Can I mark my `ViewModel` as `@Stable` ? What wud be the conditions to do that? (learning about Com...

# compose

[theapache64]

Can I mark my ViewModel as @Stable ? What wud be the conditions to do that? (learning about Compose Stability)

[dewildte]

@Stable is usually used on classes that are passed in as state to a composable.
ViewModel is not meant to be used as state, it is a state holder so I am not sure why you would need to tell compose that it is stable.

[theapache64]

@Composable s that are inside my LazyColumn has its own ViewModel . Since data comes from a SnapshotStateList, every time I add a new item to the list causing a recomposition to all the children inside the LazyColumn.

Structure

LazyColumn{
    val myDataSet : SnapshotStateList
    ...
    item(myDataSet){ item ->
        MyItem(item)
    }
}

@Composable
fun MyItem(
    item : Item,
    myViewModel : MyViewModel = createViewModel(item.key)
){
    ...
}

[theapache64]

This not happening when I remove MyViewModel from the MyItem . Since all public properties are exposed via the compose State, i was thinking why not declare it as @Stable ?

[dewildte]

Have you read the guide to state and state hoisting in the compose documentation?

[dewildte]

The you should know that using the ViewModel this way is not the proper way to use it.

[dewildte]

ViewModel is recommended to be used at the Screen Level not lower then that.

[dewildte]

Data in the ViewModel can be observed and passed down to @Composable s in the screen.

[dewildte]

LazyColumn{
    val myDataSet = viewModel.dataSet 
    ...
    items(items = myDataSet) { itemState ->
        MyItem(myItemState = itemState)
    }
}

@Composable
fun MyItem(
    myItemState: MyItemState
){
    ...
}

[dewildte]

Assuming your LazyColumn is in the screen level @Composable

[dewildte]

message has been deleted

[dewildte]

MyItemState would ideally be an immutable data class.

[theapache64]

Eric I got what you’re saying, but I think there’s a difference between “passing down page viewModel” vs composables having its own viewModel . In my case, I want my composable to have its own viewModel to control itself. The composable viewModel will be responsible to define the behaviour of the composable. This is a bit different than the MVVM arch we see.

[dewildte]

In that case you would have to make the ViewModel @Stable and ensure it upholds the contract that @Stable implies.

[dewildte]

Two instances of MyViewModel(itemA) should be structurally equal.

[dewildte]

MyViewModel(itemA) == MyViewModel(itemA) should be true

[dewildte]

That seems like it would be a lot of overhead to maintain.

[Albert Chang]

To be clear, there’s no problem in marking state holders as stable. For example, ScrollState and LazyListState are both stable state holders.
I recommend that you start with compose compiler metrics and see why it’s not inferred as stable.

[Albert Chang]

Also note that MyViewModel(itemA) == MyViewModel(itemA) is not necessary for it to be stable.

[theapache64]

@Albert Chang

…with compose compiler metrics and see why it’s not inferred as stable.

unstable viewModel: MyViewModel? = @dynamic createViewModel(..)

Yes I did that, and its happening because viewModel marked as unstable

[Albert Chang]

Yeah and what you should do is checking why MyViewModel is marked as unstable in classes.txt.