Re: How the values are handled for the instances?

[Bill McCormick]

You can never have more then one "copy" of the same object instance open 
in memory. If you opened ID 1 and ID 2 you can edit them both as much as 
you like. But reopening 1 while the original pointer is still in memory 
just creates a second reference to the in memory object.

Balaji.R wrote:
Hi All,
I am perplexed about the strategy of handling the objects in Cach�.
Let me explain my question with an example.
set objEmp1=##class(Employee.EmpDetails).%OpenId(1)
Write !,objEmp1.MiddleInitial
Output: T
set objEmp1.MiddleInitial="A"
Write !,objEmp1.MiddleInitial
Output: A
without saving this instance, I tried to create another instance for this
class and ID.
set objEmp2=##class(Employee.EmpDetails).%OpenId(1)
Write !,objEmp2.MiddleInitial
Output: A
My question is,
When I tried to change the MiddleInitial using the first instance objEmp1, I
could find the MiddleInitial updated with 'A' for the instance objEmp1. But
without saving or killing the instance objEmp1, when I tried to create
another instance of the same class and ID say objEmp2, and output the
MiddleInitial value, the value which was set for objEmp1 was displayed for
objEmp2's property also.
How could the multiple instances share the same value without any
updation/saving made to the prior instance ? How the values are handled for
the instances?
Kindly Clarify. FYI, I am using Cach� 5.0
Thanks in Advance.
-Balaji