Kotlin `IntRange`s don't hold any values in memory...

# announcements

[Manuel Pérez Alcolea]

Kotlin `IntRange`s don't hold any values in memory by themselves, is that right? I'm asking because my experience tells me they should be lazy, yet they are not `Sequence`s, so it's possible to get one from with asSequence() instead. And something like a direct map gives you a list. But I would still assume a range object doesn't contain anything, simply because "it makes sense" in my head.

>>> (0..4).asSequence()
res0: kotlin.sequences.Sequence<kotlin.Int> = kotlin.collections.CollectionsKt___CollectionsKt$asSequence$$inlined$Sequence$1@7bafb922
>>> (0..4).map { it * 2 }
res1: kotlin.collections.List<kotlin.Int> = [0, 2, 4, 6, 8]

So what's in an IntRange before applying anything? (no asSequence()) for example val myRange = 0..4

[Nir]

I haven't looked but I would assume than IntRange is simply its own type of object

[Nir]

that supports both iteration, and in

[Nir]

and probably only stores the end points

[Nir]

(so, yes, it would be lazy)

[Nir]

okay, let me rephrase

[Nir]

its lazy in the sense that its not storing the whole list of values on it

[Nir]

but when you do chain operations on it (via e.g. map) it does things eagerly

[Nir]

If you just follow IntRange's parents in the docs this is pretty easy to figure out

[Nir]

it eventually inherits from Iterable, not Sequence

[nanodeath]

IntRange's are indeed stateless, and depending on how you construct them (e.g. for (i in 1..4)) may even get compiled away.

[Manuel Pérez Alcolea]

but when you do chain operations on it (via e.g. map) it does things eagerly

thanks, this is exactly what I meant. I'm glad I didn't need to give extra information to communicate the idea properly

[Nir]

Yeah. Remember that things like map, are simply generic extension functions. There's one map for sequences, which is lazy, and one for iterables, which isn't.

[Nir]

So to figure out which map you're calling, you just need to figure out which base class you have.

[Manuel Pérez Alcolea]

(e.g. for (i in 1..4)) may even get compiled away.

ah yes, I saw that in the bytecode before. Well done compiler, well done. That's why I mentioned IntRange specifically