I don't understand.

How is your code different than the code I posted below (which behaves
"correctly")?

========================

using System;

namespace ArrayAssignment
{
 class Foo // some class to put in the array
 {
  private long myId;

  public Foo( long aId)
  {
   myId = aId;
  }

  public long ID
  {
   get { return myId;}
   set { myId = value;}
  }
 }

 class Class1
 {
  [STAThread]
  static void Main(string[] args)
  {
   Foo[] aFoo1;
   Foo[] aFoo2;
   long i;

   // initialize first array
   aFoo1 = new Foo[10];

   for (i = 0; i < 10; i++)
      {
    aFoo1[i] = new Foo(i);
   }

   // second array references first array
   aFoo2 = aFoo1;

   //check values
   Console.WriteLine(aFoo1[3].ID); // should be same as index = 3
   Console.WriteLine(aFoo2[3].ID); // should be same as aFoo1 = 3

   aFoo2[3].ID = 27;

   //check values
   Console.WriteLine(aFoo1[3].ID); // should be new value = 27
   Console.WriteLine(aFoo2[3].ID); // should still be same as aFoo1 = 27

   // make a new array
   aFoo2 = new Foo[aFoo1.Length];

   for (i = 0; i < aFoo2.Length; i++)
   {
    aFoo2[i] = new Foo(i + aFoo2.Length);
   }

   //check values
   Console.WriteLine(aFoo1[3].ID);  // orignal array has not changed = 27
   Console.WriteLine(aFoo2[3].ID);  // now have new array value is index +
length = 13

   aFoo2[3].ID = 100;

   //check values
   Console.WriteLine(aFoo1[3].ID);  // orignal array has not changed - 27
   Console.WriteLine(aFoo2[3].ID);  // should be new value = 100
  }
 }
}



----- Original Message -----
From: "Kerry Whelan" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Friday, May 24, 2002 2:53 AM
Subject: Re: [DOTNET] Array addresses


> I had tried this:-
>
> ConfigApp.Config.ConfigCollection newColl = new
> ConfigApp.Config.ConfigCollection[orgColl.Length];
>
> already, but it has the same effect as newColl = orgColl;
>
> Any other ideas?
>
> Kerry
>
> You can read messages from the DOTNET archive, unsubscribe from DOTNET, or
> subscribe to other DevelopMentor lists at http://discuss.develop.com.
>
>

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