Merging multiple case (in match/case) in Scala - scala

Question

I have the code that instance.get returns value, and based on the type I process accordingly.
instance.get match {
case v:Range => {
val sizeInBytes = util.conversion.Util.getBytesForBits(v.size)
val value = v.decode(contentByteArray.slice(index, index + sizeInBytes))
index += sizeInBytes
res(key) = value
}
case v:Encoding => {
val sizeInBytes = util.conversion.Util.getBytesForBits(v.size)
val value = v.decode(contentByteArray.slice(index, index + sizeInBytes))
index += sizeInBytes
res(key) = value
}
...
}
In the code, I have duplication for the Range and Encoding type. How can I merge the two cases?
I tried the | operator, but it doesn't work.
case v:Range | v:Encoding

Answer 1

This can't work, because Range.size and Encoding.size are two completely different methods despite the fact that they are named the same. And same is true for Range.decode and Edncoding.decode.
So, when you write v.size, the type of v has to be known, it has to be either v:Encoding or v:Range, not v:Encoding|v:Range.
How to fix this? Make a common trait like this:
trait SomethingWithDecodeAndSize {
def size: Int
def decode(bytes: Array[Byte]): Whatever
}
And then, change the definitions of Range and Encoding:
class Range extends SomethingWithDecodeAndSize { ... }
class Encoding extends SomethingWithDecodeAndSize { ... }
Now you can just do case v: SomethingWithDecodeAndSize => ... in your match clause.
Also ... Don't do instance.get, that's bad taste. Do instead
instance match {
Some(v: SomethingWithDecodeAndSize) => ...
}
Update
If you cannot modify the definitions of the original classes, you can use an extractor:
object SomethingWithDecodeAndSize {
def unapply(a: Any): Option[SomethingWithDecodeAndSize] = a match {
case r: Range => Some(new SomethingWithDecodeAndSize {
def size = r.size
def decode(bytes: Array[Byte]) = r.decode(bytes)
})
case r: Encoding => Some(new SomethingWithDecodeAndSize {
def size = r.size
def decode(bytes: Array[Byte]) = r.decode(bytes)
})
case _ => None
}
}
Now, you can do case Some(SomethingWithDecodeAndSize(v)) => ... in your match.

Answer 2

An alternate solution to #Dima's in case you can't change definition of Range and Encoding (and there is no supertype with required methods):
trait RangeOrEncoding {
def size: Int
def decode(bytes: Array[Byte]): Whatever
}
implicit def liftRange(r: Range): RangeOrEncoding = new RangeOrEncoding {
def size = r.size
def decode(bytes: Array[Byte]) = r.decode(bytes)
}
// similar conversion for Encoding
// can also be a local def
private def handleRangeOrEncoding(v: RangeOrEncoding) = {
val sizeInBytes = util.conversion.Util.getBytesForBits(v.size)
val value = v.decode(contentByteArray.slice(index, index + sizeInBytes))
index += sizeInBytes
res(key) = value
}
instance match {
case Some(v: Range) => handleRangeOrEncoding(v)
case Some(v: Encoding) => handleRangeOrEncoding(v)
...
}

Answer 3

I remember the cheerleaders in high school asking us, "How loose is your goose?"
scala> class C { def f(i: Int) = 2 * i }
defined class C
scala> class D { def f(i: Int) = 3 * i }
defined class D
scala> def test(x: Any) = x match { case y: { def f(i: Int): Int } => y.f(42) }
<console>:11: warning: a pattern match on a refinement type is unchecked
def test(x: Any) = x match { case y: { def f(i: Int): Int } => y.f(42) }
^
warning: there was one feature warning; re-run with -feature for details
test: (x: Any)Int
scala> test(new C)
res0: Int = 84
scala> test(new D)
res1: Int = 126
scala> test(42)
java.lang.NoSuchMethodException: java.lang.Integer.f(int)
at java.lang.Class.getMethod(Class.java:1786)
at .reflMethod$Method1(<console>:11)
at .test(<console>:11)
... 32 elided
I believe the answer was: "Loose, baby, loose."
Edit:
scala> import reflect.runtime._,universe._,language.reflectiveCalls
import reflect.runtime._
import universe._
import language.reflectiveCalls
scala> class C { def f(i: Int) = 2 * i }
defined class C
scala> class D { def f(i: Int) = 3 * i }
defined class D
scala> def f[A](a: A)(implicit tt: TypeTag[A]) = a match {
| case b: { def f(i: Int): Int }
| if tt.tpe <:< typeOf[{ def f(i: Int): Int }] =>
| b.f(42)
| }
<console>:19: warning: a pattern match on a refinement type is unchecked
case b: { def f(i: Int): Int }
^
f: [A](a: A)(implicit tt: reflect.runtime.universe.TypeTag[A])Int
scala> f(new C)
res0: Int = 84
scala> f(new D)
res1: Int = 126
scala> f(3) // now an ordinary MatchError
scala.MatchError: 3 (of class java.lang.Integer)
at .f(<console>:18)
... 32 elided
So you can express it as an ordinary type bounds:
scala> def f[A <: { def f(i: Int): Int }](a: A) = a.f(42)
f: [A <: AnyRef{def f(i: Int): Int}](a: A)Int
scala> f(new C)
res3: Int = 84
scala> f(17)
<console>:20: error: inferred type arguments [Int] do not conform to method f's type parameter bounds [A <: AnyRef{def f(i: Int): Int}]
f(17)
^
<console>:20: error: type mismatch;
found : Int(17)
required: A
f(17)
^
You still need to accept the cost of the reflective call, of course.