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Hello, World! A generics question that lets me pe...
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b
Hello, World! A generics question that lets me perplexed 🤔!
Copy code
fun <T> foo(a: T, block: () -> T) {}
foo("Hello") { 1 }
How come this compiles? 
T
is 
String
for 
a
but 
Int
for the 
block
lambda. Bonus question: what's a way to make this not compile and enforce the same type? 🙂 Thanks 🙏
d
T
is inferred to be 
Any
, which is the common supertype of 
String
and 
Int
, 
foo
then returns 
Any
.
😮 1
➕ 1
As for the bonus question: I don't think you can.
b
😢
n
yeah, Kotlin tends to infer Any sometimes in situations where you would really want a compiler error, but there's no magic way for the compiler to understand I guess those particular cases
Another example is passing totally unrelated types to 
listOf
v
The compiler will then fail when you try to use the result somewhere
n
when you eventually hit some non-generic, or suitably constrained code, yes 🙂 Not the end of the world but it can mean your compiler error is farther from the written error
v
Well, that's a PoV whether it is a written error or not. It could be perfectly fine and intended to have 
Any
there. :-)
n
Indeed, but it's the downside of having a top type like that. Not that there aren't benefits as well of course.
f
Copy code
fun <T> foo(a: T, block: () -> T) where T : String {}
will constrain it
m
🤫 You didn’t hear that from me.
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import kotlin.internal.*

@Suppress("INVISIBLE_MEMBER", "INVISIBLE_REFERENCE")
fun <T> foo(a: @Exact T, block: () -> T) {}

fun main() {
    foo("Hello") { 2 }
}
(evil hack™) Alternative with warning, but no less evil:
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import kotlin.internal.*

@Suppress("INVISIBLE_MEMBER", "INVISIBLE_REFERENCE")
fun <@OnlyInputTypes T> foo(a: T, block: () -> T) {}

fun main() {
    foo("Hello") { 2 }
}
🤐 3
b
To me, it seems like the original code is kinda an oxymoron. By using 
T
you're explicitly saying you don't care about the type. So it's kinda contradictory to then turn around and be surprised that any type will work.
v
@flosch that would miss the point of generics, you couldn't do 
foo(1) { 2 }
and as 
String
is final it would be pointless to use generics at all
➕ 1
f
well that depends on what the use case for the function is. there might be a reason to only constrain it to string or inheritors of string. also the bonus question was
Bonus question: what’s a way to make this not compile and enforce the same type?
and this is how you do that
v
Oh 
String
is not 
final
in Kotlin, missed that. Assumed it is 
final
like in Java. But you still miss the meaning of the bonus question. The question is how to enforce both `T`s are of the same non-
Any
type, but without restraining to just 
String
subtypes. As I said, the goal is that 
foo("Hello") { "World" }
works as well as 
foo(1) { 2 }
or any other type, while 
foo("Hello") { 1 }
does not and there seems to be no way without using internals to reach that.
d
String
is final in kotlin, because it is not 
open
😉
m
String
is definitely 
final
😅
v
Ok, thank god, so my statement was correct, using generics then would be absolutely pointless
fun foo(a: String, block: () -> String) {}
f
ok true string was a bad example, any open class then for my example I get the question now 👍
b
Thanks a lot for all the reactions. I am very surprised to see this (and to see this only now after like ... 3 or 4 years of "kotlining" professionally! but that's another story). I was thinking maybe this is due to the lambda, but now I see that this simpler example also has this issue, and now I'm even more surprised:
Copy code
fun <T> bar(a: T, b: T) {}
bar("Hello", 1)
v
Why? The inferred common supertype is 
Any
still. As you don't do anything with 
a
or 
b
where 
Any
cannot be used, this is just fine.
b
no you're right I guess I shouldn't be surprised that it compiles, but still surprised that there's no way to do this?
(not including the 
kotlin.internal
ways 😛)
n
it's an inevitable result of a) having a top type, and b) allowing type deduction to work with inheritance
C++ has neither of those things, and that example would not compile
(I should say, generic type deduction)
without b) you also wouldn't be able to do 
bar(listOf(1,2,3), mutableListOf(4,5,6))
a) is probably the more controversial one but I do think 
Any
is useful
b
Thanks for the insight! At the very least, I'm glad because TIL 🙂
n
i was coming for C++ and it was very very weird for me at first
*from
but you get used to it, and as bjorn pointed out, it usually just means you get a compiler error on the next line instead of immediately
v
I'm not called bjorn 😞 It is björn or bjoern 😉
n
i don't umlaut 🙂
but I will try to add the e
👌 1
if I remember
is that a general thing in german, that o with umlaut -> oe ?
v
Yes
n
gotcha, thanks
v
also ä -> ae, ü -> ue and ß -> ss
n
the strauss one I knew, I think
I was in germany and had to ask about how to pronunce the "beta" 🙂
v
lol
b
Here's a trick that might work if you only need to use basic (Comparable) types:
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fun <T: Comparable<T>> foo(a: T, block: () -> T) {}
At least it will cause a compiler error.
b
Interesting 🤔 Let me try that.
v
Would of course still work with a type and its subtype
☝️ 1
But 
Any
wouldn't be inferred, as it is not 
Comparable
b
indeed
amazing
v
What we miss is
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fun <T> foo(t: T) where T : !Any {}
😄
👍 3
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