Working with Interfaces in C#
Interfaces
To rectify the drawback of multiple inheritance, the creators of C# have introduced a new concept called interfaces. Java programmers may be well aware of this concept. All interfaces should be declared with the keyword interface. You can implement any number of interfaces in a single derived class, but you should provide signatures to all method definitions of the corresponding interfaces. To illustrate, Listing 1 shows how to declare interfaces and implement them in a class:
Listing 1
using System;
interface Interdemo
{
void Show();
}
class Interimp:Interdemo
{
public void Show()
{
Console.WriteLine("Show() method Implemented");
}
public static void Main(string[] args)
{
Interimp inter = new Interimp();
inter.Show();
}
}
Combining Interfaces
Two or more interfaces can be combined into a single interface and implemented in a class, as shown in Listing 2:
Listing 2
using System;
interface Interdemo
{
void Show();
}
interface Interdemo1
{
void Display();
}
interface Combineinter:Interdemo,Interdemo1
{
//Above interfaces combined
}
class Multipleinterimp:Combineinter
{
public void Show()
{
Console.WriteLine("Show() method Implemented");
}
public void Display()
{
Console.WriteLine("Display() method Implemented");
}
public static void Main(string[] args)
{
Multipleinterimp inter = new Multipleinterimp();
inter.Show();
inter.Display();
}
}
You easily can determine whether a particular interface is implemented in a class by using is and as operators. The is operator enables you to check whether one type or class is compatible with another type or class; it returns a Boolean value. Listing 3 illustrates the usage of the is operator by revisiting Listing 2.
Listing 3
using System;
interface Interdemo
{
bool Show();
}
interface Interdemo1
{
bool Display();
}
class Interimp:Interdemo
{
public bool Show()
{
Console.WriteLine("Show() method Implemented");
return true;
}
public static void Main(string[] args)
{
Interimp inter = new Interimp();
inter.Show();
if(inter is Interdemo1)
{
Interdemo1 id = (Interdemo1)inter;
bool ok = id.Display();
Console.WriteLine("Method Implemented");
}
else
{
Console.WriteLine("Method not implemented");
}
}
}
Whereas the is operator returns a boolean value as shown in the preceding listing, the as operator returns null if there is any incompatibility between types. Listing 4 examines the usage of this operator. Here we have revisited Listing 3. Notice the change in code inside the Main () method.
Listing 4
using System;
interface Interdemo
{
bool Show();
}
interface Interdemo1
{
bool Display();
}
class Interimpas:Interdemo
{
public bool Show()
{
Console.WriteLine("Show() method Implemented");
return true;
}
public static void Main(string[] args)
{
Interimpas inter = new Interimpas();
inter.Show();
Interdemo1 id = inter as Interdemo1;
if(null!=id)
{
bool ok = id.Display();
Console.WriteLine("Method Implemented");
}
else
{
Console.WriteLine("Method not implemented");
}
}
}
Avoiding Name Ambiguity
Suppose you are declaring same method definitions in two different interfaces. The compiler will naturally show an error due to the ambiguity of the implemented method. Even if you use the "is" keyword, the compiler still will show warnings. To avoid this, you have to follow the syntax as shown in Listing 5:
Listing 5
Void
{
//Body goes here
}
Listing 6 illustrates the application of the preceding concept in detail:
Listing 6
using System;
interface Interdemo
{
void Show();
}
interface Interdemo1
{
void Show();
}
class Interclash:Interdemo,Interdemo1
{
void Interdemo.Show()
{
Console.WriteLine("Show() method Implemented");
}
void Interdemo1.Show()
{
Console.WriteLine("Display() method Implemented");
}
public static void Main(string[] args)
{
Interclash inter = new Interclash();
inter.Interdemo.Show();
inter.Interdemo1.Show();
}
}
By Mey Lia
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