问题
What I mean is this:
scala> class Bounded[T <: String](val t: T)
defined class Bounded
scala> val b: Bounded[_] = new Bounded("some string")
b: Bounded[_] = Bounded@2b0a141e
scala> b.t
res0: Any = some string
Why does res0 have type Any and not String? It sure could know that b.t is at least a String. Writing
val b: Bounded[_ <: String] = new Bounded("some string")
works, but it is redundant with respect to the declaration of the class itself.
回答1:
First, I have edited the question title. You are not using dependent types, which Scala doesn't have anyway, but existential types. Second, you are not inferring anything, you are explicitly declaring the type.
Now, if you did write Bounded[Any], Scala wouldn't let you. However, one of the uses of existential types is to deal with situations where the type parameter is completely unknown -- such as Java raw types, where.
So my guess is that making an exception in a situation that seems obvious enough will break some other situation where existential type is the only way to deal with something.
回答2:
There was a lengthy discussion about this topic recently on the mailing list, Type Boundary "Stickyness" on Wildcards.
It wasn't conclusive, other than to agree that existential types, such as Bounded[_] (a shorthand for Bounded[$1] forSome { type $1 }), don't lend themselves to intuition.
@extempore did find one upside to the discussion :)
On the plus side I'm finally reading the spec cover to cover. I had no idea the complete lyrics to "yellow submarine" were in the specification! Yet I have to admit, in context it was hard to see any other way that section could have been written.
来源：https://stackoverflow.com/questions/4323140/why-are-the-bounds-of-type-parameters-ignored-when-using-existential-types-in-sc