// Solution: in many cases, we do not know whether the object supports an interface. In this case, we can try to convert the object to an interface type.
// In c #, we can use either of the following methods to obtain an object that supports an interface:
The Code is as follows:
Using System;
Using System. Collections. Generic;
Using System. Text;
Namespace ConsoleApplication6
{
Class Program
{
Static void Main (string [] args)
{
Conn cn = new conn ();
If (cn is Iconnect) // whether the interface is implemented
{
Iconnect iconn = (Iconnect) cn;
Iconn. read ();
Console. ReadLine ();
}
Else
{
Console. WriteLine ("can not connvert ");
Console. ReadLine ();
}
// The IS operator IS clear, but the operation efficiency IS not high. The other method IS
Conn cnn = new conn ();
Iconnect icnn1 = cnn as Iconnect;
If (icnn1! = Null)
{
Icnn1.write ();
Console. ReadLine ();
}
Else
{
Console. WriteLine ("can not connvert ");
Console. ReadLine ();
}
// The as operator first tests whether the conversion is legal. If the conversion is legal, the conversion is performed. Otherwise, NULL is returned. Keyword NULL indicates NULL reference
}
}
Public interface Iconnect
{
Void read ();
Void write ();
}
Public class conn: Iconnect
{
Public void read ()
{
Console. WriteLine ("read () method implemented ");
}
Public void write ()
{
Console. WriteLine (
