Infinite loop seems to confuse Scala's type system

Question

Here is an artificial toy example that demonstrates my problem:

def sscce(): Int = {
  val rand = new Random()
  var count = 0
  while (true) {   // type mismatch; found: Unit, required: Int
    count += 1
    if (rand.nextInt() == 42) return count
  }
}

How can I help the compiler understand that this method will always return an Int?

I know the above toy example could easily be refactored to get rid of the infinite loop altogether, but I really want to have the infinite loop in my actual code. Trust me on this ;)

Your code doesn't, technically speaking, return an Int. It returns an Int *if* the function returns at all. I suspect this distinction is the problem. What are you trying to solve? You say you really want an infinite loop, but perhaps your real world problem can have another solution. — Daenyth, Mar 05 '15 at 17:43
I really love infinite loops, but they take forever to finish. — Michael Zajac, Mar 05 '15 at 17:44
You can't prove that it will always be executed because it won't always be executed. It will usually be executed, but scala doesn't provide a `Usually[Int]` type. I'd be shocked if there wasn't a better way to do this instead of using the infinite loop. — Daenyth, Mar 05 '15 at 18:00
possible duplicate of [Scala while(true) type mismatch? Infinite loop in scala?](http://stackoverflow.com/questions/10802917/scala-whiletrue-type-mismatch-infinite-loop-in-scala) — fredoverflow, Mar 05 '15 at 18:19

score 6 · Answer 1 · answered Mar 05 '15 at 17:47

6

Always return an Int:

def sscce(): Int = {
  val rand = new Random()
  var count = 0
  while (true) {
    count += 1
    if (rand.nextInt() == 42) return count
  }
  count // <-- this
}

answered Mar 05 '15 at 17:47

Ionuț G. Stan

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Seth Tisue · Accepted Answer · 2015-03-06T16:21:08.403

4

You can also do:

def foo: Int = {
  ...
  while(true) {
    ... return ...
  }
  throw new IllegalStateException  // unreachable
}

this will typecheck because the type of the throw is Nothing, which is a subtype of Int.

edited Mar 06 '15 at 16:21

answered Mar 05 '15 at 19:20

Seth Tisue

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score 2 · Answer 3 · answered Mar 05 '15 at 17:57

See this question. While loops don't return a value. i.e. they return Unit which is the last statement in your function. So, the definition says it returns an Int but it actually returns Unit thus the type error. @ionut's answer fixes the type error by returning count as the last statement or here is a recursive approach.

def sscce(): Int = {
  val rand = new Random()
  def ssccer(count: Int): Int = {
    if(rand.nextInt == 42) return count
    else ssccer(count + 1)
  }
  ssccer(0)
}

score 2 · Answer 4 · answered Mar 05 '15 at 17:57

Per the SLS, a while loop is executed similarly to:

def whileLoop(cond: => Boolean)(body: => Unit): Unit  =
  if (cond) { body ; whileLoop(cond)(body) } else {}

ie., it returns Unit. So the compiler sees the while as the last statement in sscce(), and therefore assumes that you're trying to return Unit. I don't think it's smart enough to realize that return count will eventually always return an Int.

The simple solution is to follow the suggestion of @Brian or @IonutGStan, and force it to return count, whether it truly needs it or not.

score 2 · Answer 5 · answered Mar 05 '15 at 18:26

From a code quality standpoint, it would be good to ditch the while(true) loop and replace it with something more readable. As a nice side effect, it also solves your problem:

def sscce(): Int = {
  val rand = new Random()
  var count = 1
  while (rand.nextInt() != 42) {
    count += 1
  }
  count
}

Infinite loop seems to confuse Scala's type system

5 Answers5