c# - 为什么 Object.GetType() 不是虚拟的？

Question

取自 MSDN 的代码示例

public class Test {
public static void Main() {
  MyBaseClass myBase = new MyBaseClass();
  MyDerivedClass myDerived = new MyDerivedClass();
  object o = myDerived;
  MyBaseClass b = myDerived;

  Console.WriteLine("mybase: Type is {0}", myBase.GetType());
  Console.WriteLine("myDerived: Type is {0}", myDerived.GetType());
  Console.WriteLine("object o = myDerived: Type is {0}", o.GetType());
  Console.WriteLine("MyBaseClass b = myDerived: Type is {0}", b.GetType());   }}

/*
This code produces the following output.
mybase: Type is MyBaseClass
myDerived: Type is MyDerivedClass
object o = myDerived: Type is MyDerivedClass
MyBaseClass b = myDerived: Type is MyDerivedClass 
*/

那么将 GetType() 设为虚拟是否合乎逻辑，至少它可以像虚拟一样工作？任何人都可以解释一下吗？ 以及其他问题 NET 框架中是否有任何其他方法具有类似于 GetType 的行为？

Answer 1

因为 .Net 框架不希望您覆盖 GetType() 方法和关于类型的欺骗。

假设您可以覆盖该方法，除了返回实例的类型之外，您还希望它做什么。当您覆盖每个类的方法以返回实例的类型时，您是否违反了 DRY。