Vinothkumar Raman Vinothkumar Raman - 3 months ago 7
Scala Question

Why is this scala code not infering type?

I am building a aggregation framework with some code which looks like this

trait Aggregate[T <: Aggregate[T, K], K] { self: T =>
def plus(another: T): T

def show: K
}


I have couple of aggregations like this,

case class Count(value: Long = 1) extends Aggregate[Count, Long] {
def plus(another: Count) = Count(value + another.value)

def show = value
}


Once i define a aggregation which is like this,

case class By[T <: Aggregate[T, K], K, G](values: HashMap[G, T]) extends Aggregate[By[T, K, G], Map[G, K]] {
override def plus(another: By[T, K, G]): By[T, K, G] = By(values.merged(another.values){case ((k1,v1), (k2,v2)) => (k1, v1 plus v2)})

override def show: Map[G, K] = values.map{case (k,v) => k -> v.show}
}

object By {
def apply[T <: Aggregate[T,K], K, G](key:G, value:T):By[T, K, G] = By(HashMap(key -> value))
}


I cannot write something like this

By("key1", Count(100))


Instead i had to write this

By[Count, Long, String]("key1", Count(100))


Since its not figuring out the
Long
part, I hate to specify those types, is there a cleaner way to achieve that?

Answer

Trick I learned:

object By {
  def apply[T <: Aggregate[T,K], K, G](key:G, value:T with Aggregate[T,K]):By[T, K, G] = By(HashMap(key -> value))
}

The with Aggregate[T,K] refinement somehow helps the compiler figure out the types.