Overriding unapply method

Question

I have a case from a library class and I want to override unapply method to reduce the number of parameters I need to pass to

Answer 1

I would ditch the overloaded unapply and just use the following for the match:

a match {
  case MyClass(x, _, _, _) => x  // Result is: 1
  case _ => "no"
}

EDIT :

If you really want to avoid the extra underscores, I think you'll need to look at something like:

a match {
  case x:MyClass => x.a
  case _ => "no"
}

Answer 2

case class MyClass(a: Int, b: String, c: String, d: Double /* and many more ones*/)
object MyClassA {
   def unapply(x: MyClass) = Some(x.a)
}

val a = new MyClass(1, "2", "3", 55.0 /* and many more ones*/)

a match {
    case MyClassA(2) => ??? // does not match
    case MyClassA(1) => a   // matches
    case _ => ??? 
}

You cannot define your custom unapply method in the MyClass object, because it would have to take a MyClass parameter, and there's already one such method there – one generated automatically for the case class. Therefore you have to define it in a different object (MyClassA in this case).

Pattern matching in Scala takes your object and applies several unapply and unapplySeq methods to it until it gets Some with values that match the ones specified in the pattern.
MyClassA(1) matches a if MyClassA.unapply(a) == Some(1).

Note: if I wrote case m @ MyClassA(1) =>, then the m variable would be of type MyClass.

Edit:

a match {
    case MyClassA(x) => x  // x is an Int, equal to a.a
    case _ => ??? 
}