Why are the bounds of type parameters ignored when using existential types in Scala?

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予麋鹿
予麋鹿 2020-12-19 15:14

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: B         


        
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  • 2020-12-19 15:34

    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.

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  • 2020-12-19 15:34

    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.

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