Question

I know that IDataReader is a interface so I can’t create an instance of interface.

However, it is possible to get instances of IdataReader.

How can this be possible?

Answer 1

You cannot create an instance of an Interface. But you can create an instance of a class that implements an interface. In case of IDataReader SqlDataReader is an implementation.

You get an instance by executing a SqlCommand:

SqlDataReader reader = command.ExecuteReader();

For example (from MSDN):

string queryString =
    "SELECT OrderID, CustomerID FROM dbo.Orders;";

using (SqlConnection connection =
           new SqlConnection(connectionString))
{
    SqlCommand command =
        new SqlCommand(queryString, connection);
    connection.Open();

    // here is your SqlDataReader that implements IDataReader
    SqlDataReader reader = command.ExecuteReader();

    // Call Read before accessing data.
    while (reader.Read())
    {
        Console.WriteLine(String.Format("{0}, {1}",
            reader[0], reader[1]));
    }

    // Call Close when done reading.
    reader.Close();
}

Edit:

I assume that you don’t understand why it’s possible to have an instance that has the type of an interface. That is no contradiction since an interface actually is a type.

So if i would create a class, say MyDataReader, that implements IDataReader, an instance of my class would be of type MyDataReader as well as IDataReader.

If anybody would create an instance of MyDataReader, he knows that it also implements all methods of IDataReader. That is important for Polymorphism.