Put the CL in debug mode and look to see what line 73 has to offer and then end debug.
From: midrange-l-bounces@xxxxxxxxxxxx [mailto:midrange-l-bounces@xxxxxxxxxxxx] On Behalf Of optional info0rmation
Sent: Wednesday, October 09, 2013 5:48 PM
To: 'Midrange Systems Technical Discussion'
Subject: V5R4 CLP question
Here is the situation
QSTRUP in QGPL is defined in sysval
System is IPLed and the QSTRUP aborts on line 73
Message ID . . . . . . : CPA0701 Severity . . . . . . . : 99
Message type . . . . . : Inquiry
Date sent . . . . . . : 09/06/13 Time sent . . . . . . : 21:35:13
Message . . . . : CPF1010 received by QSTRUP at 7300. (C D I R)
Cause . . . . . : Control language (CL) program QSTRUP in library QGPL
detected an error at statement number 7300. Message text for CPF1010 is:
Subsystem name ZMPLUS active.
Recovery . . . : This inquiry message can be avoided by changing the
program. Monitor for the error (MONMSG command) and perform error recovery
within the program. To continue, choose a reply value.
Possible choices for replying to message . . . . . . . . . . . . . . . :
C -- Cancel the CL program.
D -- Dump the CL program variables and cancel the CL program.
I -- Ignore the failing command.
I looked at the QCLSRC in QGPL for QSTRUP and looked for line 73
There was no line 73 It ended at line 63
I know the problem, source and object are not the same
I did a RTVCLSRC and named it QSTRUP2
I went in and looked at the source code and the 2 source codes were the same.
I compiled QSTRUP2 and the printout of the compiled program only shows 63 lines
How can I have a program abort on line 73 with only 63 lines in the total program?
Thanks in advance
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