c# - 为什么对通用列表的显式转换不起作用

Question

我正尝试在构造函数中为派生类 IntersectionPath 转换对象列表，如下所示。

    public class IntersectionPath : Path<IntersectionSegment>, IEnumerable
    {          

        //Constructors
        public IntersectionPath() : base() {  Verts = null; }

        public IntersectionPath(List<Intersection> inVerts, List<Segment<Node>> inEdges) : base() 
        {
            this.Segments = (List<IntersectionSegment>) inEdges;
        }

    }

Segments 在通用基类 Path 中定义

    public class Path<T> : IEnumerable<T> where T : Segment<Node>
    {   
        //public properties
        public List<Direction> Directions {get; set; }
        public List<T> Segments  {  get; set; }
    }

我已经在 IntersectionSegment 类中为转换定义了一个显式运算符(见下文，所以我不清楚为什么这不会编译。我在 IntersectionPath 构造函数中有一个关于转换的错误消息。

public class IntersectionSegment : Segment<Intersection>
{           
    //curves which intersect the primary curve at I0(Start Node) and I1(End Node)
    public Curve C0 { get; set; }
    public Curve C1 { get; set; }

    public IntersectionSegment():base() {}

    public IntersectionSegment(Intersection n0, Intersection n1):base(n0,n1){}

    public static explicit operator IntersectionSegment(Segment<Node> s)
    {
        if ((s.Start is Intersection) && (s.End is Intersection))
        {
            return new IntersectionSegment(s.Start as Intersection,s.End as Intersection);
        }
        else return null;
    }

    public static explicit operator List<IntersectionSegment>(List<Segment<Node>> ls)
    {
        List<IntersectionSegment> lsout = new List<IntersectionSegment>();
        foreach (Segment<Node> s in ls)
        {
            if ((s.Start is Intersection) && (s.End is Intersection))
            {
                lsout.Add(new IntersectionSegment(s.Start as Intersection,s.End as Intersection));
            }
            else return null;
        }
        return lsout;
    }

段定义为:

public class Segment <T> : Shape where T : Node
{
    //generic properties
    public T Start { get; set; }
    public T End { get; set; }

 }

Answer 1

List<InteractionSegment>与 InteractionSegment 不同.将一种类型的列表转换为另一种类型的列表不会转换每个项目。
你需要做这样的事情:

this.Segments = inEdges.Select(x => (InteractionSegment)x).ToList();

这使用 LINQ to Objects 来转换 inEdges 中的每个对象到InteractionSegment对象并将结果放回列表中，然后分配给 this.Segments .