Reputation: 6852
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?
Upvotes: 38
Views: 15981
Reputation: 951
First, we all agree that we cannot assign null
to scala.Int
as documented in http://www.scala-lang.org/api/current/index.html#scala.Null
Second, why when we do println(null.asInstanceOf[Int])
, it gives null
?
This is because of the implementation of println. It eventually calls java String.valueOf
method, which is
return (obj == null) ? "null" : obj.toString();
If you do a null.asInstanceOf[Int] == null
in the shell, it will return true, but it gives an opposite warning that "comparing values of types Int and Null using `==' will always yield false". I think this might be a problem in scala's type erasure.
Doing a println
only needs a scala.Any type, so the casting of null.asInstanceOf[Int]
actually has not happen yet. So we just need to remember that when you assign null.asInstanceOf[Int]
to an Int, the cast happens at runtime based on Scala's erasure semantics, and it assigns 0 to it.
By the way, you can still do a f(null) without any compilation error because scala is doing an implicit conversion for you
null -> java.lang.Integer -> scala.Int
However, you will see it blows up at run time.
Upvotes: 0
Reputation: 297225
Here's the thing: asInstanceOf
doesn't have to make sense. What this method does is to tell the compiler to STOP MAKING SENSE, and trust what you are saying.
Now, if you want to know why it returns 0, that's because asInstanceOf
works on AnyRef
, not on AnyVal
. When applied to an AnyVal
, it will use the boxed version instead, and a boxed null
has value 0.
Upvotes: 21
Reputation: 32335
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
Upvotes: 54
Reputation: 52691
It looks like it's just automatically converting it to zero:
scala> null.asInstanceOf[Int]
res0: Int = 0
And, of course, 0, unlike null, can be an Int
.
Upvotes: 4