Generic class type usage in Kotlin

Question

I am trying to write a generic base activity, that specifies it's ViewModel type a generic parameter:

abstract class BaseActivity<T : ViewModel> : AppCompatActivity()

Now I am trying to write a lazy initialized property for my ViewModel:

val viewModel by lazy { ViewModelProviders.of(this, getFactory()).get(T) }

The error that is displayed is Type Parameter T is not an expression

Also using ::class or ::class.java did not help. Can anyone explain the problem?

EDIT: I tried to use a reified inline function like this:

inline fun <reified T : ViewModel?> AppCompatActivity.obtainViewModel(factory: ViewModelProvider.Factory): T {
    return ViewModelProviders.of(this, factory).get(T::class.java)
}

And used it like this:

abstract class BaseActivity<T> : AppCompatActivity() {
    val viewModel by lazy { obtainViewModel<T>(getFactory()) 
}

Now I get the error, that T cannot be used as a reified parameter.

EDIT2: The best solution so far seems to be this: Link, but its overhead to implement the abstract token in every extending class.

docs.oracle.com/javase/tutorial/java/generics/erasure.html

Salem
– Salem

2017-12-20 08:21:39 +00:00
Commented Dec 20, 2017 at 8:21 — Salem
– Salem, Commented Dec 20, 2017 at 8:21
Could you explain this a bit more detailed? @Mango

Florian Hansen
– Florian Hansen

2017-12-20 08:33:02 +00:00
Commented Dec 20, 2017 at 8:33 — Florian Hansen
– Florian Hansen, Commented Dec 20, 2017 at 8:33

s1m0nw1 · Accepted Answer · 2017-12-20 09:43:34Z

2

Your class has a type parameter T, which unfortunately gets erased at runtime. As a result, the call get(T) does not work (I guess the method expects a Class actually?).

You seem to have already noticed that and thus tried to fix it with encapsulating the handling into a method using reified type. Yet, you cannot pass T as a reified parameter since this type will already be erased when the reified-typed method is called. As a result, obtainViewModel<T> will not work either. What you can do, is using T for ordinary generic methods, which the following demonstrates:

class Typed<T> {
    val lazyProp by lazy {
        listOf<T>()
    }
}

edited Dec 20, 2017 at 9:43

answered Dec 20, 2017 at 9:32

s1m0nw1

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6 Comments

Florian Hansen Over a year ago

So there is no way to achieve the behaviour I was trying to implement?

s1m0nw1 Over a year ago

Not that way I'm afraid. What do you think about the suggestion I just added

Florian Hansen Over a year ago

Thank you for your answer :) Unfortunately, this would make the code I am trying to wrap away become more complicated than it was before. :/

Archie G. Quiñones Over a year ago

have you found another way to do this?

Trung Le Over a year ago

@ArchieG.Quiñones not sure if u have found the solution, but check out my suggestion here stackoverflow.com/a/53480529/5315499

|

Trung Le · Accepted Answer · 2018-11-26 11:51:34Z

2

maybe a bit late, but I guess this could be the elegant way to achieve it:

abstract class BaseActivity<T : ViewModel>(
    private var modelClass: Class<T>) : AppCompatActivity() {

    val viewModel by lazy { 
        ViewModelProviders.of(this, getFactory()).get(modelClass) 
    }

}

and then

class SampleActivity : BaseActivity<SampleViewModel>(SampleViewModel::class.java) {
    // use viewModel from here
}

answered Nov 26, 2018 at 11:51

Trung Le

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