该类型必须是引用类型,才能在通用类型或方法中将其用作参数“ T”
我正在深入研究泛型,现在遇到需要帮助的情况。如主题标题所示,我在下面的“派生”类上遇到编译错误。我看到许多其他与此类似的帖子,但是我没有看到这种关系。有人可以告诉我如何解决吗?
using System;using System.Collections.Generic;
namespace Example
{
public class ViewContext
{
ViewContext() { }
}
public interface IModel
{
}
public interface IView<T> where T : IModel
{
ViewContext ViewContext { get; set; }
}
public class SomeModel : IModel
{
public SomeModel() { }
public int ID { get; set; }
}
public class Base<T> where T : IModel
{
public Base(IView<T> view)
{
}
}
public class Derived<SomeModel> : Base<SomeModel> where SomeModel : IModel
{
public Derived(IView<SomeModel> view)
: base(view)
{
SomeModel m = (SomeModel)Activator.CreateInstance(typeof(SomeModel));
Service<SomeModel> s = new Service<SomeModel>();
s.Work(m);
}
}
public class Service<SomeModel> where SomeModel : IModel
{
public Service()
{
}
public void Work(SomeModel m)
{
}
}
}
回答:
我无法复制,但我 怀疑 您的实际代码中存在某个约束条件,例如T : class
,您需要传播该约束以使编译器满意(如果没有一个复制实例,很难确定):
public class Derived<SomeModel> : Base<SomeModel> where SomeModel : class, IModel ^^^^^
see this bit
以上是 该类型必须是引用类型,才能在通用类型或方法中将其用作参数“ T” 的全部内容, 来源链接: utcz.com/qa/430193.html