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How did the op-code version open the file? Input?

Blocking was likely used that couldn't be with the file opened for Update.

Also, the op-code version didn't by chance open the file without specifying
it was a keyed file did it?

Charles

On Mon, Sep 11, 2017 at 2:57 PM, Nathan Andelin <nandelin@xxxxxxxxx> wrote:

On Mon, Sep 11, 2017 at 2:36 PM, Charles Wilt <charles.wilt@xxxxxxxxx>
wrote:

It's not the CALL themselves that's the problem...


I'm not precisely sure what has caused such a big drag on the I/O
performance. I noted in an earlier post that runtime execution path in the
service program entails:

2 procedure calls.
2 monitor, on-error blocks.
2-3 select blocks.

before an I/O op code is invoked.

The commit keyword on the "f" spec forces the service program to run under
commitment control, which may be a factor. I didn't force that in the other
program because reading a file front to back doesn't require commitment
control.



Can't tell for sure what it is without the code, but I'd suspect
- repeated opens


No, the file is opened only once and never closed until the activation
group ends. Look at the source code.

if not %open(cusmstf);
open cusmstf;
endif;



- ACTGRP(*NEW)


No, the service program runs in the *caller activation group.



- data copying


Yes, a factor, I think.
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