Reputation: 9698
Let's take this for an example:
import scala.reflect._
def get[T](list: List[Any])(implicit tag: ClassTag[T]) = {
list.flatMap {
case element: T => Some(element)
case _ => None
}
}
I can use get()
to get values of type T
from a list (e.g. get[String](list)
will give me all strings from that list).
Now, I understand that compiler provides the value of type ClassTag[String]
automatically. I also understand that ClassTag
is a type class, and somewhere behind the curtain there's a piece of code that says implicitly[ClassTag[T]].getRuntimeClass()
or whatever.
But if that's so, how come we can pattern match without the class tag (we just can't differentiate between erased types in that case)? I mean, how is it achieved that, if I declare an implicit parameter (which is automatically provided by the compiler), I get one behavior, but if i don't I get a different behavior?
Upvotes: 7
Views: 5118
Reputation: 55569
The compiler automatically translates your code roughly to this:
def get[T](list: List[Any])(implicit tag: ClassTag[T]) = list.flatMap {
case (element @ tag(_: T)) => Some(element)
case _ => None
}
ClassTag
has an unapply(x: Any)
overload that allows it to pattern match on values. I've cleaned up the tree obtained from reify
to only show the relevant parts, but this will show you the full tree:
scala.reflect.runtime.universe.reify {
def get[T](list: List[Any])(implicit tag: ClassTag[T]) = {
list.flatMap {
case element: T => Some(element)
case _ => None
}
}
}
Also see the scaladoc:
The compiler tries to turn unchecked type tests in pattern matches into checked ones by wrapping a
(_: T)
type pattern asct(_: T)
, wherect
is theClassTag[T]
instance. Type tests necessary before calling other extractors are treated similarly.SomeExtractor(...)
is turned intoct(SomeExtractor(...))
ifT
inSomeExtractor.unapply(x: T)
is uncheckable, but we have an instance ofClassTag[T]
.
Upvotes: 12