Question

As a test, I wrote this code:

object Ambig extends App {
  def f( x:Int    ) { println("Int"   ) }
  def f( x:String ) { println("String") }
  f( null.asInstanceOf[Int   ] )
  f( null.asInstanceOf[String] )
  f(null)
}

I was expecting to get an error on that last invocation of f(), saying that it was ambiguous. The compiler accepted it, and produced this output:

Int
String
String

Now I’m guessing that this has to do with the fact that Int is not an AnyRef, so the only version of f that works for f(null) is f(x:String). But then, if an Int can’t be null, what does null.asInstanceOf[Int] mean? The repl says it’s of type Int:

scala> :type null.asInstanceOf[Int]
Int

but I don’t really see how that works. After all, if I try to cast a String to an Int, all hell breaks loose:

scala> "foo".asInstanceOf[Int]
java.lang.ClassCastException: java.lang.String cannot be cast to java.lang.Integer
    at scala.runtime.BoxesRunTime.unboxToInt(Unknown Source)
        ...

Of course that’s to be expected — “foo” can’t be made into an Int. But neither can null, so why does casting null to an Int work? Presumably boxing in some form, but the type is still Int, which can’t be null…

What am I missing?

Answer 1

The behaviour of casting null to an Int depends on the context in which it is done.

First of all, if you cast a null to an Int, it actually means a boxed integer, whose value is null. If you put the expression in a context where the expected type is Any (which is translated to Object behind the scene, because in the JVM bytecode, there is no way to refer to a primitive type and a reference type with the same reference), then this value is not converted further – that is why println(null.asInstanceOf[Int]) prints null.

However, if you use this same boxed integer value in a context where a primitive Int (Java int) is expected, it will be converted to a primitive, and null is (as a default value for reference types) converted to 0 (a default value for primitive types).

If a generic method does this cast, then, naturally, you get a null back.

However, if this method is specialized, then its return type is Int (which is a primitive integer in this case), so the null: Any value has to be converted to a primitive, as before.

Hence, running:

object Test extends App {
  println(null.asInstanceOf[Int])

  def printit(x: Int) = println(x)

  printit(null.asInstanceOf[Int])

  def nullint[T] = null.asInstanceOf[T]

  println(nullint[Int])

  def nullspecint[@specialized(Int) T] = null.asInstanceOf[T]

  println(nullspecint[Int])
}

produces:

null
0
null
0