C# - Keyword usage virtual+override vs. new

C#SyntaxOverridingMethod HidingMember Hiding

C# Problem Overview


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?

C# Solutions


Solution 1 - C#

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.

Virtual/Override explanation image

Solution 2 - C#

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.

Solution 3 - C#

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;
    }
}

Solution 4 - C#

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.

Solution 5 - C#

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.

Solution 6 - C#

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.

Solution 7 - C#

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

Solution 8 - C#

  • 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.

Solution 9 - C#

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.

Solution 10 - C#

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).

Solution 11 - C#

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");  
        }  
    }  
}  

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
Questioni3ensaysView Question on Stackoverflow
Solution 1 - C#Orion EdwardsView Answer on Stackoverflow
Solution 2 - C#alberteinView Answer on Stackoverflow
Solution 3 - C#Franci PenovView Answer on Stackoverflow
Solution 4 - C#Joseph DaigleView Answer on Stackoverflow
Solution 5 - C#tvanfossonView Answer on Stackoverflow
Solution 6 - C#WedgeView Answer on Stackoverflow
Solution 7 - C#NescioView Answer on Stackoverflow
Solution 8 - C#ChetanView Answer on Stackoverflow
Solution 9 - C#Dror WeissView Answer on Stackoverflow
Solution 10 - C#Emre TapcıView Answer on Stackoverflow
Solution 11 - C#MehdiView Answer on Stackoverflow