为什么Scala模式匹配不等于isInstanceOf

Question

I believed for a long time that this two constructions is equivalent:

if (myVar.isInstanceOf[MyType]) myVar.asInstanceOf[MyType].doSomething

and

myVar match {
  case my : MyType => my.doSomething
  case _ => {}
}

But suddenly I've found that I get type error while trying to match Number value to the Double type, but asInstanceOf[Double] works fine. WTF is happening?

---

simple example for scala REPL

val d = 3.5
val n : Number = d
n.isInstanceOf[Double]

works fine:

Boolean = true

but

n match {
  case x : Double => println("double")
  case _ => println("not a double")
}

produces type error:

:11: error: pattern type is incompatible with expected type;
found   : Double
required: Number
          case x : Double => println("double")

Answer 1

scala.Double is not inherited from java.lang.Number but from AnyVal . You want to match on java.lang.Double :

n match {
  case x : java.lang.Double => println("double")
  case _                    => println("not a double")
}

When using

val d = 3.5
val n : Number = d  // implicit conversion from scala.Double to java.lang.Double

scala.Double is implicitly converted to java.lang.Double during assignment to n