Reputation: 3838
Considering the following scala program:
val arr: Seq[String] = Seq("abc", "def")
val cls = arr.head.getClass
println(cls)
val ttg: TypeTag[Seq[String]] = typeOf[Seq[String]]
val fns = ttg.tpe
.members
val fn = fns
.filter(_.name.toString == "head")
.head // Unsafely access it for now, use Option and map under normal conditions
.asMethod // Coerce to a method symbol
val fnTp = fn.returnType
println(fnTp)
val fnCls = ttg.mirror.runtimeClass(fnTp)
assert(fnTp == cls)
Since TypeTag has both Seq and String information, I would expect that fn.returnType
give the correct result "String", but in this case I got the following program output:
cls = class java.lang.String
fnTp = A
And subsequently throw this exception:
A needed class was not found. This could be due to an error in your runpath. Missing class: no Java class corresponding to A found
java.lang.NoClassDefFoundError: no Java class corresponding to A found
at scala.reflect.runtime.JavaMirrors$JavaMirror.typeToJavaClass(JavaMirrors.scala:1258)
at scala.reflect.runtime.JavaMirrors$JavaMirror.runtimeClass(JavaMirrors.scala:202)
at scala.reflect.runtime.JavaMirrors$JavaMirror.runtimeClass(JavaMirrors.scala:65)
Obviously type String was erased, leaving only a wildcard type 'A'
Why TypeTag is unable to yield the correct erased type as intended?
Upvotes: 1
Views: 353
Reputation: 14224
Seq.head
is defined as def head: A
. And fn
is just a method symbol of the method head
from a generic class Seq[A]
, it doesn't know anything about the concrete type. So its returnType
is exactly that A
just as defined in Seq
.
If you want to know what that A
would be in some concrete Type
, you'd have to specify that explicitly. For instance, you can use infoIn
on the method symbol:
scala> val fnTp = fn.infoIn(ttg.tpe)
fnTp: reflect.runtime.universe.Type = => String
scala> val fnRetTp = fnTp.resultType
fnRetTp: reflect.runtime.universe.Type = String
Upvotes: 3