Interface in C# with Example
Interface is an another important pillar of object oriented programming. An interface only contains declaration of its members like properties, methods, delegates, and events, but not fields. Make sure, you should not provide implementation for any of its member inside the interface. To create an interface, we make use of interface keyword. By common naming convention, whenever we create an interface, we put I as prefix. It helps in quickly recognizing that we are using an interface.
All the interface members are public by default, so make sure you should not put any kind of access modifier in front of any of its member. A class and structure both can inherit from an interface, so in this case, we must provide implementation for all of its members.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Hello
{
interface IEmployee
{
void Print();
}
class Employee : IEmployee
{
public string Name { get; set; }
public void Print()
{
Console.WriteLine("Employee Name: {0}", Name);
}
}
class Program
{
static void Main(string[] args)
{
Employee emp = new Employee();
emp.Name = "Albert Pinto";
emp.Print();
}
}
}
Output:
Employee Name: Albert Pinto
Press any key to continue . . .
In C#, multiple interface inheritance is allowed. Due to this, if your class or structure is inheriting from multiple interfaces, you must provide implementation for all the members available in those interfaces, failing to do so will result in compile time error.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Hello
{
interface IEmployee
{
void PrintName();
}
interface IEmployee2
{
void PrintID();
}
class Employee : IEmployee, IEmployee2
{
public string Name { get; set; }
public string ID { get; set; }
public void PrintName()
{
Console.WriteLine("Employee Name: {0}", Name);
}
public void PrintID()
{
Console.WriteLine("Employee ID: {0}", ID);
}
}
class Program
{
static void Main(string[] args)
{
Employee emp = new Employee();
emp.Name = "Albert Pinto";
emp.ID = "100";
emp.PrintName();
emp.PrintID();
}
}
}
Output:
Employee Name: Albert Pinto
Employee ID: 100
An interface can also inherit from another interface or multiple interfaces. In this case, you again have to provide implementation for all the interface members available in that inheritance chain.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Hello
{
interface IEmployee
{
void PrintName();
}
interface IEmployee2
{
void PrintID();
}
interface IEmployee3 : IEmployee, IEmployee2
{
void PrintDesignation();
}
//class
class Employee : IEmployee3
{
public string Name { get; set; }
public string ID { get; set; }
public string Designation { get; set; }
public void PrintName()
{
Console.WriteLine("Employee Name: {0}", Name);
}
public void PrintID()
{
Console.WriteLine("Employee ID: {0}", ID);
}
public void PrintDesignation()
{
Console.WriteLine("Employee Designation: {0}", Designation);
}
}
class Program
{
static void Main(string[] args)
{
Employee emp = new Employee();
emp.Name = "Albert Pinto";
emp.ID = "100";
emp.Designation = "Supervisor";
emp.PrintName();
emp.PrintID();
emp.PrintDesignation();
}
}
}
Output:
Employee Name: Albert Pinto
Employee ID: 100
Employee Designation: Supervisor
The most important thing which you have to remember is that you cannot create an instance of interface like a class or a structure because interface only contains declaration, not implementation. However, from class inheritance, we know that a base class reference variable can point to the derived class object. Same goes in this case, if your class or structure in inheriting from an interface, you can create a reference variable of that interface and point it to the derived class or structure object.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Hello
{
interface IEmployee
{
void PrintName();
}
//class
class Employee : IEmployee
{
public void PrintName()
{
Console.WriteLine("Employee Name: Albert");
}
}
class Program
{
static void Main(string[] args)
{
//Interface reference variable
IEmployee emp = new Employee();
emp.PrintName();
}
}
}
Output:
Employee Name: Albert
