How can I obtain the default value for a type in Scala?

Question

I'm trying to write a Scala function that returns the default value of a type (0, 0.0, false, '\0', etc. for value types and null for reference types). I came up with this:

def defaultValue[U]: U = {
  class Default[U] { var default: U = _ }
  new Default[U].default
}

and while this works well if called directly, it returns null even for value types when called through a function that itself is generic, as shown in this REPL session:

Welcome to Scala version 2.8.1.final (Java HotSpot(TM) 64-Bit Server VM, Java 1.6.0_24).
Type in expressions to have them evaluated.
Type :help for more information.

scala> def defaultValue[U]: U = { class Default[U] {var default: U = _ }; new Default[U].default }
defaultValue: [U]U

scala> defaultValue[Boolean] // direct call works
res0: Boolean = false

scala> var res: Any = 0
res: Any = 0

scala> def setRes[U] = { res = defaultValue[U]; defaultValue[U] }
setRes: [U]U

scala> setRes[Boolean] // returns a Boolean, but...
res1: Boolean = false

scala> res
res2: Any = null // ... sets the res variable to null.

Can someone explain to me:

1. why this happens (and why the compiler/interpreter doesn't complain if there is not enough information for it to return a true Boolean); and
2. how I can fix it?

Solution 1

You can create your own Default type-class to handle this. Here is what the code looks like. I added special handling for scala collections that returns an empty collection instead of null.

import scala.collection.immutable

class Default[+A](val default: A)

trait LowerPriorityImplicits {
  // Stop AnyRefs from clashing with AnyVals
  implicit def defaultNull[A <: AnyRef]:Default[A] = new Default[A](null.asInstanceOf[A])  
}

object Default extends LowerPriorityImplicits {
  implicit object DefaultDouble extends Default[Double](0.0)
  implicit object DefaultFloat extends Default[Float](0.0F)
  implicit object DefaultInt extends Default[Int](0)
  implicit object DefaultLong extends Default[Long](0L)
  implicit object DefaultShort extends Default[Short](0)
  implicit object DefaultByte extends Default[Byte](0)
  implicit object DefaultChar extends Default[Char]('\u0000')
  implicit object DefaultBoolean extends Default[Boolean](false)
  implicit object DefaultUnit extends Default[Unit](())

  implicit def defaultSeq[A]: Default[immutable.Seq[A]] = new Default[immutable.Seq[A]](immutable.Seq())
  implicit def defaultSet[A]: Default[Set[A]] = new Default[Set[A]](Set())
  implicit def defaultMap[A, B]: Default[Map[A, B]] = new Default[Map[A, B]](Map[A, B]())
  implicit def defaultOption[A]: Default[Option[A]] = new Default[Option[A]](None)

  def value[A](implicit value: Default[A]): A = value.default
}

These are the results of using this in the repl. Notice that the default value for String can be overriden by creating a new implicit Default[String].

scala> Default.value[Int]
res0: Int = 0

scala> Default.value[Boolean]
res1: Boolean = false

scala> Default.value[String]
res2: String = null

scala> Default.value[Set[Int]]
res3: Set[Int] = Set()

scala> Default.value[immutable.Seq[Int]]
res4: scala.collection.immutable.Seq[Int] = List()

scala> Default.value[String]
res5: String = null

scala> Default.value[AnyRef]
res6: AnyRef = null

scala> implicit val emptyStringAsDefault:Default[String] = new Default[String]("")
emptyStringAsDefault: Default[String] = Default@7d78d7b4

scala> Default.value[String]
res7: String = ""

Solution 2

Here is a more condensed version of your issue:

scala> defaultValue[Boolean]: Any
res0: Any = null

scala> defaultValue[Boolean]: Boolean
res1: Boolean = false

The first version is what applies when you call res = defaultValue[U] because even though U is of type Boolean, res is of type Any

If you compile this little program using the -Xprint:all option

object Test {
  def defaultValue[U]: U = { class Default[U] {var default: U = _ }; new Default[U].default }

  def main(args:Array[String]) {
    val any = defaultValue[Boolean]: Any
    println(any)
    val bool = defaultValue[Boolean]: Boolean
    println(bool)
  }
}

You'll see that right before the erasure phase, you have:

val any: Any = (Test.this.defaultValue[Boolean](): Any);
scala.this.Predef.println(any);
val bool: Boolean = (Test.this.defaultValue[Boolean](): Boolean);
scala.this.Predef.println(bool)

Then at the end of the erasure phase:

val any: java.lang.Object = (Test.this.defaultValue(): java.lang.Object);
scala.this.Predef.println(any);
val bool: Boolean = (scala.Boolean.unbox(Test.this.defaultValue()): Boolean);
scala.this.Predef.println(scala.Boolean.box(bool))

So what happens is that under the hood defaultValue[Boolean] returns null in both cases, but then null is unboxed into false when the return type is a Boolean. You can verify that in the REPL:

scala> Boolean.unbox(null)
res0: Boolean = false

scala> null.asInstanceOf[Boolean]
res1: Boolean = false

Edit: I had an idea - not that I'm recommending it. Not sure what your use case is (res = false seems easier to me..)

scala> def f[@specialized U] = { class X { var x: U = _ }; (new X).x }
f: [U]U

scala> var res: Any = _
res: Any = null

scala> def g[@specialized U] = { res = f[U]; f[U] }
g: [U]U

scala> g[Boolean]
res0: Boolean = false

scala> res
res1: Any = false

Solution 3

I know there is already "best answer", but what about the really simple:

def defaultValue[U: ClassManifest]: U = new Array[U](1)(0)

It seems to work, although it's somewhat expensive due to creating a temporary array object. Does anyone know of any cases where it gives the wrong value?

I was looking for a "cheaper" alternative, but this question tells me there probably isn't one.

Solution 4

For the record, here's the only I've found (yet) to make this work reliably. Improvements are welcome.

def defaultValue[T: ClassManifest]: T = classManifest[T].erasure.toString match {
  case "void" => ().asInstanceOf[T]
  case "boolean" => false.asInstanceOf[T]
  case "byte" => (0: Byte).asInstanceOf[T]
  case "short" => (0: Short).asInstanceOf[T]
  case "char" => '\0'.asInstanceOf[T]
  case "int" => 0.asInstanceOf[T]
  case "long" => 0L.asInstanceOf[T]
  case "float" => 0.0F.asInstanceOf[T]
  case "double" => 0.0.asInstanceOf[T]
  case _ => null.asInstanceOf[T]
}

I'm aware that I get null even if T <: NotNull, which is a problem. Then again, there is a problem with initialization of vars with _ for NotNull subclasses.

Solution 5

I have written a blog post on building a defaulting mechanism for Scala. You can find it here.

If you do not want Option[_] to default to None, String to "" etc then get rid of the respective implicits from the object Default.

View more solutions

Comments

[Jean-Philippe Pellet]

Thanks, that's a very nice explanation for the Why. Now why can't the compiler tell me more at compile time? Like, look man, you want a Boolean, but here I may give you null, sorry?

[huynhjl]

Not sure there are other use cases of def f[T]: T other than the one you have. It's hard to materialize a T without any passed parameter. So may be there is no warning because nobody ever thought of doing that.

[Jean-Philippe Pellet]

The @specialized trick is nice, but dangerous because it breaks down as soon as you wrap the call to f from a context which is not specialized. This could cause bugs that are hard to track down.

[Puterdo Borato]

So simply calling null.asInstanceOf[B] should perfectly do the job.