RE: Service program / Activation group problem

[NigsRpgId@xxxxxxx]

Hi Martin,

Thanks for your response.

You got it spot on ! - when I said the activation group was being deleted, I 
should have said it was deliberate - ie there was a RCLACTGRP ACTGRP(*ELIGIBLE) 
coded in Pgm A

>From what you say, the RCLACTGRP doesnt really work if you are using the 
>default activation group - especially if your initial program is running in it 
>!

I can only assume that the reason for the default activation here group is for 
historical reasons - there may have been a time when the system was a mixture 
of OPM and ILE objects.

I would also assume that the RCLACTGRP was put there for a specific reason, so 
I am very wary about removing it as it may be required for another purpose 
unknown to me. (eg ensuring that when a data library is changed all procedures 
use the new library and none are left pointing to the old one ?). Its too much 
of an unknown quantity for me at the moment.

hmmmmm..  some thought is required !

Anyway thanks for the explanation, at least I know why its happening now!

Rgds
Nigel

> My problem is when the user executes the function the first
> time, all works well. However if the process returns to CL
> Pgm A, to set/reset library list, the named activation group
> for the service program D is deleted.

That shouldn't be the case unless a RCLACTGRP command is executed, or the API 
(CEETREC?) is called from within the service program.  If either of these is 
happening, stopping it should remove the problem.

However, the real cause of the problem is program P running in the default 
activation group.  The default activation group is there to provide ILE/OPM 
compatibility, and if you are developing in ILE, you
should really steer well clear of it, as it often leads to problems like this.  
If, in a case like the above, you run RCLACTGRP *ELIGIBLE, for example 
(strongly NOT recommended outside of develpment/testing) the
service program's activation group will be destroyed, but the default one can't 
be.  Then, when you call program A again, the system expects everything 
downstream to be activated (because it doesn't have to create the first 
activation group, I assume).  And when it isn't, you get the behaviour you see.

HTH.

Cheers,

Martin.