Scala：定义类型的构造函数中可以有协方差吗？

Question

I have the class Variable[X <: SeqVal[_]](initialState:Calc[X])

which I instantiate with new Variable[SeqVal[Float]](Max()) where Max is

case class Max(seq: Int = 0, value: Float = .0f) extends SeqVal[Float] with Calc[SeqVal[Float]] , and there are other case classes other than Max .

This does not compile even though Max does implement a variant of the trait Calc[SeqVal[_]] .

[error] ../Variable.scala:14: type mismatch;
[error]  found   : com.quasiquant.calc.Max
[error]  required: com.quasiquant.calc.Calc[com.quasiquant.calc.Price]
[error] Note: com.quasiquant.messages.SeqVal[Float] >: com.quasiquant.calc.Price (and com.quasiquant.calc.Max <: com.quasiquant.calc.Calc[com.quasiquant.messages.SeqVal[Float]]), but trait Calc is invariant in type X.
[error] You may wish to define X as -X instead. (SLS 4.5)
[error]   extends Variable[Price](Max(), initChildren)

I need help in trying to work out how I can change the bounding of initialState:Calc[X] so the initialState can be set to anything that implements Calc[X] (not just Max ). I would prefer it if i didn't have to add a second type parameter to all the instantiations of Variable

Answer 1

Here's the code I tried compiling:

trait SeqVal[T]
trait Calc[T]

class Variable[X <: SeqVal[_]](initialState: Calc[X])

case class Max(seq: Int = 0, value: Float = .0f)
    extends SeqVal[Float] with Calc[SeqVal[Float]]

object test {
    new Variable[SeqVal[Float]](Max())
}

Compiles without errors.