英文:
Will I get the same instance of ViewModel() when I use Hilt as DI in Android Studio project?
问题
I use Hilt as DI in Android Studio project, I know there are two ways to create the instance of ViewModel().
Way 1
way1: RecordSoundViewModel = hiltViewModel()
Way 2
val way2: RecordSoundViewModel by viewModels()
1: Coudl you tell me if viewModel1, viewModel2, viewModel3 and viewModel4 is the same instance of RecordSoundViewModel?
2: If the four instances are not the same instance of RecordSoundViewModel, how can I get the same instance?
@AndroidEntryPoint
class ActivityMain : ComponentActivity() {
override fun onCreate(savedInstanceState: Bundle?) {
super.onCreate(savedInstanceState)
val viewModel3: RecordSoundViewModel by viewModels()
}
override fun onNewIntent(intent: Intent?) {super.onNewIntent(intent)val viewModel4: RecordSoundViewModel by viewModels()}
}
@Composable
fun ScreenHome_Line(
viewModel1: RecordSoundViewModel = hiltViewModel()
) {
}
@Composable
fun ScreenHome_Recording(
viewModel2: RecordSoundViewModel = hiltViewModel()
) {
}
@HiltViewModel
class RecordSoundViewModel @Inject constructor(
...
): ViewModel()
{
...
}
英文:
I use Hilt as as DI in Android Studio project, I know there are two ways to create the instance of ViewModel().
Way 1
way1: RecordSoundViewModel = hiltViewModel()
Way 2
val way2: RecordSoundViewModel by viewModels()
1: Coudl you tell me if viewModel1, viewModel2, viewModel3 and viewModel4 is the same instance of RecordSoundViewModel?
2: If the four instances are not the same instance of RecordSoundViewModel, how can I get the same instance?
@AndroidEntryPointclass ActivityMain : ComponentActivity() {override fun onCreate(savedInstanceState: Bundle?) {super.onCreate(savedInstanceState)val viewModel3: RecordSoundViewModel by viewModels()}override fun onNewIntent(intent: Intent?) {super.onNewIntent(intent)val viewModel4: RecordSoundViewModel by viewModels()}}@Composablefun ScreenHome_Line(viewModel1: RecordSoundViewModel = hiltViewModel()) {}@Composablefun ScreenHome_Recording(viewModel2: RecordSoundViewModel = hiltViewModel()) {}@HiltViewModelclass RecordSoundViewModel @Inject constructor(...): ViewModel(){...}
答案1
得分: 1
Yes, you can use the same viewmodel instance.
It is understood when the source code is examined.
public class ViewModelLazy
private val viewModelClass: KClass
private val storeProducer: () -> ViewModelStore,
private val factoryProducer: () -> ViewModelProvider.Factory,
private val extrasProducer: () -> CreationExtras = { CreationExtras.Empty }
) : Lazy
private var cached: VM? = null
override val value: VMget() {val viewModel = cachedreturn if (viewModel == null) {val factory = factoryProducer()val store = storeProducer()ViewModelProvider(store,factory,extrasProducer()).get(viewModelClass.java).also {cached = it}} else {viewModel}}override fun isInitialized(): Boolean = cached != null
}
英文:
Yes you can use the same viewmodel instance.
It is understood when the source code is examined.
public class ViewModelLazy<VM : ViewModel> @JvmOverloads constructor(private val viewModelClass: KClass<VM>,private val storeProducer: () -> ViewModelStore,private val factoryProducer: () -> ViewModelProvider.Factory,private val extrasProducer: () -> CreationExtras = { CreationExtras.Empty }) : Lazy<VM> {private var cached: VM? = nulloverride val value: VMget() {val viewModel = cachedreturn if (viewModel == null) {val factory = factoryProducer()val store = storeProducer()ViewModelProvider(store,factory,extrasProducer()).get(viewModelClass.java).also {cached = it}} else {viewModel}}override fun isInitialized(): Boolean = cached != null}
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