如何正确使用 StateFlow 更新屏幕的 UIState

我看到这篇文章

https://proandroiddev.com/loading-initial-data-in-launchedeffect-vs-viewmodel-f1747c20ce62

其中作者描述了数据初始化方法的优点和缺点，得出的结论是最好通过StateFlow.我决定尝试一下，前提是我使用UiState包装器作为屏幕的一般状态：

sealed class UiState<out T> {
    object Loading : UiState<Nothing>()
    data class Error(val code: Int? = null, val message: String? = null) : UiState<Nothing>()
    data class Content<T>(val data: T) : UiState<T>()
}

也就是说，我在更改之前的实现：

class MyRepo {
    fun getMyData(): Flow<List<String>> {
        return flow {
            delay(1000)
            emit(listOf("1", "2"))
        }
    }
}

class MyViewModel1(repo: MyRepo): ViewModel() {
    data class ScreenStateUI(
        val data: List<String> = emptyList(),
        val title: String = "Title"
    )

    private val _screenUiState: MutableStateFlow<UiState<ScreenStateUI>> = MutableStateFlow(UiState.Loading)
    val screenUiState: StateFlow<UiState<ScreenStateUI>> = _screenUiState.asStateFlow()

    init {
        viewModelScope.launch {
            repo.getMyData()
                .map<List<String>, UiState<ScreenStateUI>> { UiState.Content(ScreenStateUI(data = it)) }
                .collectLatest {
                    if (it is UiState.Content) {
                        _screenUiState.emit(
                            UiState.Content(data = it)
                        )
                }
        }
    }

    fun updateTitle(title: String) {
        _screenUiState.update {
            if (it is UiState.Content) {
                it.copy(data = it.data.copy(title = title))
            } else {
                it
            }
        }
    }
}

初始化期间，状态被加载，UI可以通过订阅变化screenUiState，也就是说，如果需要使用它fun updateTitle(title: String)，没有问题。

现在，我更改了实现以摆脱init块中的初始化，并在 UI 订阅事件后立即执行所有操作：

    class MyRepo {
        fun getMyData(): Flow<List<String>> {
            return flow {
                delay(1000)
                emit(listOf("1", "2"))
            }
        }
    }

class MyViewModel2(repo: MyRepo): ViewModel() {
    data class ScreenStateUI(
        val data: List<String> = emptyList(),
        val title: String = "Title"
    )

    val screenUiState: StateFlow<UiState<ScreenStateUI>> by lazy {
        repo.getMyData()
            .map<List<String>, UiState<ScreenStateUI>> { UiState.Content(ScreenStateUI(data = it)) }
            .onStart { emit(UiState.Loading) }
            .catch { emit(UiState.Error(message = it.message)) }
            .stateIn(
                scope = viewModelScope,
                started = SharingStarted.WhileSubscribed(5000),
                initialValue = UiState.Loading
            )
    }

    fun updateTitle(title: String) {
        ...
    }
}

也就是说，一切似乎都不错，实际上现在 UI 仅在订阅时才会发起请求，但现在状态无法更改，因此该方法fun updateTitle(title: String)无法更新title。

我在这里缺少什么？