类型不带参数Scala

Question

I have case class State and want to extend Variable class from it; but in Variable class only need value in constructor. Where can I put run function?

case class State[S, +A](run: S => (A, S)) {
//.....has `map` function
  def map[B, X >: State[S, B]](f: A => B): X =
    State(state => {
      val (a, s2) = run(state)
      (f(a), s2)
    })
}

class Variable[+A](value: A) extends State[A, A] { // ERROR

  def get: Variable[A] =
    map(x => x)
  def set(newValue: A): Variable[A] =
    map(_ => newValue)
}

UPDATE I've changed to something like this:

class Variable[+A](value: A, run: A => (A, A)) extends State[A, A](run) {
  def get: Variable[A] =
    map(x => x) // ERROR HERE
  def set(newValue: A): State[A, A] =
    map(_ => newValue)
}
object Variable {
  def create[A](value: A): Variable[A] = new Variable[A](value, x => (x, x))
}

But I've gotten error:

type mismatch; found : com.libs.State[A,A] required: com.libs.Variable[A] Variable.scala /scala/src/com/libs line 4 Scala Problem

Answer 1

The problem is that you cannot define a Variable using map , since map defines a State which is only a super-type of Variable . How can your program know how to set the additional information of your subclass, using only map ?

However, if you define type Variable[+A] = State[A, A] , so that it is not a subclass but an alias for the same class, you will have some variance errors, since State is invariant in its type parameter S , so Variable must be too.