I've created a Kotlin equivalent of TypeReference<T> like so:
abstract class TypeReference<T> : Comparable<T> {
val type: Type get() = getGenericType()
val arguments: List<Type> get() = getTypeArguments()
final override fun compareTo(other: T): Int {
return 0
}
private fun getGenericType(): Type {
val superClass = javaClass.genericSuperclass
check(superClass !is Class<*>) {
"TypeReference constructed without actual type information."
}
return (superClass as ParameterizedType).actualTypeArguments[0]
}
private fun getTypeArguments(): List<Type> {
val type = getGenericType()
return if (type is ParameterizedType) {
type.actualTypeArguments.toList()
} else emptyList()
}
}
In order to obtain Class<*> of the generic type and its arguments, I've also created the following extension function (and this is where I believe the problem lies, since this is where the stack trace fails).
fun Type.toClass(): Class<*> = when (this) {
is ParameterizedType -> rawType.toClass()
is Class<*> -> this
else -> Class.forName(typeName)
}
I'm unit testing this like so:
@Test
fun `TypeReference should correctly identify the List of BigDecimal type`() {
// Arrange
val expected = List::class.java
val expectedParameter1 = BigDecimal::class.java
val typeReference = object : TypeReference<List<BigDecimal>>() {}
// Act
val actual = typeReference.type.toClass()
val actualParameter1 = typeReference.arguments[0].toClass()
// Assert
assertEquals(expected, actual)
assertEquals(expectedParameter1, actualParameter1)
}
The problem I think, lies in the extension function else -> Class.forName(typeName) as it throws:
java.lang.ClassNotFoundException: ? extends java.math.BigDecimal
Is there a better way to obtain the Class<*> of a Type, even when they're generic type parameters?