Is there any way to create generic type trait without passing parameters

Question

I have trait like this I can create pipe objects like Pipe[String,Int], Pipe[String,String],Pipe[(String,Double),String]

trait Pipe[In,Out] {
  def apply(rdd: RDD[In]): RDD[Out]
}

Is there any way to create trait Pipe like this

trait Pipe {
  def apply[In,Out](rdd: RDD[In]): RDD[Out]
}

then I can use Pipe objects without giving any parameters.But I couldn't create this trait objects.

Answer 1

In order to produce an RDD[Out] from an RDD[In] you need to know how to transform one into the other. This transformation is what would hold the Out parameter, eg:

trait Pipe[In,Out] {
  def apply(rdd: RDD[In]): RDD[Out] = rddTransformer(rdd)
  private def rddTransformer: RDD[In] => RDD[Out]
}

If you know at compile time what Out type you want for some In types, then look into Functional Dependencies: https://milessabin.com/blog/2011/07/16/fundeps-in-scala/