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Avoid unnecessary Jetpack compose recomposition(避免不必要的Jetpack Compose重新组合)

翻译 作者：bug小助手 更新时间：2023-10-26 22:25:32 25 4
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So I have this structure.

所以我有这样的结构。


sealed class ViewState {
    object Loading : ViewState()

    data class Content(
        val toolbarUiModel: MyToolbarUiModel? = null,
        val listUiModel : ListUiModel? = null,
        ...
    ) : ViewState()
}

And in activity

在活动中


val uiState: ViewState by viewModel.viewState.collectAsState()

when (uiState) {
  is Content -> {
    Content(
       uiState
    )
  }

  is EmptyState -> ...
}

And Content looks like this :

内容如下所示：


Content(
   val state : ViewState
) {
  ...
  state.listUiModel?.let { uiModel ->
    MyListComponent()
  }
  ...
}

Now what happens is even if I make changes only in ListUiModel or any single Ui Model, the entire Content composable is getting recomposed, because in ViewModel, I am doing

现在发生的情况是，即使我只在ListUiModel或任何单个Ui Model中进行更改，整个可合成的内容也会被重新组合，因为在ViewModel中，我正在


fun updateListUiMode() {
   _state = _state.copy(listUiModel = newModel) // This triggers recomposition of the entire screen. However, only ListComponent is expected to be recomposed.
}

Is my code structure wrong? Because I don’t want to recompose entire screen when I update only a single component. Am I missing anything?

我的代码结构是否错误？因为当我只更新单个组件时，我不想重新组合整个屏幕。我错过了什么吗？


Thank you for all the help!

谢谢你的帮助！


更多回答

You have passed the state as a parameter in Content Composable to it will Recompose Entire Content

您已将状态作为可合成内容中的参数传递给它，它将重新合成整个内容

Thanks for the answer. I did try passing individual UI models to Content composable but still I am getting same result. I'll try one more time though

谢谢你的回答。我确实尝试了将单独的UI模型传递给Content Composable，但仍然得到了相同的结果。不过，我会再试一次

@Kaustubh you best approach would be having individual states for every single thing then that would mean you recompose only the necessary composables

@Kaustubh你最好的方法是对每一件事都有单独的状态，这意味着你只重新组合必要的可组合组件

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