads-up by a colleague, we figured that adding types to arrayOf() fixes some of my problems, i.e. the following does work now:

fun callArrayOfArrays() {
    arrayOfArrays(arrayOf(1), arrayOf(1))
    val a = arrayOf<Any>(1)
    arrayOfArrays(a, a)
    val aoa = arrayOf<Array<Any>>(a)
    arrayOfArrays(*aoa)
    arrayOfArrays(aoa.toList().toTypedArray())
    arrayOfArrays(*(aoa.toList().toTypedArray()))
}

I still believe that the former should be fine, too. And I do yearn for an understandable explanation for this behavior.

I appreciate that case 0 works, but I fail to understand all the other cases.

For case 1, I would expect that assigning the arrayOf(1) to a variable does not change the semantics, but here we are.

For case 2, I would expect that it works like I would expect the first case to work, just the "Any" being an Array here.

For case 3, I can see the difference, but I do not understand it, and certainly do not know how to ever make this work.

For case 4, I believe this is the vararg taking a single array. However, I cannot spread it either.

Joffrey · Accepted Answer · 2019-04-26 16:36:52Z

2

I believe this has to do with what T is inferred to be in the calls to arrayOf<T>(...).

The way I understand it is the following:

In case 0, the param type expected by arrayOfArrays() is Array<Any>, and thus it makes the compiler infer Array<Any> as the type of the expression arrayOf(1). The compiler then checks that 1 is an instance of Any, which is true (because Int is a subtype of Any), and therefore 1 is a valid parameter for arrayOf<Any>().

In case 1, though, arrayOf(1) has no other type information than the parameter 1, which is an Int, and so the expression as a whole is inferred to have the type Array<Int>, which is not a subtype of Array<Any> (because of the invariance of the generic type parameter here). The error message is not that great, though, I have to admit.

Same goes for case 2, where the type of aoa is Array<Array<Int>> instead of Array<Any>. Again, the error message is not very helpful.

Case 3 is really the same as case 1, where each element in aoa is expected to be of type Array<Any> but is of type Array<Int>. It could work if Array was covariant in T, that is probably where the "expected Array<out Any>" comes from, but I really don't fully understand the error message here.

Case 4 is similar to case 0 in the sense that the compiler has more local type information (the type expected by arrayOfArrays()), and so it guides the way it will pick the T when interpreting the declaration of toTypedArray().

edited Apr 26, 2019 at 16:36

answered Apr 26, 2019 at 16:25

Joffrey

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2 Comments

Matthias Over a year ago

Thank you for taking the time! This does make sense to me. In my actual code, I only had to insert a few type parameters and all errors were gone. I do not yet understand why vararg T implies Array<out T>, but I will get to that eventually, I hope ;-)

Joffrey Over a year ago

@Matthias I guess you have already given a look, but just in case you want to find out more about these out and in keywords, you can read more about generics and variance in Kotlin here

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