How to get a type parameter's class name?-CodePudding

I have a base class:

abstract class JSONDeserializationStrategy<T : Any>: DeserializationStrategy<T> {
    protected abstract fun parse(json: JsonObject): T

    protected abstract fun getSerializationException(): SerializationException

}

and then a derived class

class MyClassParserDeserializationStrategy : JSONDeserializationStrategy<MyClass>() {

    override val descriptor: SerialDescriptor 
               = buildClassSerialDescriptor("MyClass")

    override fun getSerializationException(): SerializationException 
               = throw SerializationException("Invalid JSON received for MyClass.")

How could I move the property descriptor and the method getSerializationException from the derived class into the base class, since they only "adapt" by providing their name as String? I was trying to do something in the direction of T::class.java.simpleName as String but it didnt work. What is the best way to do this?

CodePudding user response：

As @Tenfour04 explained, T is erased, so it is not directly accessible. However, as long as the subclass of JSONDeserializationStrategy provides the T as a specific class/type, it can be acquired with a bit of reflection voodoo:

fun main() {
    val strategy = MyClassParserDeserializationStrategy()
    println(strategy.descriptor.serialName)  // MyClass
}

abstract class JSONDeserializationStrategy<T : Any>: DeserializationStrategy<T> {
    protected val type: KType = this::class.supertypes
        .first { it.classifier == JSONDeserializationStrategy::class }
        .arguments[0].type!!

    @Suppress("UNCHECKED_CAST")
    protected val typeClass = requireNotNull(type.classifier as? KClass<T>) {
        "T is unknown"
    }

    override val descriptor: SerialDescriptor = buildClassSerialDescriptor(typeClass.simpleName!!)

    fun getSerializationException(): SerializationException =
        throw SerializationException("Invalid JSON received for ${typeClass.simpleName!!}.")

    ...
}

I'm not 100% sure, but I believe both type!! and simpleName!! are safe, they can't be null.

Also, it won't work if T is really fully erased and unknown, e.g.:

val strategy = GenericParserDeserializationStrategy<MyClass>() // exception

For this reason it makes sense to open type/typeClass properties for overriding, so generic non-abstract subclasses could provide their own means to acquire T. However, then we would probably need to move the initialization of most of properties outside of the constructor.

CodePudding user response：

Because of type erasure, T's class is not accessible. Work-around could be to add it as a constructor property that returns the type, and then the subclasses must pass the type. The property needs to be in the constructor, rather than provided as an abstract property for subclasses to override because you need it to initialize descriptor at instantiation time. (It's highly discouraged to call an open property at class initialization time.)

abstract class JSONDeserializationStrategy<T : Any>(protected val typeClass: KClass<out T>): DeserializationStrategy<T> {
    protected abstract fun parse(json: JsonObject): T

    override val descriptor: SerialDescriptor = buildClassSerialDescriptor(typeClass.simpleName!!)

    fun getSerializationException(): SerializationException =
        throw SerializationException("Invalid JSON received for ${typeClass.simpleName}.")
}

class TodaySaleParserDeserializationStrategy : JSONDeserializationStrategy<TodaySale>(TodaySale::class) {

}