“new”关键字不会重写，它表示一个与基类方法无关的新方法。

public class Foo
{
     public bool DoSomething() { return false; }
}

public class Bar : Foo
{
     public new bool DoSomething() { return true; }
}

public class Test
{
    public static void Main ()
    {
        Foo test = new Bar ();
        Console.WriteLine (test.DoSomething ());
    }
}

它输出false，如果你使用override，它会输出true。

(基本代码取自Joseph Daigle)

所以，如果你在做真正的多态性，你应该总是OVERRIDE。唯一需要使用“new”的地方是当方法与基类版本没有任何关联时。

using System;  
using System.Text;  
  
namespace OverrideAndNew  
{  
    class Program  
    {  
        static void Main(string[] args)  
        {  
            BaseClass bc = new BaseClass();  
            DerivedClass dc = new DerivedClass();  
            BaseClass bcdc = new DerivedClass();  
  
            // The following two calls do what you would expect. They call  
            // the methods that are defined in BaseClass.  
            bc.Method1();  
            bc.Method2();  
            // Output:  
            // Base - Method1  
            // Base - Method2  
  
            // The following two calls do what you would expect. They call  
            // the methods that are defined in DerivedClass.  
            dc.Method1();  
            dc.Method2();  
            // Output:  
            // Derived - Method1  
            // Derived - Method2  
  
            // The following two calls produce different results, depending
            // on whether override (Method1) or new (Method2) is used.  
            bcdc.Method1();  
            bcdc.Method2();  
            // Output:  
            // Derived - Method1  
            // Base - Method2  
        }  
    }  
  
    class BaseClass  
    {  
        public virtual void Method1()  
        {  
            Console.WriteLine("Base - Method1");  
        }  
  
        public virtual void Method2()  
        {  
            Console.WriteLine("Base - Method2");  
        }  
    }  
  
    class DerivedClass : BaseClass  
    {  
        public override void Method1()  
        {  
            Console.WriteLine("Derived - Method1");  
        }  
  
        public new void Method2()  
        {  
            Console.WriteLine("Derived - Method2");  
        }  
    }  
}  

我总是觉得这样的事情用图片更容易理解:

再一次，用joseph daigle的密码，

public class Foo
{
     public /*virtual*/ bool DoSomething() { return false; }
}

public class Bar : Foo
{
     public /*override or new*/ bool DoSomething() { return true; }
}

如果你像这样调用代码:

Foo a = new Bar();
a.DoSomething();

注意:重要的是，我们的对象实际上是一个Bar，但我们将它存储在类型为Foo的变量中(这类似于强制转换它)。

那么结果将如下所示，这取决于您在声明类时使用的是virtual/override还是new。

new关键字用于隐藏。-意味着你在运行时隐藏你的方法。输出将基于基类方法。 Override用于重写。-表示您正在调用派生类方法引用基类。输出将基于派生类方法。

Beyond just the technical details, I think using virtual/override communicates a lot of semantic information on the design. When you declare a method virtual, you indicate that you expect that implementing classes may want to provide their own, non-default implementations. Omitting this in a base class, likewise, declares the expectation that the default method ought to suffice for all implementing classes. Similarly, one can use abstract declarations to force implementing classes to provide their own implementation. Again, I think this communicates a lot about how the programmer expects the code to be used. If I were writing both the base and implementing classes and found myself using new I'd seriously rethink the decision not to make the method virtual in the parent and declare my intent specifically.

下面是一些代码来理解虚拟方法和非虚拟方法行为的区别:

class A
{
    public void foo()
    {
        Console.WriteLine("A::foo()");
    }
    public virtual void bar()
    {
        Console.WriteLine("A::bar()");
    }
}

class B : A
{
    public new void foo()
    {
        Console.WriteLine("B::foo()");
    }
    public override void bar()
    {
        Console.WriteLine("B::bar()");
    }
}

class Program
{
    static int Main(string[] args)
    {
        B b = new B();
        A a = b;
        a.foo(); // Prints A::foo
        b.foo(); // Prints B::foo
        a.bar(); // Prints B::bar
        b.bar(); // Prints B::bar
        return 0;
    }
}