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C# program to implement multiple interfaces with the same method in the same class

Here, we are going to learn how to implement multiple interfaces with the same method in the same class in C#?
Submitted by Nidhi, on October 14, 2020

Here, we will implement two interfaces with the same method in the same class. Each interface contains a method declaration.

Program:

The source code to implement multiple interfaces with the same method in the same class is given below. The given program is compiled and executed successfully on Microsoft Visual Studio.

//C# program to implement multiple interfaces 
//with the same method in the same class.

using System;

interface MyInf1
{
    //Method Declaration
    void Method();
}

interface MyInf2
{
    //Method Declaration
    void Method();
}

class Sample : MyInf1,MyInf2
{
    //Method definitions
    void MyInf1.Method()
    {
        Console.WriteLine("MyInf1:Method() called");
    }
    void MyInf2.Method()
    {
        Console.WriteLine("MyInf2:Method() called");
    }    
}

class Program
{
    public static void Main(String[] args)
    {
        MyInf1 M1;
        MyInf2 M2;

        M1 = new Sample();
        M2 = new Sample();

        M1.Method();
        M2.Method();
    }
}

Output:

MyInf1:Method() called
MyInf2:Method() called
Press any key to continue . . .

Explanation:

Here, we created the two interfaces MyInf1 and MyInf2. Both Interfaces contains the declaration of Method(). After that, we implemented both interfaces into class Sample with method definitions.

void MyInf1.Method()
{
    Console.WriteLine("MyInf1:Method() called");
}
void MyInf2.Method()
{
    Console.WriteLine("MyInf2:Method() called");
}

Here we have to specify the interface name with method name to define the method inside the class.

Now look to the Program class, It contains the Main() method, the Main() method is the entry point for the program. Here we created two references M1 and M2.

M1 = new Sample();
M2 = new Sample();

Here, both references initialized with the object of Sample class. But we called Method() using both references that will print the corresponding message on the console screen.

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