is this behavior expected of reified parameters? ...

# getting-started

[Asq]

is this behavior expected of reified parameters?  Given (this is as simple as I could drive down my example):

inline fun <reified R, reified S> reifiedGenericIsSame(r: R, s: S): Boolean = (r is S) && (s is R)
interface FooBar<out A: Any, out B: Any>
typealias FBOI<A> = FooBar<Int,A>
typealias FBOS<A> = FooBar<String,A>

and the following:

open class Base<out A: Any, out B: Any>(val a: A, val b: B): FooBar<A, B> 

class SpecialOfInt<out B: Any>(b: B) : Base<Int, B>(1, b)
class SpecialOfStr<out B: Any>(b: B) : Base<String, B>("1", b)

// val soitd: FBOI<Int> = SpecialOfInt(2)
// val sostd: FBOS<Int> = SpecialOfStr(2)
val soi = SpecialOfInt(2)
val sos = SpecialOfStr(2)
//
// EDIT
//
val soitd: FBOI<Int> = soi
val sostd: FBOS<Int> = sos

then this holds:

reifiedGenericIsSame(soi, sos) shouldBe false     // passes
reifiedGenericIsSame(soi, sos) shouldBe true      // FAILS
reifiedGenericIsSame(soitd, sostd) shouldBe false // FAILS
reifiedGenericIsSame(soitd, sostd) shouldBe true  // passes

which 🤞is VERY counterintuitive, as it appears a typealias (allegedly syntax sugar) is changing a result semantically. I suppose type erasure side effects may be invoked as the mechanics, but this question is about the intrinsic behavior. Could please anyone provide me with great kotlin wisdom?

[mkrussel]

The is operator doesn't take into account the generics of the type being checked so listOf("foo") is List<Int> will be true. That's why you get true for soitd and sostd.

On the other hand soi and sos are different types, so the is check fails as expected.

[mkrussel]

The type alias is not changing anything. Both soi and sos are FooBar so the is check passes. It doesn't mater that one is FooBar<Int,A> and the other FooBar<String, A>

[Asq]

Hm. Am I misreading your reply? reifiedGenericIsSame(soi, sos) is false

[Asq]

They are different types, conforming to FooBar . Also soitd and sosts are different types. If I set

val soitd: FBOI<Int> = soi
val sostd: FBOS<Int> = sos

the outcome is identical.

[mkrussel]

I do think I got confused. soi and sos are different types so the method correct returns false.

soitd and sostd are also different types, but are declared as the same type FooBar. Since the typealias is just syntax sugar, both R and S in reifiedGenericIsSame are both FooBar. Bot soitd and sostd are of types that are subtypes of FooBar so the function returns true

[mkrussel]

The types of S and R are determined at compile time and not runtime.

[Asq]

I hear you. The type of soi as well as the type of soitd IIUC is SpecialOfInt, i.e. Foobar<Int, A> . The type of sos and sostd is SpecialOfStr, i.e. FooBar<String,A> . The difference between soi and sos is correctly resolved. The difference between soitd and sostd is not. The only delta is the typedef FBOI or FBOS . Nothing else is different. I have difficulty reconciling your explanation with my understanding.

[mkrussel]

reifiedGenericIsSame(soitd, sostd) gets inlined as the following

(soitd is FBOS<Int>) && (sostd is FBOI<Int>)

which gets translated to (soitd is FooBar<String,A>) && (sostd is FooBar<Int,A>)

after type erasure it becomes (soitd is FooBar) && (sostd is FooBar)

That is true because all the variables are of type FooBar

[mkrussel]

Remember the generic parameters for S and R are determined at compile time

[Asq]

Do I gather correctly then that reifiedGenericIsSame(soi, sos) gets inlined (eventually?) as (soi is FooBar<String,A>) && (sos is FooBar<Int,A>) ? If so, why would the eventual result be different? (soi and sos are correctly recognized as different types, soitd and sostd are incorrectly recognized as the same type). Or could it possibly be that reifiedGenericIsSame(soi, sos) gets inlined (through some different mechanism?) as (soi is SpecialOfStr<A>) && (sos is SpecialOfInt<A>) ? If so, the typealias is at some fault, in addition to type erasure.

Q: Should the typealias be (always?) ignored by the compiler in a is comparison? It should be, after all, just sugar.

[mkrussel]

Yes the typealias will be ignored by the is and the real value will be used.

The difference between reifiedGenericIsSame(soi, sos) and reifiedGenericIsSame(soitd, sostd) are the declared types of the references (not the objects they point to).

reifiedGenericIsSame(soitd, sostd) will use is FBOI and is FBOS , These are type aliases to the same type so the checks always pass.

reifiedGenericIsSame(soi, sos) will use is SpecialOfInt and is SpecialOfStr

because those are the types of the references at compile time.

[mkrussel]

type aliases have nothing to do with it.

[Asq]

So it is possible, by means of a typealias, to assign a reference so that the reference will have (formally) a different type than that of the underlying (pointed-to) object. That statement is wildly interesting IMHO and fraught with consequences.

Does this conflict with the principle that an object has exactly one type, namely the type assigned at instantiation? It appears that, if a typealias can change the type of a reference, thinking of it as sugar as intuition suggested is VERY wrong. Also, how does this reconcile with

fun <A, B> isSameType(a: A, b: B): Boolean = a?.let{ outer -> outer::class == b?.let{ it::class } } ?: false

the above, in the example provided, behaves in a manner sometimes contradicting reifiedGenericIsSame to the extent that isSameType(soi, sos) and isSameType(soitd, sostd) behave in a consistent manner.

[mkrussel]

An object is a single type but the is operator is checking is a relationship. So a SpecialOfInt is a FooBar so is operator returns true. Type alias will not let you change the type of the object.

val foo = 5
foo is Int // true
foo is Number // true
foo is Any // true
foo is Int? // true
foo is String // false

[mkrussel]

foo::class == Int

[mkrussel]

In your example the soitd is a reference that can refer to any FooBar<Int,A> It happens to be referencing an object of type SpecialOfInt

[Asq]

Hm.