Disclaimer: the code snippet below relates to one of ongoing Coursera courses. Let's consider it's posted just for a learning purpose and should not be used for submitting as a solution for one's homework assignment.
As the comment below states, we need to transform a list of Futures to a single Future of a list. More than that, the resulting Future should fail if at least one of input futures failed.
I met the following implementation and I don't understand it completely.
/** Given a list of futures `fs`, returns the future holding the list of values of all the futures from `fs`.
* The returned future is completed only once all of the futures in `fs` have been completed.
* The values in the list are in the same order as corresponding futures `fs`.
* If any of the futures `fs` fails, the resulting future also fails.
*/
def all[T](fs: List[Future[T]]): Future[List[T]] =
fs.foldRight(Future(Nil:List[T]))((f, fs2) =>
for {
x <- f
xs <- fs2
} yield (x::xs))
In particular, I don't understand the next things in it:
Future[T] -> T
transformation happen? It looks like xs <- fs2
is the only place we touch initial Futures
, and each of xs
type should be Future[T]
(but somehow it becomes just T
).Future
object does fail when one of the input Futures
fails. 1) Say f is a Future[T]
, then writing
for {
t <- f
} yield List(t)
will store the result of the Future f in t - therefor t is of type T. The yield turns it into a List[T], and the type of the whole for-comprehension ends up being Future[List[T]]. So the for-comprehension is where you extract your Ts from your Futures
, do something with them, and put them back in a Future (OK, I'm simplifying a little bit here).
It's equivalent to
f.map(t => List(t))
2) If your Future f contains a Failure, then the for-comprehension will just return this failed Future instead of executing the yield.
As a general note, for-comprehension in Scala is just sugar that can be rewritten with map, flatMap, filter, foreach
.
I'm an English-speaking right-hander, so normally I foldLeft, but each step of the fold looks like:
Fn flatMap ((x: T) => Fs map (xs => x :: xs))
Your value is x
.
The function is applied on success, which explains why a failure stops you cold:
scala> timed(Await.ready(all(List(Future{Thread sleep 5*1000; 1},Future(2),Future{Thread sleep 10*1000; 3})), Duration.Inf))
res0: (Long, scala.concurrent.Awaitable[List[Int]]) = (10002419021,scala.concurrent.impl.Promise$DefaultPromise@2a8025a0)
scala> timed(Await.ready(all(List(Future{Thread sleep 5*1000; 1},Future(???),Future{Thread sleep 10*1000; 3})), Duration.Inf))
res1: (Long, scala.concurrent.Awaitable[List[Int]]) = (5000880298,scala.concurrent.impl.Promise$DefaultPromise@3750d517)
Notice that the failing version short-circuits.
See the ScalaDoc for flatMap for both bits of information.
Edit: I was speaking cautiously because it is Coursera work, but more plainly, this requirement is not met: "The returned future is completed only once all of the futures in fs
have been completed."
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