Why does this code ``` (1..10).forEach { println(it) return@forEach ...

# getting-started

[Karlo Lozovina]

Why does this code

(1..10).forEach {
            println(it)
            return@forEach
        }

print all of 10 numbers? what does even "return@forEach" mean here? It seems to work more like "continue", instead of "break"?

[bezrukov]

Yes it behaves like continue because it's an early return from a lambda's scope. forEach is not a language construct, but just a function.
You need to either use native for-loop (for (it in 1..10) ) or non-labeled return. It's impossible to get break behavior with forEach (you need to use something like takeWhile)

[Karlo Lozovina]

Thanks!

[Karlo Lozovina]

I'm guesing it's not the problem with forEach, this is in general impossible? To return from a lambda block?

[Karlo Lozovina]

Without having a separate predicate like takeWhile, of course

[ephemient]

you can think of this as translating into

val lambda: (Int) -> Unit = forEach@{
    println(it)
    return@forEach
}
for (i in 1..10) {
    lambda(i)
}

which makes it clear why return can't stop the forEach

[Youssef Shoaib [MOD]]

BTW, you can get break behaviour from forEach, but it requires a bit of nesting. It looks like this:

run {
    (1..10).forEach { 
        println(it)
        if(it == 5){ //Break condition 
            return@run
        }
    }
}

[Karlo Lozovina]

Thanks, it might come in handy sometimes, even though that extra nesting is a bit inelegant 🙂 ... having some kind of nonlocal return would be useful here and there ...