仿制药无法在飞镖/扑朔迷离中使用超类的功能
我对飞镖的语法有问题。我希望能够在通用类上使用构造函数。因此，我让通用类扩展一个具有指定构造函数的抽象类。但是代码仍然向我表明它不起作用。有人有主意吗？
T fetchItem<T extends JsonModel>(){
return T.fromJson();
// This line shows the error
// The method 'fromJson' isn't defined for the type 'Type'.
}
abstract class JsonModel {
JsonModel.fromJson();
}
以下解决方案有效，但我认为这非常丑陋：
T fetchItem<T extends JsonModel<T>>(T t){
return t.fromJson();
}
abstract class JsonModel<T> {
T fromJson();
}
I have a problem with the syntax in Dart. I want to be able to use a constructor on a generic class. So I let the generic class extend an abstract class which has the specified constructor. But the Code still shows me that it's not working. Does anyone have an idea?
T fetchItem<T extends JsonModel>(){
return T.fromJson();
// This line shows the error
// The method 'fromJson' isn't defined for the type 'Type'.
}
abstract class JsonModel {
JsonModel.fromJson();
}
The following solution works, but I think it's extremly ugly:
T fetchItem<T extends JsonModel<T>>(T t){
return t.fromJson();
}
abstract class JsonModel<T> {
T fromJson();
}
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（1）
构造函数没有继承。仅仅因为基类具有某个构造函数，并不意味着派生类具有该构造函数。
它与仿制药无关。如果您从JSONMODEL扩展了X类，则根本没有该名称的构造函数。
Constructors are not inherited. Just because a base class has a certain constructor, does not mean the derived class has that constructor.
It doesn't have anything to do with generics. If you extended a class X from your JsonModel, it simply would not have a constructor of that name.