具有通用接口的通用方法

dev*_*irl 3 c# generics

我正在尝试创建一个泛型方法,其中类型是通用接口.

private void ShowView<T>(string viewName) where T : IView<Screen>
{ 
    IRegion mainRegion = _regionManager.Regions[RegionNames.MainRegion];
    T view = (T)mainRegion.GetView(viewName);
    if (view == null)
    {
        view = _container.Resolve<T>();
        mainRegion.Add(view, viewName);
    }
    mainRegion.Activate(view);
    view.LoadData();
    view.ViewModel.IsActive = true;
}
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界面是IView<T> where T : Screen.

所以我有Screen ConcreteView : IView<ConcreteViewModel>,ConcreteViewModel : Screen哪里是基类.当我尝试这样做时,ShowView<ConcreteView>("concrete");我得到一个UnknownMethod错误.

是因为ConcreteView使用ConcreteViewModel代替Screen来实现它的IView吗?有没有办法重写方法,以便它的工作原理?

Ant*_*ram 7

IView<ConcreteViewModel>不是IView<Screen>没有提供接口的变化

interface IView<out T>
{
}
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(以上可以从C#4.0开始)

否则,您应该能够编写如下所示的方法签名

void ShowView<T, U>(string viewName) where T : IView<U> where U : Screen
{
     // code
}
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并调用它 ShowView<ConcreteView, ConcreteViewModel>("blah");