c# - 为什么Object.GetType()不是虚拟的？

来自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
*/

那么使GetType()至少是虚拟的是否合乎逻辑，这是否合乎逻辑？有人可以解释吗？
还有其他问题。NET框架中是否还有其他类似GetType的行为？

最佳答案

因为.Net框架不希望您覆盖有关类型的GetType()方法和spoof。

假设您可以重写该方法，除了返回实例的类型之外，您还希望它做什么。当您为每个类重写该方法以返回实例的类型时，那么您就不会违反DRY了。

关于c# - 为什么Object.GetType()不是虚拟的？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/3253329/