How is it possible to initialize an interface?

Question

What I have learned so far is that we can not create an instance of an interface.

    interface IFood
    {
        string Color { get; set; }
    }
    class Apple : IFood
    {
        public string Color { get ; set; }
    }

IFood food = new IFood(); //this gives compile error

IFood food = new Apple(); //this will work

Upto here everything were okay. But when I work with Microsoft.Office.Interop.Excel I have seen something like below

Application excel = new Application();// Application is an interface

What am I missing here?

COM uses a hyper-pure interface-based programming style. You only ever use an interface, the implementation is completely hidden. Basic reason why it is usable in practically any language. The only place where the notion of a "class" ever comes up is when you need to create an object that implements the interface you want to use. Done hyper-pure style with a factory method. That function only needs a {guid} that identifies the class, the C# compiler can dig it out of the declaration. — Hans Passant
– Hans Passant, Commented Dec 31, 2019 at 11:56

nyconing · Accepted Answer · 2020-01-20 06:35:44Z

10

It was an interop ability of COM

Microsoft.Office.Excel API including the Application class, are written in C++

Due to architectural in C++ are more freedom, initialize an interface is needed in some case

.

.NET uses CoClass attribute on a COM object to workaround with initiate an interface

C# wont allow to initiate an interface, but with a CoClass attribute, the interface initialization can be routed to the class CoClass

(example code worth thousand words) So lets reproduce this workaround:

[CoClass(typeof(SugarGlider))] 
[ComImport] // pretend as a COM class
[Guid("000208D5-0000-0000-C000-000000000046")] // put it randomly just to fool the ComImport
public interface ISquirrel
{
     string Foo();
}

[ClassInterface(ClassInterfaceType.None)]
public class SugarGlider : ISquirrel
{
    public string Foo(){ return "Bar"; }
}

You can now initiate the interface by new ISquirrel()

Full example and runs online: https://rextester.com/ORAZQ51751

edited Jan 20, 2020 at 6:35

answered Dec 31, 2019 at 5:36

nyconing

1,1361 gold badge10 silver badges33 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Beingnin Over a year ago

Nice explanation

JohanP · Accepted Answer · 2019-12-31 05:33:21Z

2

It uses CoClass attribute, which is a COM concept. That attribute allows you tell the compiler that your interface will be implemented by Application class, thus allows you to instantiate the interface like that.

answered Dec 31, 2019 at 5:33

JohanP

5,5423 gold badges27 silver badges37 bronze badges

Comments

NGambit · Accepted Answer · 2019-12-31 05:42:44Z

2

The magic is happening because of the CoClass attribute. It declares that the interface Application is to be implemented by ApplicationClass

That’s why the compiler allows Application excel = new Application(); since it can infer what class to instantiate (i.e. ApplicationClass)

What does the C# CoClass attribute do?

How does the C# compiler detect COM types?

edited Dec 31, 2019 at 5:42

answered Dec 31, 2019 at 5:30

NGambit

1,19114 silver badges29 bronze badges

Comments

Indar · Accepted Answer · 2019-12-31 05:24:08Z

-3

An interface can't be instantiated directly.

"Interfaces Summary" section for more details - https://learn.microsoft.com/en-us/dotnet/csharp/programming-guide/interfaces/

answered Dec 31, 2019 at 5:24

Indar

3152 silver badges12 bronze badges

Collectives™ on Stack Overflow

How is it possible to initialize an interface?

4 Answers 4

1 Comment

Comments

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

1 Comment

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related