小编典典

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”

c#

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)
        {

        }
    }
}

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2020-05-19

共1个答案

小编典典

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
2020-05-19