我遇到了从这几个类和接口中获得所需行为的问题. 这是我的问题, //Inside a Unit Test that has access to internal methods and propertiesINode firstNode, secondNode;INodeId id = new NodeId (4);first = new Node (id, "no
这是我的问题,
//Inside a Unit Test that has access to internal methods and properties INode firstNode, secondNode; INodeId id = new NodeId (4); first = new Node (id, "node"); second = new Node (id, "node"); Assert.IsTrue (first == second);
上面的断言失败了,因为它似乎是在对象类的equals方法而不是Node和NodeId类中的重载运算符.
如果您对如何获得所需行为有任何建议,那将是非常棒的.
以下是我正在研究的框架的一部分:
public interface IIdentifier<T> where T : class { TKeyDataType GetKey<TKeyDataType> (); bool Equals (IIdentifier<T> obj; } public interface INode { string name { get; } INodeId id { get; } } public interface INodeId : IIdentifier<INode> { } public class Node : INode { internal Node(INodeId id, string name) { //Work } public static bool operator == (Node n1, Node n2) { return n1.equals(n2); } public static bool operator != (Node n1, Node n2) { return !n1.equals(n2); } public bool Equals (INode node) { return this.name == node.name && this.id = node.id; } #region INode Properties } public class NodeId : INodeId { internal NodeId(int id) { //Work } public static bool operator == (NodeId n1, NodeId n2) { return n1.equals(n2); } public static bool operator != (NodeId n1, NodeId n2) { return !n1.equals(n2); } public override bool Equals (object obj) { return this.Equals ((IIdentifier<INode>) obj); } public bool Equals (IIdentifier<INode> obj) { return obj.GetKey<int>() == this.GetKey<int>(); } public TKeyDataType GetKey<TKeyDataType> () { return (TKeyDataType) Convert.ChangeType ( m_id, typeof (TKeyDataType), CultureInfo.InvariantCulture); } private int m_id; }运算符重载在编译时根据声明的操作数类型解析,而不是在运行时的实际对象类型上解析.另一种说法是运算符重载不是虚拟的.所以你上面做的比较是INode.operator ==,而不是Node.operator ==.由于未定义INode.operator ==,因此重载将解析为Object.operator ==,它只是进行引用比较.
这没有什么好办法.最正确的做法是在操作数可能是对象的任何地方使用Equals()而不是==.如果你真的需要虚假的虚拟运算符重载,你应该在你的对象继承的根基类中定义operator ==,并让该重载调用Equals.但请注意,这不适用于接口,这就是您所拥有的.