问题
Suppose I have this monadic class:
case class Foo[A](xs: List[A]) {
def map[B](f: A => B) = Foo(xs map f)
def flatMap[B](f: A => Foo[B]) = Foo(xs flatMap f.andThen(_.xs))
def withFilter(p: A => Boolean) = {
println("Filtering!")
Foo(xs filter p)
}
}
The following is from a 2.10.0 REPL session:
scala> for { (a, b) <- Foo(List(1 -> "x")) } yield a
res0: Foo[Int] = Foo(List(1))
And here's the same thing in 2.10.1:
scala> for { (a, b) <- Foo(List(1 -> "x")) } yield a
Filtering!
res0: Foo[Int] = Foo(List(1))
This is completely unexpected (to me), and leads to particularly confusing errors in cases where filtering requires additional constraints (such as Scalaz's \/ or EitherT).
I wasn't able to find any discussion of this change in the 2.10.1 release notes. Can someone point out where and why this new desugaring behavior was introduced?
回答1:
The story is more complex than that, and it's in fact a 2.10.0 regression that was plugged there.
The "no-withFilter
" behavior was introduced in c82ecab, and because of things like SI-6968, this was reverted partially #1893. Further adaptations followed (SI-6646, SI-7183)
The takeaway sentence you're looking for is :
The parser can't assume that a pattern (a, b) will match, as results of .isInstanceOf[Tuple2] can't be statically known until after the typer.
来源:https://stackoverflow.com/questions/17424763/new-desugaring-behavior-in-scala-2-10-1