Jetpack Compose and ViewModel-Instantiation

Question

What does the following line of code do?

fun NotesApp(noteViewModel: NoteViewModel = viewModel()) {

The default parameter confuses me. The class, which inherits from ViewModel, is called NoteViewModel. Wouldn't the default parameter be: NoteViewModel()?

But it works nevertheless. How is that possible?

Is this a Composable? – Richard Onslow Roper Aug 24 '22 at 06:40 — Richard Onslow Roper, Aug 24 '22 at 06:40
@Richard Onslow Roper: Yep. It's within a Composable. – cluster1 Aug 24 '22 at 07:16 — cluster1, Aug 24 '22 at 07:16

Richard Onslow Roper · Accepted Answer · 2022-08-24T08:02:51.417

2

It is the default initializer for view models inside a Composable. It takes in a type parameter, so I don't think what you've posted above would be enough.

For example, if you wanted to initialise a MainViewModel, it'll be something like

val vm = viewModel<MainViewModel>()

It is supposed to return the same instance of a viewmodel, if it had been created in the past.

edited Aug 24 '22 at 08:02

answered Aug 24 '22 at 06:43

Richard Onslow Roper

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Your answer with the default initializer makes sense to me. I guess in the case, I have shown, it takes the type-hint from "noteViewModel: NoteViewModel". Means: The "... : NoteViewModel ..." might be equivalent to . – cluster1 Aug 24 '22 at 07:12

score 1 · Answer 2 · answered Aug 24 '22 at 06:45

It works because the function viewModel() provides that viewmodel. I assume it's from the androidx.lifecycle.viewmodel.compose package.

Returns an existing ViewModel or creates a new one in the given owner (usually, a fragment or an activity), defaulting to the owner provided by LocalViewModelStoreOwner.

Jetpack Compose and ViewModel-Instantiation

2 Answers2