如何將列表轉換爲隱式運算符類型

Question

我想學習MVVM，並在我的ViewModel中添加了隱式運算符，它將模型轉換爲ViewModel，反之亦然，但現在的問題是如何將模型列表轉換爲列表ViewModel的？

下面是我試圖使用列表轉換結構代碼，但它不工作：

Person.cs

class Person 
{ 
    public string FirstName { get; set; } 
    public string LastName { get; set; } 
} 

PersonViewModel.cs

class PersonViewModel : Person 
{ 
    public string FullName { get; set; } 

    public static implicit operator List<PersonViewModel>(IList<Person> person) 
    { 
     if (person == null) 
      return null; 

     return person.Select(c => new PersonViewModel(c)).ToList(); // This is not working 

    } 

    public static implicit operator PersonViewModel(Person person) 
    { 
     return new PersonViewModel 
     { 
      FirstName = person.FirstName, 
      LastName = person.LastName, 
     }; 
    } 

    public static implicit operator Person(PersonViewModel personViewModel) 
    { 
     return new Person 
     { 
      FirstName = personViewModel.FirstName, 
      LastName = personViewModel.LastName, 
     }; 
    } 

} 

Answer 1

嘗試使用鑄鐵

person.Cast<PersonViewModel>().ToList(); 

已更新

這會給你其他錯誤。您可以使用擴展方法，並在需要時明確調用它們。這樣的事情（請確保你最終檢查爲空）

Answer 2

我很抱歉，但你的「MVVM模式」似乎對我來說很奇怪。
這是我MVVM模式的理解

//Model 
class Person 
{ 
    public string FirstName { get; set; } 
    public string LastName { get; set; } 
} 

//ModelWrapper which implements INotifyPropertyChanged if Model doesn't 
//Here you can add additional properties which will serve a View's needs 
class PersonModelWrapper 
{ 
    private Person _Model; 

    //Through this constructor you will create instance of modelwrapper 
    //without any conversions 
    public PersonModelWrapper (Person model) 
    { 
     _Model = model; 
    } 

    public string FirstName 
    { 
     get 
     { 
      return _Model.FirstName; 
     } 
     set 
     { 
      _Model.FirstName = value; 
     } 
    } 
    public string LastName 
    { 
     get 
     { 
      return _Model.LastName; 
     } 
     set 
     { 
      _Model.LastName = value; 
     } 
    } 
    public string FullName 
    { 
     get 
     { 
      return _Model.FirstName + " " _Model.LastName; 
     } 
    } 

    //Through this property you will always access current model property 
    //without conversions 
    public Person Model { get { return _Model; }} 
} 

//ViewModel which used as DataContext in the View 
class PersonViewModel 
{ 
    private PersonModelWrapper _Wrapper; 
    public string Wrapper 
    { 
     get 
     { 
      return _Wrapper; 
     } 
     set 
     { 
      _Wrapper = value; 
     } 
    } 
} 

編輯的答案@馬克·費爾德曼評論
如果包裝看起來不必要或代碼膨脹，然後「ModelWrapper」可以跳過INotifyPropertyChanged可以實現由模型本身。 如果您不關心通知有關模型屬性更改的視圖，則不需要將模型集合轉換爲ViewModels集合。
您的視圖模型將包含類型的ObservableCollection

class PersonViewModel 
{ 
    public ObservableCollection<Person> Persons { get; set; } 

    //if you don't care about tracking changes of collection(Add, Remove) 
    //then use only List<Person> without any conversions 
    public List<Person> Persons { get; set; } 

} 

視圖模型的屬性，而不包裝

public class PersonViewModel: INotifyPropertyCahnged 
{ 
    private Person _Model; 
    public Person Model 
    { 
     get { return _Model; } 
     set 
     { 
      if(Equals(_Model, value) == true) 
       return; 
      _Model = value; 
      this.RaisePropertyChanged(); 
     } 

     public String FullName {get {return _ModelFirstName + " " + _Model.LastName; }} 

     public PersonViewModel(Person model) 
     { 
      _Model = model; 
     } 
    } 
} 

然後將模型的清單視圖模型的收集發生在相同的方式，通過構造