The Code A is from a Android offical sample project here.

The author create a val uiState, it's MutableStateFlow, I know that MutableStateFlow is hot flow, it will occupy system resource when it is created.

Do I need to destory a StateFlow object created in view model by myself? will the system release it automatically when the app doesn't need it again?

Code A

class InterestsViewModel(
    private val interestsRepository: InterestsRepository
) : ViewModel() {

    // UI state exposed to the UI
    private val _uiState = MutableStateFlow(InterestsUiState(loading = true))
    val uiState: StateFlow<InterestsUiState> = _uiState.asStateFlow()


    ...
}

CodePudding user response：

It's a normal object and as such will be cleaned up once there are no references to it (so when both observing view and ViewModel seizes to exist)

CodePudding user response：

You don't need to manually destroy the flow after you stop using it.

You mentioned that MutableStateFlow is a "hot" flow. When we talk about "hot" and "cold" flows we're really talking about how the values get emitted from the flow.

• In a hot flow, there's typically a second "producer" coroutine that's doing the work to emit values. When you try to collect the flow, you won't receive any values unless the producer is actively emitting them.
• In a cold flow, collecting the flow also does the work that produces the values. There's no need for a separate producer coroutine; all the work is done by the consumer.

In that sense, it's correct to call a MutableStateFlow a hot flow. To collect a value from it, you have to first emit a value to it from somewhere else.

However, that doesn't need to imply that the flow holds any resources. The flow is really just acting as a communication channel between the coroutines that are producing and consuming values. Once those coroutines are no longer using or referencing the flow, it becomes eligible for garbage collection just like any normal object.