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C# new virtual method and as/is casts [duplicate]

What are differences between declaring a method in a base type "virtual" and then overriding it in a child type using the "override" keyword as opposed to simply using the "new" keyword when declaring the matching method in the child type?

I always find things like this more easily understood with pictures:
Again, taking joseph daigle's code,
public class Foo
{
public /*virtual*/ bool DoSomething() { return false; }
}
public class Bar : Foo
{
public /*override or new*/ bool DoSomething() { return true; }
}
If you then call the code like this:
Foo a = new Bar();
a.DoSomething();
NOTE: The important thing is that our object is actually a Bar, but we are storing it in a variable of type Foo (this is similar to casting it)
Then the result will be as follows, depending on whether you used virtual/override or new when declaring your classes.

The "new" keyword doesn't override, it signifies a new method that has nothing to do with the base class method.
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 ());
}
}
This prints false, if you used override it would have printed true.
(Base code taken from Joseph Daigle)
So, if you are doing real polymorphism you SHOULD ALWAYS OVERRIDE. The only place where you need to use "new" is when the method is not related in any way to the base class version.

Here's some code to understand the difference in the behavior of virtual and non-virtual methods:
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;
}
}

The new keyword actually creates a completely new member that only exists on that specific type.
For instance
public class Foo
{
public bool DoSomething() { return false; }
}
public class Bar : Foo
{
public new bool DoSomething() { return true; }
}
The method exists on both types. When you use reflection and get the members of type Bar, you will actually find 2 methods called DoSomething() that look exactly the same. By using new you effectively hide the implementation in the base class, so that when classes derive from Bar (in my example) the method call to base.DoSomething() goes to Bar and not Foo.

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.

virtual / override tells the compiler that the two methods are related and that in some circumstances when you would think you are calling the first (virtual) method it's actually correct to call the second (overridden) method instead. This is the foundation of polymorphism.
(new SubClass() as BaseClass).VirtualFoo()
Will call the SubClass's overriden VirtualFoo() method.
new tells the compiler that you are adding a method to a derived class with the same name as a method in the base class, but they have no relationship to each other.
(new SubClass() as BaseClass).NewBar()
Will call the BaseClass's NewBar() method, whereas:
(new SubClass()).NewBar()
Will call the SubClass's NewBar() method.

The difference between the override keyword and new keyword is that the former does method overriding and the later does method hiding.
Check out the folllowing links for more information...
MSDN and Other

new keyword is for Hiding. - means you are hiding your method at runtime. Output will be based base class method.
override for overriding. - means you are invoking your derived class method with the reference of base class. Output will be based on derived class method.

My version of explanation comes from using properties to help understand the differences.
override is simple enough, right ? The underlying type overrides the parent's.
new is perhaps the misleading (for me it was). With properties it's easier to understand:
public class Foo
{
public bool GetSomething => false;
}
public class Bar : Foo
{
public new bool GetSomething => true;
}
public static void Main(string[] args)
{
Foo foo = new Bar();
Console.WriteLine(foo.GetSomething);
Bar bar = new Bar();
Console.WriteLine(bar.GetSomething);
}
Using a debugger you can notice that Foo foo has 2 GetSomething properties, as it actually has 2 versions of the property, Foo's and Bar's, and to know which one to use, c# "picks" the property for the current type.
If you wanted to use the Bar's version, you would have used override or use Foo foo instead.
Bar bar has only 1, as it wants completely new behavior for GetSomething.

Not marking a method with anything means: Bind this method using the object's compile type, not runtime type (static binding).
Marking a method with virtual means: Bind this method using the object's runtime type, not compile time type (dynamic binding).
Marking a base class virtual method with override in derived class means: This is the method to be bound using the object's runtime type (dynamic binding).
Marking a base class virtual method with new in derived class means: This is a new method, that has no relation to the one with the same name in the base class and it should be bound using object's compile time type (static binding).
Not marking a base class virtual method in the derived class means: This method is marked as new (static binding).
Marking a method abstract means: This method is virtual, but I will not declare a body for it and its class is also abstract (dynamic binding).

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");
}
}
}

How do I call a super class method from sub class object which has overridden?

I've been reading about Polymorphism and I've been through something interesting.
I have created two classes: a super class and a derived class. As you can see derived class inherits from the super class and using its method.
In the derived class I have overridden the method of super class.
So my question is: I want to call the super class method from an object of a derived class
For example:
derivedClass a = new derivedClass()
a.method();
I want a.methood() to execute the superclass method.
Is that possible?

Summing up your question: You have a base class and a derived class. The derived class overrides a virtual method of the base class. Given an instance of the derived class, is it possible to force a call to the base class version of the method, instead of the derived class?
Short answer: From inside the class, yes. From outside the class, no.
Essentially what you wish to do here is a non virtual call to a virtual method. It is legal to do a non-virtual call to a base class method, but only from inside the class:
class B
{
public virtual void M() { Console.WriteLine("B.M"); }
}
class D : B
{
public override void M() { Console.WriteLine("D.M"); }
public void BM() { base.M(); }
}
...
D d = new D();
d.M(); // D.M
d.BM(); // B.M
Now, for advanced players there are sneaky ways to call B.M with an instance of D as the receiver, and I'm not going to tell you what they are. You should not attempt to do so. The CLR may flag such attempts as violations of its "verified code" rules. The rules of virtual overriding are there for your convenience and safety. Do not attempt to do an end-run around those rules.

From outside the derived class it is not possible. From within the derived class it is.
If you have these classes defined:
class A
{
public virtual void Method1() { Console.WriteLine("A"); }
}
class B : A
{
override public void Method1() { Console.WriteLine("B"); }
public void BaseMethod1() { base.Method1(); }
}
You can execute the following:
B b = new B();
b.Method1(); //Outputs "B"
b.BaseMethod1(); //Outputs "A"
((A)b).Method1(); //***Outputs "B" (even though you cast it as A)
If you change overide to new then the output of the last line is "A"

You can try calling base method as per your requirement, if there is some condition, according to which you want to call it, you can use that condition as well.
public class SubClass : SuperClass
{
public bool IsCallBaseClassMethod { get; set; }
public override void Method(){
if (IsCallBaseClassMethod)
{
base.Method();
}
}
}
Set the condition according to your requirement for the object and call the method for same object.
SubClass testObject = new SubClass();
testObject.IsCallBaseClassMethod = true;
testObject.Method();

C# didnt call drived class function in polymorphism when pass parameter [duplicate]

Wondering what the difference is between the following:
Case 1: Base Class
public void DoIt();
Case 1: Inherited class
public new void DoIt();
Case 2: Base Class
public virtual void DoIt();
Case 2: Inherited class
public override void DoIt();
Both case 1 and 2 appear to have the same effect based on the tests I have run. Is there a difference, or a preferred way?

The override modifier may be used on
virtual methods and must be used on
abstract methods. This indicates for
the compiler to use the last defined
implementation of a method. Even if
the method is called on a reference to
the base class it will use the
implementation overriding it.
public class Base
{
public virtual void DoIt()
{
}
}
public class Derived : Base
{
public override void DoIt()
{
}
}
Base b = new Derived();
b.DoIt(); // Calls Derived.DoIt
will call Derived.DoIt if that overrides Base.DoIt.
The new modifier instructs the
compiler to use your child class implementation
instead of the parent class
implementation. Any code that is not
referencing your class but the parent
class will use the parent class
implementation.
public class Base
{
public virtual void DoIt()
{
}
}
public class Derived : Base
{
public new void DoIt()
{
}
}
Base b = new Derived();
Derived d = new Derived();
b.DoIt(); // Calls Base.DoIt
d.DoIt(); // Calls Derived.DoIt
Will first call Base.DoIt, then Derived.DoIt. They're effectively two entirely separate methods which happen to have the same name, rather than the derived method overriding the base method.
Source: Microsoft blog

virtual: indicates that a method may be overriden by an inheritor
override: overrides the functionality of a virtual method in a base class, providing different functionality.
new: hides the original method (which doesn't have to be virtual), providing different functionality. This should only be used where it is absolutely necessary.
When you hide a method, you can still access the original method by up casting to the base class. This is useful in some scenarios, but dangerous.

In the first case you are hiding the definition in the parent class. This means that it will only be invoked when you are dealing with the object as the child class. If you cast the class to its parent type, the parent's method will be invoked. In the second instance, the method is overridden and will be invoked regardless of whether the object is cast as the child or parent class.

new means respect your REFERENCE type(left-hand side of =) , thereby running reference types's method. If redefined method doesn't have new keyword, it is behaved as it has. Moreover, it also known as non-polymorphic inheritance. That is,
“I’m making a brand new method in the derived class that has
absolutely nothing to do with any methods by the same name in the base
class.” - by said Whitaker
override, which must be used with virtual keyword in its base class, means respect your OBJECT type(right-hand side of =), thereby
running method overriden in regardless of reference type. Moreover, it also known as polymorphic inheritance.
My way to bear in mind both keywords that they are opposite of each other.
override: virtual keyword must be defined to override the method. The method using override keyword that regardless of reference type(reference of base class or derived class) if it is instantiated with base class, the method of base class runs. Otherwise, the method of derived class runs.
new: if the keyword is used by a method, unlike override keyword, the reference type is important. If it is instantiated with derived class and the reference type is base class, the method of base class runs. If it is instantiated with derived class and the reference type is derived class, the method of derived class runs. Namely, it is contrast of override keyword. En passant, if you forget or omit to add new keyword to the method, the compiler behaves by default as new keyword is used.
class A
{
public string Foo()
{
return "A";
}
public virtual string Test()
{
return "base test";
}
}
class B: A
{
public new string Foo()
{
return "B";
}
}
class C: B
{
public string Foo()
{
return "C";
}
public override string Test() {
return "derived test";
}
}
Call in main:
A AClass = new B();
Console.WriteLine(AClass.Foo());
B BClass = new B();
Console.WriteLine(BClass.Foo());
B BClassWithC = new C();
Console.WriteLine(BClassWithC.Foo());
Console.WriteLine(AClass.Test());
Console.WriteLine(BClassWithC.Test());
Output:
A
B
B
base test
derived test
New code example,
Play with code by commenting in one-by-one.
class X
{
protected internal /*virtual*/ void Method()
{
WriteLine("X");
}
}
class Y : X
{
protected internal /*override*/ void Method()
{
base.Method();
WriteLine("Y");
}
}
class Z : Y
{
protected internal /*override*/ void Method()
{
base.Method();
WriteLine("Z");
}
}
class Programxyz
{
private static void Main(string[] args)
{
X v = new Z();
//Y v = new Z();
//Z v = new Z();
v.Method();
}

try following: (case1)
((BaseClass)(new InheritedClass())).DoIt()
Edit: virtual+override are resolved at runtime (so override really overrides virtual methods), while new just create new method with the same name, and hides the old, it is resolved at compile time -> your compiler will call the method it 'sees'

All combinations of none, virtual, override, new and abstract:

In case 1 if you used call the DoIt() method of the inherited class while the type is declared as the base class you will see the action of the base class even.
/* Results
Class1
Base1
Class2
Class2
*/
public abstract class Base1
{
public void DoIt() { Console.WriteLine("Base1"); }
}
public class Class1 : Base1
{
public new void DoIt() { Console.WriteLine("Class1"); }
}
public abstract class Base2
{
public virtual void DoIt() { Console.WriteLine("Base2"); }
}
public class Class2 : Base2
{
public override void DoIt() { Console.WriteLine("Class2"); }
}
static void Main(string[] args)
{
var c1 = new Class1();
c1.DoIt();
((Base1)c1).DoIt();
var c2 = new Class2();
c2.DoIt();
((Base2)c2).DoIt();
Console.Read();
}

The difference between the two cases is that in case 1, the base DoIt method does not get overridden, just hidden. What this means is that depending on the type of the variable depends on which method will get called. For example:
BaseClass instance1 = new SubClass();
instance1.DoIt(); // Calls base class DoIt method
SubClass instance2 = new SubClass();
instance2.DoIt(); // Calls sub class DoIt method
This can be really confusing and results in non expected behaviour and should be avoided if possible. So the preferred way would be case 2.

I had the same question and it's really confusing,
you should consider that override and new keywords working only with objects of type base class and value of derived class. In this case only you'll see the effect of override and new:
So if you have class A and B, B inherits from A, then you instantiate an object like this:
A a = new B();
Now on calling methods will take its state into consideration.
Override: means that it extends the function of the method, then it uses the method in the derived class, whereas new tell the compiler to hide the method in the derived class and use the method in the base class instead.
Here is a very good sight to that subject:
https://msdn.microsoft.com/EN-US/library/ms173153%28v=VS.140,d=hv.2%29.aspx?f=255&MSPPError=-2147217396

The article below is in vb.net but I think the explanation about new vs overrides is very easy to grasp.
https://www.codeproject.com/articles/17477/the-dark-shadow-of-overrides
At some point in the article, there is this sentence:
In general, Shadows assumes the function associated with the type is
invoked, while Overrides assumes the object implementation is
executed.
The accepted answer to this question is perfect but I think this article provide good examples to add better meaning about the differences between these two keywords.

If keyword override is used in derive class then its override the parent method.
If Keyword new is used in derive class then derive method hided by parent method.

Out of all those, new is the most confusing. Through experimenting, the new keyword is like giving developers the option to override the inheriting class implementation with the base class implementation by explicitly defining the type. It is like thinking the other way around.
In the example below, the result will return "Derived result" until the type is explicitly defined as BaseClass test, only then "Base result" will be returned.
class Program
{
static void Main(string[] args)
{
var test = new DerivedClass();
var result = test.DoSomething();
}
}
class BaseClass
{
public virtual string DoSomething()
{
return "Base result";
}
}
class DerivedClass : BaseClass
{
public new string DoSomething()
{
return "Derived result";
}
}

The functional difference will not be show in these tests:
BaseClass bc = new BaseClass();
bc.DoIt();
DerivedClass dc = new DerivedClass();
dc.ShowIt();
In this exmample, the Doit that is called is the one you expect to be called.
In order to see the difference you have to do this:
BaseClass obj = new DerivedClass();
obj.DoIt();
You will see if you run that test that in the case 1 (as you defined it), the DoIt() in BaseClass is called, in case 2 (as you defined it), the DoIt() in DerivedClass is called.

new : It just hides the method of the base class, but you can access it if you want.
override: It's overriding the base class's method and you can't access it even if you want to.
Example
using System;
public class Program
{
public static void Main()
{
BaseClass test = new DerivedClass();
var result = test.DoSomething();
Console.WriteLine(result);
}
class BaseClass
{
public string DoSomething()
{
return "Base result";
}
}
class DerivedClass : BaseClass
{
public new string DoSomething()
{
return "Derived result";
}
}
}
Result: Base result
P.S. I copied and slightly changed the above example from the following link:
https://stackoverflow.com/a/45822233/10995103

Is it possible to override a non-virtual method?

Is there any way to override a non-virtual method? or something that gives similar results (other than creating a new method to call the desired method)?
I would like to override a method from Microsoft.Xna.Framework.Graphics.GraphicsDevice with unit testing in mind.

No, you cannot override a non-virtual method. The closest thing you can do is hide the method by creating a new method with the same name but this is not advisable as it breaks good design principles.
But even hiding a method won't give you execution time polymorphic dispatch of method calls like a true virtual method call would. Consider this example:
using System;
class Example
{
static void Main()
{
Foo f = new Foo();
f.M();
Foo b = new Bar();
b.M();
}
}
class Foo
{
public void M()
{
Console.WriteLine("Foo.M");
}
}
class Bar : Foo
{
public new void M()
{
Console.WriteLine("Bar.M");
}
}
In this example both calls to the M method print Foo.M. As you can see this approach does allow you to have a new implementation for a method as long as the reference to that object is of the correct derived type but hiding a base method does break polymorphism.
I would recommend that you do not hide base methods in this manner.
I tend to side with those who favor C#'s default behavior that methods are non-virtual by default (as opposed to Java). I would go even further and say that classes should also be sealed by default. Inheritance is hard to design for properly and the fact that there is a method that is not marked to be virtual indicates that the author of that method never intended for the method to be overridden.
Edit: "execution time polymorphic dispatch":
What I mean by this is the default behavior that happens at execution time when you call virtual methods. Let's say for example that in my previous code example, rather than defining a non-virtual method, I did in fact define a virtual method and a true overridden method as well.
If I were to call b.Foo in that case, the CLR would correctly determine the type of object that the b reference points to as Bar and would dispatch the call to M appropriately.

No you can't.
You can only override a virtual method - see the MSDN here:
In C#, derived classes can contain methods with the same name as base class methods.
The base class method must be defined virtual.

You can't override non-virtual method of any class in C# (without hacking CLR), but you can override any method of interface the class implements.
Consider we have non-sealed
class GraphicsDevice: IGraphicsDevice {
public void DoWork() {
Console.WriteLine("GraphicsDevice.DoWork()");
}
}
// with its interface
interface IGraphicsDevice {
void DoWork();
}
// You can't just override DoWork in a child class,
// but if you replace usage of GraphicsDevice to IGraphicsDevice,
// then you can override this method (and, actually, the whole interface).
class MyDevice: GraphicsDevice, IGraphicsDevice {
public new void DoWork() {
Console.WriteLine("MyDevice.DoWork()");
base.DoWork();
}
}
And here's demo
class Program {
static void Main(string[] args) {
IGraphicsDevice real = new GraphicsDevice();
var myObj = new MyDevice();
// demo that interface override works
GraphicsDevice myCastedToBase = myObj;
IGraphicsDevice my = myCastedToBase;
// obvious
Console.WriteLine("Using real GraphicsDevice:");
real.DoWork();
// override
Console.WriteLine("Using overriden GraphicsDevice:");
my.DoWork();
}
}

If the base class isn't sealed then you can inherit from it and write a new method that hides the base one (use the "new" keyword in the method declaration). Otherwise no, you cannot override it because it was never the original authors intent for it to be overridden, hence why it isn't virtual.

I think you're getting overloading and overriding confused, overloading means you have two or more methods with the same name but different sets of parameters while overriding means you have a different implementation for a method in a derived class (thereby replacing or modifying the behaviour in it's base class).
If a method is virtual, you can override it using the override keyword in the derrived class. However, non-virtual methods can only hide the base implementation by using the new keyword in place of the override keyword. The non-virtual route is useless if the caller accesses the method via a variable typed as the base type as the compiler would use a static dispatch to the base method (meaning the code in your derrived class would never be called).
There is never anything preventing you from adding an overload to an existing class, but only code that knows about your class would be able to access it.

The answer to this question is not entirely No. It depends on how you structure your inheritance and access the instances of your classes. If your design meets the following, you would be able to override non-virtual method from base class with the new modifier:
The method is declared in an interface that your classes inherit from
You are accessing the class instances using the interface
Take example of the following:
interface ITest { void Test(); }
class A : ITest { public void Test(){Console.WriteLine("A");} }
class B : A { public new void Test(){Console.WriteLine("B");} }
ITest x = new B();
x.Test(); // output "A"
calling x.Test() will output "A" to the console. However if you re-declare the interface in the definition of class B, x.Test() will output B instead.
interface ITest { void Test(); }
class A : ITest { public void Test(){Console.WriteLine("A");} }
class B : A, ITest { public new void Test(){Console.WriteLine("B");} }
ITest x = new B();
x.Test(); // output "B"

In the case you are inheriting from a non-derived class, you could simply create an abstract super class and inherit from it downstream instead.

Is there any way to override a non-virtual method? or something that gives similar results (other than creating a new method to call the desired method)?
You cannot override a non-virtual method. However you can use the new modifier keyword to get similar results:
class Class0
{
public int Test()
{
return 0;
}
}
class Class1 : Class0
{
public new int Test()
{
return 1;
}
}
. . .
// result of 1
Console.WriteLine(new Class1().Test());
You will also want to make sure that the access modifier is also the same, otherwise you will not get inheritance down the line. If another class inherits from Class1 the new keyword in Class1 will not affect objects inheriting from it, unless the access modifier is the same.
If the access modifier is not the same:
class Class0
{
protected int Test()
{
return 0;
}
}
class Class1 : Class0
{
// different access modifier
new int Test()
{
return 1;
}
}
class Class2 : Class1
{
public int Result()
{
return Test();
}
}
. . .
// result of 0
Console.WriteLine(new Class2().Result());
...versus if the access modifier is the same:
class Class0
{
protected int Test()
{
return 0;
}
}
class Class1 : Class0
{
// same access modifier
protected new int Test()
{
return 1;
}
}
class Class2 : Class1
{
public int Result()
{
return Test();
}
}
. . .
// result of 1
Console.WriteLine(new Class2().Result());
As pointed out in a previous answer, this is not a good design principle.

There is a way of achieving this using abstract class and abstract method.
Consider
Class Base
{
void MethodToBeTested()
{
...
}
void Method1()
{
}
void Method2()
{
}
...
}
Now, if you wish to have different versions of method MethodToBeTested(), then
change Class Base to an abstract class and method MethodToBeTested() as an abstract method
abstract Class Base
{
abstract void MethodToBeTested();
void Method1()
{
}
void Method2()
{
}
...
}
With abstract void MethodToBeTested() comes an issue; the implementation is gone.
Hence create a class DefaultBaseImplementation : Base to have the default implementation.
And create another class UnitTestImplementation : Base to have unit test implementation.
With these 2 new classes, the base class functionality can be overridden.
Class DefaultBaseImplementation : Base
{
override void MethodToBeTested()
{
//Base (default) implementation goes here
}
}
Class UnitTestImplementation : Base
{
override void MethodToBeTested()
{
//Unit test implementation goes here
}
}
Now you have 2 classes implementing (overriding) MethodToBeTested().
You can instantiate the (derived) class as required (i.e. either with base implementation or with unit test implementation).

Difference between shadowing and overriding in C#?

What's difference between shadowing and overriding a method in C#?

Well inheritance...
suppose you have this classes:
class A {
public int Foo(){ return 5;}
public virtual int Bar(){return 5;}
}
class B : A{
public new int Foo() { return 1;} //shadow
public override int Bar() {return 1;} //override
}
then when you call this:
A clA = new A();
B clB = new B();
Console.WriteLine(clA.Foo()); // output 5
Console.WriteLine(clA.Bar()); // output 5
Console.WriteLine(clB.Foo()); // output 1
Console.WriteLine(clB.Bar()); // output 1
//now let's cast B to an A class
Console.WriteLine(((A)clB).Foo()); // output 5 <<<-- shadow
Console.WriteLine(((A)clB).Bar()); // output 1
Suppose you have a base class and you use the base class in all your code instead of the inherited classes, and you use shadow, it will return the values the base class returns instead of following the inheritance tree of the real type of the object.
Run code here
Hope I'm making sense :)

Shadowing is actually VB parlance for what we would refer to as hiding in C#.
Often hiding (shadowing in VB) and overriding are shown as in answer by Stormenet.
A virtual method is shown to be overridden by a sub-class and calls to that method even on the super-class type or from inside code of the super-class will call the replacement implementation from the sub-class.
Then a concrete method is shown (one not marked as virtual or abstract) being hidden by using the new keyword when defining a method with an identical signature on the sub-class. In this case when the method is called on the super-class type the original implementation is used, the new implementation is only available on the sub-class.
However what is often missed is that it is also possible to hide a virtual method.
class A
{
public virtual void DoStuff() { // original implementation }
}
class B : A
{
public new void DoStuff() { //new implementation }
}
B b = new B();
A a = b;
b.DoStuff(); //calls new implementation
a.DoStuff(); //calls original implementation.
Note in the above example DoStuff becomes concrete and can not be overriden. However it is also possible to use both the virtual and new keywords together.
class A
{
public virtual void DoStuff() { // original implementation }
}
class B : A
{
public new virtual void DoStuff() { //new implementation }
}
class C : B
{
public override void DoStuff() { //replacement implementation }
}
C c = new C();
B b = c;
A a = b;
c.DoStuff(); //calls replacement implementation
b.DoStuff(); //calls replacement implementation
a.DoStuff(); //calls original implementation.
Note that despite all the methods involved being virtual, the override on C does not affect the virtual method on A because of the use of new in B hides the A implementation.
Edit: Its been noted in the comments to this answer that the above may be dangerous or at least not particularly useful. I would say yes it can be dangerous and would be out there if it were at all useful.
In particular you could get into all sorts of trouble if you also change the accessability modifiers. For example:-
public class Foo
{
internal Foo() { }
protected virtual string Thing() { return "foo"; }
}
public class Bar : Foo
{
internal new string Thing() { return "bar"; }
}
To an external inheritor of Bar, Foo's implementation of Thing() remains accesible and overridable. All legal and explainable according to .NET type rules neverless quite unintuative.
I've posted this answer to deepen an understanding of how things work not as a suggestion of techniques that can be used freely.

I think the main difference is that with shadowing, you're essentially reusing the name, and just ignoring the superclass use. With overriding, you're changing the implementation, but not the accessibility and signature (e.g. parameter types and return). See http://www.geekinterview.com/question_details/19331 .

Shadowing is a VB.NET concept. In C#, Shadowing is known as Hiding. It hides the derived class method. It is accomplished using the ‘new’ keyword.
Override keyword is used to provide a completely new implementation of a base class method (which is marked ‘Virtual’) in the derived class.

The answers above do not describe how to shadow/hide a base class' constructor. For completeness adding that special case here.
Say if you want to wrap ApplicationException in your own class MyApplicationException here is the syntax....
public class MyApplicationException : ApplicationException
{
public MyApplicationException(string message) : base(message)
{
// NOOP
} // constructor
public MyApplicationException(string message, Exception innerException) : base(message, innerException)
{
// NOOP
} // constructor
} // class MyApplicationException

Basically if you have something like below,
Class A
{
}
Class B:A
{
}
A a = new B();
Any method you call on the object 'a' will be made on the type of 'a'(Here the type is 'A')
But if you implement the same method in class B that is already present in Class A, the compiler will give you a warning to use a "New" keyword. If you use "New", the warning will disappear. Other than this there is no difference between using "New" or not using it in the inherited class.
In some situations you may need to call a method of the reference class the particular instance holds at that moment instead of calling a method on the object type. In the above case the reference it holds is 'B', but the type is 'A'. So if you want the method call should happen on 'B', then you use Virtual and override to achieve this.
Hope this helps...
Daniel Sandeep.

If there is a case in which you cannot alter the contents of a class, let's say A, but you want to use its some methods along with you have a method which name is common, you can use your own method implementation by the new keyword.
The crux point is to use it that both the reference and object must be of the same type.
class A
{
public void Test()
{
Console.WriteLine("base");
}
}
class B : A
{
public new void Test()
{
Console.WriteLine("sub");
}
public static void Main(string[] args)
{
A a = new A();
B aa = new B();
a.Test();
aa.Test();
}
}