存在类型的Scala类型推断

Question

Consider the following code snippet, which is a reduced version of my original problem:

case class RandomVariable[A](values: List[A])
case class Assignment[A](variable: RandomVariable[A], value: A)

def enumerateAll(vars: List[RandomVariable[_]], evidence: List[Assignment[_]]): Double = 
  vars match {
    case variable :: tail =>
      val enumerated = for {value <- variable.values
        extendedEvidence = evidence :+ Assignment(variable, value)
      } yield enumerateAll(tail, extendedEvidence)
      enumerated.sum
    case Nil => 1.0
  }

This fails with the compile-time error that variable was inferred to have type RandomVariable[_0] when Assignment required type Any . Why is value not also inferred to have type _0 ? I tried giving the existential type a name in order to give a hint to the compiler by using case (variable: RandomVariable[T forSome {type T}]) :: tail => but that also would not compile (saying it could not find type T, which I'd be interested in an explanation for as well).

For further motivation, consider when we capture the type parameter as follows:

case variable :: tail =>
  def sum[A](variable: RandomVariable[A]): Double = {
    val enumerated = for {value <- variable.values
      extendedEvidence = evidence :+ Assignment(variable, value)
      } yield enumerateAll(tail, extendedEvidence)
    enumerated.sum
  }
  sum(variable)

This compiles without warnings/errors. Is there something I can modify in the first example to not require this extra function?

EDIT : To be more explicit, I want to know why value is not inferred to be of type _0 even though variable is of type _0 and every value comes from a List[_0] in variable . Also I would like to know if there are any additional ways to tell the compiler of this fact (aside from capturing the type in a function as I gave above).

Answer 1

Another compiling solution, that is cleaner(?) than using a function to capture the type. However, it makes it even more puzzling as to why type inference fails in the original case.

def enumerateAll(vars: List[RandomVariable[_]], evidence: List[SingleAssignment[_]]): Double = vars match {
  case (variable@RandomVariable(values)) :: tail =>
    val enumeration = for {value <- values
      assignment = SingleAssignment(variable, value)
      extendedEvidence = evidence :+ assignment
    } yield enumerateAll(tail, extendedEvidence)
    enumeration.sum
  case Nil => 1.0
}

It also returns the following warning:

scala: match may not be exhaustive.
It would fail on the following input: List((x: questions.RandomVariable[?] forSome x not in questions.RandomVariable[?]))
  def enumerateAll(vars: List[RandomVariable[_]], evidence: List[SingleAssignment[_]]): Double = vars match {

Which I'm unable to decipher as of this posting. Also, running it with a few test cases produces the desired result without a match error using RandomVariable s of int, double, and string in the parameter list.

Answer 2

Shouldn't you bind the types of RandomVariable and Assignment together?

 def [A] enumerateAll(vars: List[RandomVariable[A]], evidence: List[Assignment[A]]): Double = 

actually, you can be more permissive and just say

 def [A] enumerateAll(vars: List[RandomVariable[A]], evidence: List[Assignment[_ <: A]]): Double = 

Answer 3

The error code gives some indication of a solution.

<console>:15: error: type mismatch;
 found   : RandomVariable[_0] where type _0
 required: RandomVariable[Any]
Note: _0 <: Any, but class RandomVariable is invariant in type A.
You may wish to define A as +A instead. (SLS 4.5)
           extendedEvidence = evidence :+ Assignment(variable, value)

It's telling you that it saw a more specific type than it inferred and even suggested to make RandomVariable allow a covariant A. This would allow it to vary the type downward when required.

case class RandomVariable[+A](values: List[A])

Alternatively, you can explicitly set the generic type in enumerateAll for both parameters. In that way it can infer the appropriate type instead of being forced to infer Any. This definition doesn't require the RandomVariable covariant change as both parameters are of the same type.

def enumerateAll[A](vars: List[RandomVariable[A]], evidence: List[Assignment[A]]): Double = 

This question may help with the explanation. Why doesn't the example compile, aka how does (co-, contra-, and in-) variance work?