Scala: abstract type pattern A is unchecked since it is eliminated by erasure

问题

I am writing the function that can catch exceptions of the certain type only.

def myFunc[A <: Exception]() {
    try {
        println("Hello world") // or something else
    } catch {
        case a: A => // warning: abstract type pattern A is unchecked since it is eliminated by erasure
    }
}

What is the corrent way to bypass jvm type erasure in such case?

回答1:

You could use ClassTag like in this answer.

But I'd prefer this approach:

def myFunc(recover: PartialFunction[Throwable, Unit]): Unit = {
  try {
    println("Hello world") // or something else
  } catch {
    recover
  }
}

Usage:

myFunc{ case _: MyException => }

Using ClassTag:

import scala.reflect.{ClassTag, classTag}

def myFunc[A <: Exception: ClassTag](): Unit = {
  try {
    println("Hello world") // or something else
  } catch {
    case a if classTag[A].runtimeClass.isInstance(a) =>
  }
}

Note also that in general you should use Try with recover method: Try will catch only NonFatal exceptions.

def myFunc(recover: PartialFunction[Throwable, Unit]) = {
  Try {
    println("Hello world") // or something else
  } recover {
    recover
  }.get // you could drop .get here to return `Try[Unit]`
}

回答2:

For every type check (e.g. case a: A) the JVM needs the corresponding class object to perform the check. In your case the JVM does not have the class object, because A is a variable type parameter. However, you can add additional information about A by implicitly passing a Manifest[A] to myFunc. As a shorthand you can just add : Manifest to your type declaration of A:

def myFunc[A <: Exception : Manifest]() {
    try {
        println("Hello world") // or something else
    } catch {
        case a: A => // warning: abstract type pattern A is unchecked since it is eliminated by erasure
    }
}

来源：https://stackoverflow.com/questions/21180973/scala-abstract-type-pattern-a-is-unchecked-since-it-is-eliminated-by-erasure