Scala: Using supertype operator in pattern matching

Question

I have made my own framework of traits and classes that extend my traits. The parent of all classes is a trait named 'Contract'. 'Combinator' and 'ElementaryContract' are two immediate children of Contract.

def disintegrateContract[T](element: T): Elem = 
{

    element match
    {
      case com <: Combinator =>  matchCombinator(com)
      case e <:ElementaryContract =>matchElementaryContract(e)
    }
}

I want to make a match class that recognizes whether a passed 'Contract' is a subtype of 'Combinator' or 'ElementaryContract' and then pass it to some other functions. This is the compilation error I get:

 '=>' expected but '<:' found

Probably it does not recognize the subtype operator. How can I make this happen?

lukass · Accepted Answer · 2023-08-24 10:27:46Z

3

If understood you correctly the usual pattern matching should be fine -- if Bar extends Foo, it is Foo (plus something else):

class SuperA
class ImplA extends SuperA

class SuperB
class ImplB extends SuperB

def disintegrateContract[T](element: T) = element match {
   case a: SuperA =>  println("I extend SuperA")
   case b: SuperB =>  println("I extend SuperB")
}

disintegrateContract(new ImplA)
// I extend SuperA
disintegrateContract(new ImplB)
// I extend SuperB

To have exact your situation there should be Super which SuperA and SuperB will extend, but it changes nothing.

edited Aug 24, 2023 at 10:27

lukass

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answered Dec 5, 2013 at 15:25

om-nom-nom

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2 Comments

Core_Dumped Over a year ago

So, you are saying I do not need the <: operator?

om-nom-nom Over a year ago

@Core_Dumped yes, you don't need it and no, AFAIK, bounds are not supported in pattern matching.

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