How can I get the concrete type of a super class type parameters from a subclass?
Assuming that I have a generic super class, an intermediate class and a subclass as follow:
class SuperClass[A, B]
class IntClass[A] extends SuperClass[A, Int]
class MyClass extends SuperClass[String, Int]
I'd like a function "getTypeArgs" that returns type parameters of SuperClass:
val superClassType = typeOf[SuperClass[_, _]]
val superClassOfStringInt = appliedType(superClassType.typeConstructor, typeOf[String], typeOf[Int])
getTypeArgs(superClassOfStringInt, superClassType)
// [TEST1] should return List(String, Int)
getTypeArgs(typeOf[MyClass], superClassType)
// [TEST2] should return List(String, Int)
val intClassOfLong = appliedType(typeOf[IntClass[_]].typeConstructor, typeOf[Long])
getTypeArgs(intClassOfLong, superClassType)
// [TEST3] should return List(Long, Int)
I've tried the solution In Scala Reflection, How to get generic type parameter of a concrete subclass?:
def getTypeArgs(t: Type, fromType: Type): List[Type] = {
internal
.thisType(t.dealias.typeSymbol)
.baseType(fromType.typeSymbol.asClass)
.typeArgs
}
It works for TEST2, but TEST1 returns List(A, B) and TEST2 returns List(A, Int).
I can fix the TEST1 by adding a test to check equality of symbols:
def getTypeArgs(t: Type, fromType: Type): List[Type] = {
if (t.erasure.typeSymbol == fromType.typeSymbol)
t.typeArgs
else
internal
.thisType(t.dealias.typeSymbol)
.baseType(fromType.typeSymbol.asClass)
.typeArgs
}
I don't know how to make TEST2 work.
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