How do I get around type erasure on Scala? Or, why can't I get the type parameter of my collections?

Question

It's a sad fact of life on Scala that if you instantiate a List[Int], you can verify that your instance is a List, and you can verify that any individual element of it is an Int, but not that it is a List[Int], as can be easily verified:

scala> List(1,2,3) match {
     | case l : List[String] => println("A list of strings?!")
     | case _ => println("Ok")
     | }
warning: there were unchecked warnings; re-run with -unchecked for details
A list of strings?!

The -unchecked option puts the blame squarely on type erasure:

scala>  List(1,2,3) match {
     |  case l : List[String] => println("A list of strings?!")
     |  case _ => println("Ok")
     |  }
<console>:6: warning: non variable type-argument String in type pattern is unchecked since it is eliminated by erasure
        case l : List[String] => println("A list of strings?!")
                 ^
A list of strings?!

Why is that, and how do I get around it?

Zarla

Scala 2.8 Beta 1 RC4 just made some changes to how type erasure works. I'm not sure if this directly affects your question. – Snider

That's just what types erasure to, that has changed. The short of it can be summed as "Proposal: The erasure of "Object with A" is "A" instead of "Object"." The actual specification is rather more complex. It's about mixins, at any rate, and this question is concerned about generics. – Zarla

Thanks for the clarification -- I'm a scala newcomer. I feel like right now is a bad time to jump into Scala. Earlier, I could have learnt the changes in 2.8 from a good base, later I'd never have to know the difference! – Snider

Here's a somewhat related question about TypeTags. – Airlee

Running scala 2.10.2, I saw this warning instead: <console>:9: warning: fruitless type test: a value of type List[Int] cannot also be a List[String] (but still might match its erasure) case list: List[String] => println("a list of strings?") ^ I find your question and answer to be very helpful, but I'm not sure if this updated warning is useful to readers. – Ornamentation

@KevinMeredith It really needs updating, though. :( – Zarla

Scala 2.12.0-M3 still have this warning <console>:13: warning: fruitless type test: a value of type List[Int] cannot also be a List[String] (the underlying of List[String]) (but still might match its erasure) case l : List[String] => println( "Int list") ^ Int list – Lunch

Answer 1

This answer uses the Manifest-API, which is deprecated as of Scala 2.10. Please see answers below for more current solutions.

Scala was defined with Type Erasure because the Java Virtual Machine (JVM), unlike Java, did not get generics. This means that, at run time, only the class exists, not its type parameters. In the example, JVM knows it is handling a scala.collection.immutable.List, but not that this list is parameterized with Int.

Fortunately, there's a feature in Scala that lets you get around that. It’s the Manifest. A Manifest is class whose instances are objects representing types. Since these instances are objects, you can pass them around, store them, and generally call methods on them. With the support of implicit parameters, it becomes a very powerful tool. Take the following example, for instance:

object Registry {
  import scala.reflect.Manifest
  
  private var map= Map.empty[Any,(Manifest[_], Any)] 
  
  def register[T](name: Any, item: T)(implicit m: Manifest[T]) {
    map = map.updated(name, m -> item)
  }
  
  def get[T](key:Any)(implicit m : Manifest[T]): Option[T] = {
    map get key flatMap {
      case (om, s) => if (om <:< m) Some(s.asInstanceOf[T]) else None
    }     
  }
}

scala> Registry.register("a", List(1,2,3))

scala> Registry.get[List[Int]]("a")
res6: Option[List[Int]] = Some(List(1, 2, 3))

scala> Registry.get[List[String]]("a")
res7: Option[List[String]] = None

When storing an element, we store a "Manifest" of it too. A Manifest is a class whose instances represent Scala types. These objects have more information than JVM does, which enable us to test for the full, parameterized type.

Note, however, that a Manifest is still an evolving feature. As an example of its limitations, it presently doesn't know anything about variance, and assumes everything is co-variant. I expect it will get more stable and solid once the Scala reflection library, presently under development, gets finished.

Zarla

I agree that answering these kinds of questions is a good idea : I had read about this somewhere before, but it is much easier to find on stack overflow. – Crosscountry

The get method can be defined as for ((om, v) <- _map get key if om <:< m) yield v.asInstanceOf[T]. – Gasify

@Aaron Very good suggestion, but I fear it might obscure the code for people relatively new to Scala. I wasn't very experience with Scala myself when I wrote that code, which was sometime before I put it in this question/answer. – Zarla

@MatthieuF Good question. It is no longer experimental, but it isn't complete yet either. Right now Odersky himself is working on a Scala reflection library. I expect manifests will evolve some out of that effort. I edited the answer to reflect (pun intended) the present status. – Zarla

Looking forward to a new TypeTag answer ;) – Overpass

@KimStebel You know that TypeTag are actually automatically used on pattern matching? Cool, eh? – Zarla

Cool! Maybe you should add that to the answer. – Overpass

When you call Registry.register(...), you neither pass in the implicit m: Manifest, nor creates any implicit local variable. — I take it that the compiler generates and "appends" that additional parameter? – Augean

To answer my own question just above: Yes, the compiler generates the Manifest param itself, see: https://mcmap.net/q/87967/-how-to-create-a-typetag-manually "the [manifest/type-tag] instance [...] is being created implicitly by the compiler" – Augean

This is supposed to be easier in Scala 2.10. I would like to see a comparison of both 2.10 and < 2.10. – Gravel

Answer 2

You can do this using TypeTags (as Daniel already mentions, but I'll just spell it out explicitly):

import scala.reflect.runtime.universe._
def matchList[A: TypeTag](list: List[A]) = list match {
  case strlist: List[String @unchecked] if typeOf[A] =:= typeOf[String] => println("A list of strings!")
  case intlist: List[Int @unchecked] if typeOf[A] =:= typeOf[Int] => println("A list of ints!")
}

You can also do this using ClassTags (which saves you from having to depend on scala-reflect):

import scala.reflect.{ClassTag, classTag}
def matchList2[A : ClassTag](list: List[A]) = list match {
  case strlist: List[String @unchecked] if classTag[A] == classTag[String] => println("A List of strings!")
  case intlist: List[Int @unchecked] if classTag[A] == classTag[Int] => println("A list of ints!")
}

ClassTags can be used so long as you don't expect the type parameter A to itself be a generic type.

Unfortunately it's a little verbose and you need the @unchecked annotation to suppress a compiler warning. The TypeTag may be incorporated into the pattern match automatically by the compiler in the future: https://issues.scala-lang.org/browse/SI-6517

Verrucose

What about removing unnecessary [List String @unchecked] as it does not add anything to this pattern match (Just using case strlist if typeOf[A] =:= typeOf[String] => will do it, or even case _ if typeOf[A] =:= typeOf[String] => if the bound variable is not needed in body of the case). – Terminal

I guess that would work for the given example but I think most real usages would benefit from having the type of the elements. – Verrucose

In the examples above, doesn't the unchecked part in front of the guard condition do a cast? Wouldn't you get a class cast exception when going through the matches on the first object that cant' be cast to a string? – Expense

Hm no I believe there is no cast before applying the guard - the unchecked bit is sort of a no-op until the code to the right of the => is executed. (And when the code on the rhs is executed, the guards provide a static guarantee on the type of the elements. There might be a cast there, but it's safe.) – Verrucose

Does this solution produce significant runtime overhead? – Parada

I'm going to venture the short answer is "no" (the long answer is probably "it depends" on what you mean by significant runtime overhead). Notably there is no reflection (which would probably count as significant runtime overhead). The classtag's are immediately provided by the compiler. So there's just the instanceof check for List (which happens to be redundant in this case) and the classTag equality check, which I imagine is fast. – Verrucose

the problem I see here is that if I want to call this definition with a List of an unknown type (aka List[_]) I get an "No ClassTag available for _" error. – Eads

@AndrewNorman Right. The OP asks why we can't write code that knows the type of A in List[A] when we clearly know its type at compile-time. For List[_] we're in a different situation: We don't know the type at compile-time. In that case, the best you can do is get an element, if one exists, and then do a runtime type check using isInstanceOf or pattern match case x: String => ... case y: Int => ... etc – Verrucose

Answer 3

You can use the Typeable type class from shapeless to get the result you're after,

Sample REPL session,

scala> import shapeless.syntax.typeable._
import shapeless.syntax.typeable._

scala> val l1 : Any = List(1,2,3)
l1: Any = List(1, 2, 3)

scala> l1.cast[List[String]]
res0: Option[List[String]] = None

scala> l1.cast[List[Int]]
res1: Option[List[Int]] = Some(List(1, 2, 3))

The cast operation will be as precise wrt erasure as possible given the in-scope Typeable instances available.

Absolutism

It should be noted that the "cast" operation will recursively go through the whole collection and its subcollections and check whether all involved value are of the right type. (I.e., l1.cast[List[String]] does roughly for (x<-l1) assert(x.isInstanceOf[String]) For large datastructures or if the casts happen very often, this may be an inacceptable overhead. – Arrange

Answer 4

I came up with a relatively simple solution that would suffice in limited-use situations, essentially wrapping parameterized types that would suffer from the type erasure problem in wrapper classes that can be used in a match statement.

case class StringListHolder(list:List[String])

StringListHolder(List("str1","str2")) match {
    case holder: StringListHolder => holder.list foreach println
}

This has the expected output and limits the contents of our case class to the desired type, String Lists.

More details here: http://www.scalafied.com/?p=60

Forbidden

Answer 5

There is a way to overcome the type erasure issue in Scala. In Overcoming Type Erasure in matching 1 and Overcoming Type Erasure in Matching 2 (Variance) are some explanation of how to code some helpers to wrap the types, including Variance, for matching.

Valedictory

This doesn't overcome type erasure. In his example, doing val x:Any = List(1,2,3); x match { case IntList(l) => println( s"Match ${l(1)}" ); case _ => println( s"No match" ) } produces "No match" – Centromere

Answer 6

I found a slightly better workaround for this limitation of the otherwise awesome language.

In Scala, the issue of type erasure does not occur with arrays. I think it is easier to demonstrate this with an example.

Let us say we have a list of (Int, String), then the following gives a type erasure warning

x match {
  case l:List[(Int, String)] => 
  ...
}

To work around this, first create a case class:

case class IntString(i:Int, s:String)

then in the pattern matching do something like:

x match {
  case a:Array[IntString] => 
  ...
}

which seems to work perfectly.

This will require minor changes in your code to work with arrays instead of lists, but should not be a major problem.

Note that using case a:Array[(Int, String)] will still give a type erasure warning, so it is necessary to use a new container class (in this example, IntString).

Meteoric

"limitation of the otherwise awesome language" it's less a limitation of Scala and more a limitation of the JVM. Perhaps Scala could have been designed to include type information as it ran on the JVM, but I don't think a design like that would have preserved interoperability with Java (i.e., as designed, you can call Scala from Java.) – Bolten

As a followup, support for reified generics for Scala in .NET/CLR is an ongoing possibility. – Bolten

Answer 7

Since Java does not know the actual element type, I found it most useful to just use List[_]. Then the warning goes away and the code describes reality - it is a list of something unknown.

Lilia

Answer 8

I'm wondering if this is a suited workaround:

scala> List(1,2,3) match {
     |    case List(_: String, _*) => println("A list of strings?!")
     |    case _ => println("Ok")
     | }

It does not match the "empty list" case, but it gives a compile error, not a warning!

error: type mismatch;
found:     String
requirerd: Int

This on the other hand seems to work....

scala> List(1,2,3) match {
     |    case List(_: Int, _*) => println("A list of ints")
     |    case _ => println("Ok")
     | }

Isn't it kinda even better or am I missing the point here?

Nuris

Doesn't work with List(1, "a", "b"), which has type List[Any] – Herbie

Although sullivan's point is correct and there are related problems with inheritance, I still found this useful. – Encompass

Answer 9

Not a solution but a way to live with it without sweeping it under the rug altogether: Adding the @unchecked annotation. See here - http://www.scala-lang.org/api/current/index.html#scala.unchecked

Melanoid

Answer 10

I wanted to add an answer which generalises the problem to: How do a get a String representation of the type of my list at runtime

import scala.reflect.runtime.universe._

def whatListAmI[A : TypeTag](list : List[A]) = {
    if (typeTag[A] == typeTag[java.lang.String]) // note that typeTag[String] does not match due to type alias being a different type
        println("its a String")
    else if (typeTag[A] == typeTag[Int])
        println("its a Int")

    s"A List of ${typeTag[A].tpe.toString}"
}

val listInt = List(1,2,3)
val listString = List("a", "b", "c")

println(whatListAmI(listInt))
println(whatListAmI(listString))

Unlace

Answer 11

Using pattern match guard

    list match  {
        case x:List if x.isInstanceOf(List[String]) => do sth
        case x:List if x.isInstanceOf(List[Int]) => do sth else
     }

Orgell

The reason why this one will not work is that isInstanceOf does a runtime check based on the type information available to the JVM. And that runtime information will not contain the type argument to List (because of type erasure). – Arrange