Develop Reference

Polymorphism misunderstanding; accidentally typed 'virtual' instead of 'override' with unexpected result [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");
}
}
}

Why in this code the output of the virtual functions is like this? [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");
}
}
}

C# Casting to the base class and using it's methods

I tried casting a new variable/object to it's base class but when I debug I notice it has the class members of the derived class. So when I call an over ridden method it calls the derived class instead of the base even though the object is declared as the base.
Person person = new Student("Bill", 3.5, "chem");
Console.WriteLine(person.ToString());
....
public class Person
{
string name;
public override string ToString() => this.name + " age:" + this.age;
}
public class Student : Person
{
string major;
double GPA;
public override string ToString() => Name + " major:" + major + " GPA: " + GPA;
}
To remedy this I need to define the ToString() method in the derived class as new. But I don't understand why as I'm casting to the base so it should use the base method? Is it due to the way the compiler treats reference objects? The new object of base class still references the derived class memory instead of copying the appropriate data to it's own memory block.

Yes and that's known as Polymorphism or run time polymorphism. It's happening so cause the actual type is Student even though the declared type is person in your below code line. So the method call happens on the actual type and not on the declared type.
Person person = new Student("Bill", 3.5, "chem");

There are three keywords needed to control this behaviour with polymophism. Virtual, Override, and New. These are explained well here, and from MSDN:
Override:
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.
New:
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.
An Example:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace BindingExample
{
public class SomeBaseClass
{
public virtual string CustomString() => "From SomeBaseClass";
}
public class FirstSpecialisation : SomeBaseClass
{
public override string CustomString() => "From FirstSpecialisation";
}
public class SecondSpecialisation : SomeBaseClass
{
public new string CustomString() => "From SecondSpecialisation";
}
class Program
{
static void Main(string[] args)
{
// Base Example, as expected
SomeBaseClass a1 = new SomeBaseClass();
Console.WriteLine(a1.CustomString());
// First Example, both output the same result
FirstSpecialisation b1 = new FirstSpecialisation();
SomeBaseClass b2 = b1;
Console.WriteLine(b1.CustomString());
Console.WriteLine(b2.CustomString());
// Second Example, output different results
SecondSpecialisation c1 = new SecondSpecialisation();
SomeBaseClass c2 = c1;
Console.WriteLine(c1.CustomString());
Console.WriteLine(c2.CustomString());
Console.ReadLine();
}
}
}
Output:
From SomeBaseClass
From FirstSpecialisation
From FirstSpecialisation
From SecondSpecialisation
From SomeBaseClass
Note the above behaviour with the SecondSpecialisation class on lines 4 and 5.
This is particularly interesting in your example, because you are overriding a method with 2 depths. If you use overide twice, an unusual behaviour is evoked; the last override expressed takes precedence. Hence, you must use the new keyword.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace BindingExample
{
public class SomeBaseClassA
{
public override string ToString() => "From SomeBaseClassA";
}
public class SpecialisationA : SomeBaseClassA
{
public override string ToString() => "From SpecialisationA";
}
public class SomeBaseClassB
{
public new string ToString() => "From SomeBaseClassB";
}
public class SpecialisationB : SomeBaseClassB
{
public new string ToString() => "From SpecialisationB";
}
class Program
{
static void Main(string[] args)
{
// First Example
SomeBaseClassA a1 = new SomeBaseClassA();
Console.WriteLine(a1);
SpecialisationA a2 = new SpecialisationA();
Console.WriteLine(a2);
SomeBaseClassA a3 = a2;
Console.WriteLine(a3);
// Second Example
SomeBaseClassB b1 = new SomeBaseClassB();
Console.WriteLine(b1.ToString());
SpecialisationB b2 = new SpecialisationB();
Console.WriteLine(b2.ToString());
SomeBaseClassB b3 = b2;
Console.WriteLine(b3.ToString());
Console.ReadLine();
}
}
}
Output:
From SomeBaseClassA
From SpecialisationA
From SpecialisationA
From SomeBaseClassB
From SpecialisationB
From SomeBaseClassB
I did observe when overriding the ToString method that it was called on Object. I would assume this is because the Console.WriteLine method can take an object, and because the new keyword was used on the 'ToString' methods within the specialisations, the compiler calls Object.ToString instead. This is something you may wish to be mindful of.

Sealed method in C#

I am a newbie in C#.I am reading about Sealed keyword.I have got about sealed class.I have read a line about Sealed method where we can make Sealed method also.The line was (By declaring method as sealed, we can avoid further overriding of this method.)
I have created a demo but didn't understand that the meaning of above line and sealed method use. Below is my code:-
using System;
namespace ConsoleApplication2
{
class Program:MyClass
{
public override sealed void Test()
{
Console.WriteLine("My class Program");
}
static void Main(string[] args)
{
Program obj = new Program();
obj.Test();
Console.ReadLine();
}
}
class MyClass
{
public virtual void Test()
{
Console.WriteLine("My class Test");
}
}
}
Please tell me why we use Sealed methods and what are the advantages of Sealed methods.

Well, you are testing with only two levels of inheritance, and you are not getting to the point that you're "further overriding" a method. If you make it three, you can see what sealed does:
class Base {
public virtual void Test() { ... }
}
class Subclass1 : Base {
public sealed override void Test() { ... }
}
class Subclass2 : Subclass1 {
public override void Test() { ... } // Does not compile!
// If `Subclass1.Test` was not sealed, it would've compiled correctly.
}

A sealed class is a class which cannot be a base class of another more derived class.
A sealed method in an unsealed class is a method which cannot be overridden in a derived class of this class.
Why do we use sealed methods? What are the advantages of sealed methods?
Well, why do you use virtual methods? To provide a point at which behaviour of a class can be customized. So why do you use sealed methods? To provide a point at which you are guaranteed that no further changes will occur in the behaviour of any derived class with respect to this method.
Points where the behaviour of a class can be customized are useful but dangerous. They are useful because they enable derived classes to change behaviour of the base class. They are dangerous... wait for it... because they enable derived classes to change behaviour of the base class. Virtual methods basically allow third parties to make your classes do crazy things that you never anticipated or tested.
I like writing code that does what I anticipate and what I tested. Sealing a method allows you to continue to allow parts of the class to be overridden while making the sealed methods have guaranteed, testable, stable behaviour that cannot be further customized.

Well, you'd only use "sealed" on a method if you didn't want any derived classes to further override your method. Methods are sealed by default, if they're not declared as virtual and not overriding another virtual method. (In Java, methods are virtual by default - to achieve "default" C# behaviour you have to mark a method as final.)
Personally I like to control inheritance carefully - I prefer whole classes to be sealed where possible. However, in some cases you still want to allow inheritance, but ensure that some methods aren't overridden further. One use might be to effectively template the method, e.g. for diagnostics:
public sealed override void Foo(int x)
{
Log("Foo called with argument: {0}", x);
FooImpl(x);
Log("Foo completed");
}
protected abstract void FooImpl(int x);
Now subclasses can't override Foo directly - they'd have to override FooImpl, so our behaviour will always be executed when other code calls Foo.
The templating could be for other reasons of course - for example to enforce certain aspects of argument validation.
In my experience sealed methods aren't used terribly often, but I'm glad the ability is there. (I just wish classes were sealed by default, but that's another conversation.)

Now if you declare a subclass of Program you won't be able to override the Test method, that's the point.
In most cases you won't need sealed methods but they can prove themselves useful when you're developing a base class for some plugin system that third-party developers have to extend in order to create their own plugin. You can keep some service data like plugin name and whether it is enabled and use sealed methods and properties to do it, so plugin developers won't mess with it. That's one case when sealed members come handy

Sealed Methods
Sealed method is used to define the overriding level of a virtual method.
Sealed keyword is always used with override keyword.
Practical demonstration of sealed method
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace sealed_method
{
class Program
{
public class BaseClass
{
public virtual void Display()
{
Console.WriteLine("Virtual method");
}
}
public class DerivedClass : BaseClass
{
// Now the display method have been sealed and can;t be overridden
public override sealed void Display()
{
Console.WriteLine("Sealed method");
}
}
//public class ThirdClass : DerivedClass
//{
// public override void Display()
// {
// Console.WriteLine("Here we try again to override display method which is not possible and will give error");
// }
//}
static void Main(string[] args)
{
DerivedClass ob1 = new DerivedClass();
ob1.Display();
Console.ReadLine();
}
}
}

First of all, let's start with a definition; sealed is a modifier which if applied to a class make it non-inheritable and if applied to virtual methods or properties makes them non-ovveridable.
public sealed class A { ... }
public class B
{
...
public sealed string Property { get; set; }
public sealed void Method() { ... }
}
An example of its usage is specialized class/method or property in which potential alterations can make them stop working as expected (for example, the Pens class of the System.Drawing namespace).
...
namespace System.Drawing
{
//
// Summary:
// Pens for all the standard colors. This class cannot be inherited.
public sealed class Pens
{
public static Pen Transparent { get; }
public static Pen Orchid { get; }
public static Pen OrangeRed { get; }
...
}
}
Because a sealed class cannot be inherited, it cannot be used as base class and by consequence, an abstract class cannot use the sealed modifier. It's also important to mention that structs are implicitly sealed.
Example
public class BaseClass {
public virtual string ShowMessage()
{
return "Hello world";
}
public virtual int MathematicalOperation(int x, int y)
{
return x + y;
}
}
public class DerivedClass : BaseClass {
public override int MathematicalOperation(int x, int y)
{
// since BaseClass has a method marked as virtual, DerivedClass can override it's behavior
return x - y;
}
public override sealed string ShowMessage()
{
// since BaseClass has a method marked as virtual, DerivedClass can override it's behavior but because it's sealed prevent classes that derive from it to override the method
return "Hello world sealed";
}
}
public class DerivedDerivedClass : DerivedClass
{
public override int MathematicalOperation(int x, int y)
{
// since BaseClass has a method marked as virtual, DerivedClass can override it's behavior
return x * y;
}
public override string ShowMessage() { ... } // compile error
}
public sealed class SealedClass: BaseClass {
public override int MathematicalOperation(int x, int y)
{
// since BaseClass has a method marked as virtual, DerivedClass can override it's behavior
return x * y;
}
public override string ShowMessage()
{
// since BaseClass has a method marked as virtual, DerivedClass can override it's behavior but because it's sealed prevent classes that derive from it to override the method
return "Hello world";
}
}
public class DerivedSealedClass : SealedClass
{
// compile error
}
Microsoft documentation
https://learn.microsoft.com/en-us/dotnet/csharp/language-reference/keywords/sealed

C# has it because Java has an identical feature (final methods). I've never seen a legitimate use.
If you don't want to permit extension, the whole class should be marked sealed and not just one method.

C# - Keyword usage virtual+override vs. new

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");
}
}
}