为什么要将Base对象引用到派生类型引用工作？ [英] why casting Base object to derived type reference working?

问题描述

AFAIK，将一个基类对象转换为派生类型引用会抛出一个运行时异常。但在下面的类中，这种方法完全正常。

 code> public class Node< T> 
 {
 //私有成员变量
私有T数据; 
 private NodeList< T> neighbors = null; 
 
 public Node（）{} 
 public Node（T data）：this（data，null）{} 
 public Node（T data，NodeList& b $ b {
 this.data = data; 
 this.neighbors = neighbors; 
} 
 
 public T值
 {
 get 
 {
返回数据; 
} 
 set 
 {
 data = value; 
} 
} 
 
受保护的节点列表< T>邻居
 {
 get 
 {
 return neighbors; 
} 
 set 
 {
 neighbors = value; 
} 
} 
} 
} 
 
 public class BinaryTreeNode< T> ：节点< T> 
 {
 public BinaryTreeNode（）：base（）{} 
 public BinaryTreeNode（T data）：base（data，null）{} 
 public BinaryTreeNode（T data，BinaryTreeNode& T> left，BinaryTreeNode< T> right）
 {
 base.Value = data; 
 NodeList< T> children = new NodeList< T>（2）; 
 children [0] = left; 
 children [1] = right; 
 
 base.Neighbors = children; 
} 
 
 public BinaryTreeNode< T> left 
 {
 get 
 {
 if（base.Neighbors == null）
 return null; 
 else 
 return（BinaryTreeNode< T>）base.Neighbors [0]; 
} 
 set 
 {
 if（base.Neighbors == null）
 base.Neighbors = new NodeList< T>（2） 
 
 base.Neighbors [0] = value; 
} 
} 
  

这是从MSDN - http://msdn.microsoft.com/根据我的理解，在属性左中，按照我的理解，我们可以看到en-us / library / ms379572（v = vs.80）.aspx

 （BinaryTreeNode< T>）base.Neighbors [0] 
  

将（基本）节点转换为（派生）BinartTreeNode。这是如何工作的？

解决方案

正是因为它的工作原理，我们可以得出结论， c> BinaryTreeNode< T> （或其子类）。对象从不改变类型一旦创建，所以某处（你的例子中没有显示）你创建一个 BinaryTreeNode< T> ，存储它作为邻居它是一个 Node< T> - 这很好：说的是任何 BinaryTreeNode< T> a Node< T> ;类继承意味着is a），然后再次成功地将其转回。这是合法的，但如果节点变为不是 BinaryTreeNode< T> ，则会在运行时失败。

AFAIK, casting a base class object to derived type reference throws a run time exception.But in the below class, this works perfectly fine.

public class Node<T>
{
        // Private member-variables
        private T data;
        private NodeList<T> neighbors = null;

        public Node() {}
        public Node(T data) : this(data, null) {}
        public Node(T data, NodeList<T> neighbors)
        {
            this.data = data;
            this.neighbors = neighbors;
        }

        public T Value
        {
            get
            {
                return data;
            }
            set
            {
                data = value;
            }
        }

        protected NodeList<T> Neighbors
        {
            get
            {
                return neighbors;
            }
            set
            {
                neighbors = value;
            }
        }
    }
}   

public class BinaryTreeNode<T> : Node<T>
{
    public BinaryTreeNode() : base() {}
    public BinaryTreeNode(T data) : base(data, null) {}
    public BinaryTreeNode(T data, BinaryTreeNode<T> left, BinaryTreeNode<T> right)
    {
        base.Value = data;
        NodeList<T> children = new NodeList<T>(2);
        children[0] = left;
        children[1] = right;

    base.Neighbors = children;
}

public BinaryTreeNode<T> Left
{
    get
    {
        if (base.Neighbors == null)
            return null;
        else
            return (BinaryTreeNode<T>) base.Neighbors[0];
    }
    set
    {
        if (base.Neighbors == null)
            base.Neighbors = new NodeList<T>(2);

        base.Neighbors[0] = value;
    }
}

This is the code taken from one of the sample from MSDN - http://msdn.microsoft.com/en-us/library/ms379572(v=vs.80).aspx

as per my understanding, In property Left

(BinaryTreeNode<T>) base.Neighbors[0]

casts the (base) Node to (derived) BinartTreeNode. How is this working ?

解决方案

Precisely because it works, we can conclude that the object in question is actually a BinaryTreeNode<T> (or a subclass of that). Objects never change type once created, so somewhere (not shown in your example) you are creating a BinaryTreeNode<T>, storing it as the neighbor (where all we know is that it is a Node<T> - which is fine: it is true to say that any BinaryTreeNode<T> is a Node<T>; class inheritance means "is a"), and then successfully casting it back again. That is legal, but will fail at runtime if the node turns out to not be a BinaryTreeNode<T>.

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