如何防止丢弃StatefulWidget？

Question

在树构建逻辑中添加条件构造会导致挂载/卸载小部件。如果小部件内部带有控制器，这反过来又会导致恼人的动画问题。

您如何告诉Flutter阻止widget状态被释放，而是在树的另一部分中搜索它？

build(BuildContext context) {
  var value = controller.value;

  /// Let's avoid blinking at high opacity, save CPU at the same time

  /// commenting this line will result in smooth animation
  /// when it is uncommented, Flutter will dispose the State which was previously lying under Opacity() widget  
  if(value >= 0.95) return MyAnimatedWidget();

  return Opacity(
   opacity: value,
   child: MyAnimatedWidget();
  );
}

Answer 1

当您在代码中编写MyAnimatedWidget()时，这意味着您将再次启动一个新的窗口小部件实例，这就是为什么旧的窗口小部件被释放，因为它不再被使用。

您可以在类中创建一个MyAnimatedWidget()类型的字段，并将其与不透明或不透明一起返回。所以我将只是你的小部件的一个实例，我不会被处理掉。

MyAnimatedWidget animatedWidget = MyAnimatedWidget();

build(BuildContext context) {
  var value = controller.value;

  /// Let's avoid blinking at high opacity, save CPU at the same time

  /// commenting this line will result in smooth animation
  /// when it is uncommented, Flutter will dispose the State which was previously lying under Opacity() widget  
  if(value >= 0.95) return animatedWidget ;

  return Opacity(
   opacity: value,
   child: animatedWidget ;
  );
}