Return/break out of infinite foreach in kotlin

Question

For class I have to make a program that calculates the birthday problem Now I'm having trying to learn kotlin at the same time and I'm having trouble with a little snippet of code:

        val checkSet = mutableSetOf<Int>()
        generateSequence{ Random.nextInt(n)}.forEach {
            if(!checkSet.add(it)) {
                return@outForeach
            }
        }
        outForeach@
        sum += checkSet.size

As you can see I'm trying to do this with an infinite sequence. Kotlin doesn't accept this as outForeach is an unresolved reference. But this doesn't work either:

        val checkSet = mutableSetOf<Int>()
        generateSequence{ Random.nextInt(n)}.forEach {
            if(!checkSet.add(it)) {
                return@forEach
            }
        }
        sum += checkSet.size

This will just start the forEach loop again. Is there a way to implement something as a forEachUntil or so?

p.s. I'm aware that this looks a lot like this question: 'return' doesn't jump out of forEach in Kotlin It's just that I don't really get the answers and I don't know if its applicable here. Also a way to implement forEachUntil seems for me to be far more elegant

Roland · Answer 1 · 2018-12-12T15:27:39.350

Alternatives you may want to consider instead of first:

using a simple while without body:

while (checkSet.add(Random.nextInt(n))); // <- that semicolon is required! otherwise you execute what is coming next within the while

using run with a label:

run outForeach@{
  generateSequence{ Random.nextInt(n)}.forEach {
    if(!checkSet.add(it)) {
      return@outForeach
    }
  }
}

maybe also takeWhile might be helpful. In this specific case however it is surely not (as it would check against the checkSet and leave us with a sequence that isn't consumed... but if the condition would be different, it may make sense to consider something like take, takeWhile, takeLast, etc.):
```
generateSequence { Random.nextInt(n) }
    .takeWhile(checkSet::add) // as said: for this specific condition it doesn't make sense... 
    .forEach { /* do nothing except consume the sequence */ } // the same values you added to the set would be available in this step of course
```

the while loop is definitely a more elegant solution as it produces no unused sequence or reference. — Dean, Dec 12 '18 at 15:14
that's true and while it works quite well for this specific example it may not in other cases where a real ~return is wanted.... (although I don't know any such cases that really require a goto-label-construct, that aren't also solveable using other means) — Roland, Dec 12 '18 at 15:21

score 1 · Answer 2 · answered Dec 12 '18 at 14:02

1

I think I found the solution myself:

        val checkSet = mutableSetOf<Int>()
        generateSequence{ Random.nextInt(n)}.first { !checkSet.add(it) }
        sum += checkSet.size

Basically use the function first() and keep returning false until you want to get out of the loop. And just drop the return of the function first()

answered Dec 12 '18 at 14:02

Typhaon

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It says I can do that in two days – Typhaon Dec 12 '18 at 15:12

Return/break out of infinite foreach in kotlin

2 Answers2