Create an instance from a generic parameter

Question

I want to refactor the following code so that I can reuse it with different ViewHolder types:

class TestsAdapter(
    private val clickListener: ClickListener
) : PagedListAdapter<RecyclerViewTestInfo, TestsViewHolder>(diffCallback) {

    override fun getItemCount(): Int {
        return super.getItemCount()
    }

    override fun onBindViewHolder(holder: TestsViewHolder, position: Int) {
        val testInfo = getItem(position) as RecyclerViewTestInfo

        with(holder) {
            bindTo(testInfo)
            testInfo.let {
                itemView.setOnClickListener {
                    clickListener(testInfo)
                }
            }
        }
    }

    override fun onCreateViewHolder(parent: ViewGroup, viewType: Int): TestsViewHolder =
        TestsViewHolder(parent)
}

It isn't clear to me though how you handle the creation of an instance for a generic type. In the code, onCreateViewHolder is initialized with a specific ViewHolder. How would I do this with a generic?

Daniel · Accepted Answer · 2019-12-13 10:41:40Z

The problem you are facing here is that you can't directly initialize an instance from a generic type, you need to have at least a Class object. There is a solution, however I wouldn't recommend to use it, since it adds an undesirable layer of complexity.

Pass class type to constructor:

   class Test<A : DiffUtil.Callback, B : RecyclerView.ViewHolder?>(type: Class<B>): PagedListAdapter<A, B>(diff)

There using reflection you will be able to create a new instance, however you need to know exactly the constructor, for example in your case you have a constructor with a single parameter of ViewGroup type:

override fun onCreateViewHolder(parent: ViewGroup, viewType: Int): B =
    type.getConstructor(ViewGroup::class.java).newInstance(parent)

This solution is undesirable, since there are no compile-time checks and when someone will create a new ViewHolder with a different constructor he will get a runtime error.

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