How do I make a function involving futures tail recursive? - scala

Question

In my Scala app, I have a function that calls a function which returns a result of type Future[T]. I need to pass the mapped result in my recursive function call. I want this to be tail recursive, but the map (or flatMap) is breaking the ability to do that. I get an error "Recursive call not in tail position."
Below is a simple example of this scenario. How can this be modified so that the call will be tail recursive (without subverting the benefits of Futures with an Await.result())?
import scala.annotation.tailrec
import scala.concurrent.{Await, Future}
import scala.concurrent.duration._
implicit val ec = scala.concurrent.ExecutionContext.global
object FactorialCalc {
def factorial(n: Int): Future[Int] = {
#tailrec
def factorialAcc(acc: Int, n: Int): Future[Int] = {
if (n <= 1) {
Future.successful(acc)
} else {
val fNum = getFutureNumber(n)
fNum.flatMap(num => factorialAcc(num * acc, num - 1))
}
}
factorialAcc(1, n)
}
protected def getFutureNumber(n: Int) : Future[Int] = Future.successful(n)
}
Await.result(FactorialCalc.factorial(4), 5.seconds)

Answer 1

I might be mistaken, but your function doesn't need to be tail recursive in this case.
Tail recursion helps us to not consume the stack in case we use recursive functions. In your case, however, we are not actually consuming the stack in the way a typical recursive function would.
This is because the "recursive" call will happen asynchronously, on some thread from the execution context. So it is very likely that this recursive call won't even reside on the same stack as the first call.
The factorialAcc method will create the future object which will eventually trigger the "recursive" call asynchronously. After that, it is immediately popped from the stack.
So this isn't actually stack recursion and the stack doesn't grow proportional to n, it stays roughly at a constant size.
You can easily check this by throwing an exception at some point in the factorialAcc method and inspecting the stack trace.
I rewrote your program to obtain a more readable stack trace:
object Main extends App {
import scala.concurrent.{Await, Future}
import scala.concurrent.duration._
implicit val ec = scala.concurrent.ExecutionContext.global
def factorialAcc(acc: Int, n: Int): Future[Int] = {
if (n == 97)
throw new Exception("n is 97")
if (n <= 1) {
Future.successful(acc)
} else {
val fNum = getFutureNumber(n)
fNum.flatMap(num => factorialAcc(num * acc, num - 1))
}
}
def factorial(n: Int): Future[Int] = {
factorialAcc(1, n)
}
protected def getFutureNumber(n: Int) : Future[Int] = Future.successful(n)
val r = Await.result(factorial(100), 5.seconds)
println(r)
}
And the output is:
Exception in thread "main" java.lang.Exception: n is 97
at test.Main$.factorialAcc(Main.scala:16)
at test.Main$$anonfun$factorialAcc$1.apply(Main.scala:23)
at test.Main$$anonfun$factorialAcc$1.apply(Main.scala:23)
at scala.concurrent.Future$$anonfun$flatMap$1.apply(Future.scala:278)
at scala.concurrent.Future$$anonfun$flatMap$1.apply(Future.scala:274)
at scala.concurrent.impl.CallbackRunnable.run(Promise.scala:29)
at scala.concurrent.impl.ExecutionContextImpl$$anon$3.exec(ExecutionContextImpl.scala:107)
at scala.concurrent.forkjoin.ForkJoinTask.doExec(ForkJoinTask.java:262)
at scala.concurrent.forkjoin.ForkJoinPool$WorkQueue.runTask(ForkJoinPool.java:975)
at scala.concurrent.forkjoin.ForkJoinPool.runWorker(ForkJoinPool.java:1478)
at scala.concurrent.forkjoin.ForkJoinWorkerThread.run(ForkJoinWorkerThread.java:104)
So you can see that the stack is actually short. If this was stack recursion you should have seen about 97 calls to the factorialAcc method. Instead, you see only one.

Answer 2

How about using foldLeft instead?
def factorial(n: Int): Future[Int] = future {
(1 to n).foldLeft(1) { _ * _ }
}

Answer 3

Here's a foldLeft solution that calls another function that returns a future.
def factorial(n: Int): Future[Int] =
(1 to n).foldLeft(Future.successful(1)) {
(f, n) => f.flatMap(a => getFutureNumber(n).map(b => a * b))
}
def getFutureNumber(n: Int) : Future[Int] = Future.successful(n)

Answer 4

Make factorialAcc return an Int and only wrap it in a future in the factorial function.
def factorial(n: Int): Future[Int] = {
#tailrec
def factorialAcc(acc: Int, n: Int): Int = {
if (n <= 1) {
acc
} else {
factorialAcc(n*acc,n-1)
}
}
future {
factorialAcc(1, n)
}
}
should probably work.