________________________________
From: "McKown, John"<[email protected]>
To: [email protected]
Sent: Tuesday, June 12, 2012 2:53 PM
Subject: Re: CEETERM and register 15
The reason for the L in CEETERM is because you can do CEETERM
RC=VARNAME,MODIFIER=MODNAME where VARNAME and MODNAME are fullword areas.
In general, in z/OS, you should __never__ use R0 (impossible actually), R1,
R14, or R15 for a CSECT/RSECT base register. Many (most) IBM macros use these
registers. And occassionally other registers as well.
For LE enabled assembler (I have written some), when I use a base register at all (not often any
more), I use R11 for my first, then R10, R9, and so as as the module grows in size. In
"modern" programming, you can avoid the use of a "base register" for code in
almost every case by using relative instructions instead of based instructions. E.g. use LARL
instead of LA for areas in the CSECT, and Jxx instead of Bxx for branching. If you load constant
values, try to use LHI if possible. Do try to write reentrant code. If you do that, you can keep
the modifiable values in the DSA, which is addressed by R13 and which is set up for you with the
CEEENTRY and CEEDSA macros. For a constant which does not fit into a halfword, or maybe is packed
decimal or characters, you can replace somethin like: L R7,FULLWORD with 2 instructions: LARL
R7,FULLWORD followed by L R7,0(,R7), and the same with packed and characters. OK, that is a bit
ugly, I admit.
--
John McKown
Systems Engineer IV
IT
Administrative Services Group
HealthMarkets®
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-----Original Message-----
From: IBM Mainframe Assembler List
[mailto:[email protected]] On Behalf Of Frank Swarbrick
Sent: Tuesday, June 12, 2012 3:25 PM
To: [email protected]
Subject: CEETERM and register 15
First off please realize that I'm an assembler beginner.
I've fiddled with it here and there for 15 years, but I'm
really just a COBOL programmer. So please be gentle! :-)
We have a vendor written, code supplied, subroutine that does
for us a few things that COBOL can't (bit fiddling; posting
of an ECB; etc.). This same exact routine is called both by
batch COBOL program and CICS COBOL programs. It is not
reentrant. In order to make it reentrant I have been working
on "LE-enabling" it. This seems to work quite well! But I
ran in to something "weird" with CEETERM. CEETERM RC=0
generates the following:
58F0 F098 013A0 1827+ L 15,=A(0)
5800 F098 013A0 1828+ L 0,=A(0)
58DD 0004 00004 1829+ L 13,4(13)
58E0 D00C 0000C 1830+ L 14,12(,13)
981C D018 00018 1831+ LM 1,12,24(13)
07FE 1832+ BR 14
Problem? The program uses register 15 as the base register.
Funny thing is that the code actually "works". But it really
doesn't. The first L sets R15 to 0. Then the second L loads
in to R0 from 152(,R15), but of course R15 is now 0, not the
real base address it needs to be. It's only coincidence that
152 bytes in to low core contains x'00000000'. If I tried to
use an RC of some other value it would still be set to 0.
I changed the program to use 14 instead of 15 for the base
register and all is now fine. But I don't see it documented
that CEETERM should not be used when the base is R15. Or is
it just common practice to not use R15 as the base? Other
than this one program I don't see that R15 is generally used
for that purpose even though it starts out with the correct
address, which is probably why it was used in this case.
Specifically, the original code was this:
SUBR CSECT
USING *,RF
B START
DC CL8'SUBR'
START DS 0H
SAVE (14,12)
Anyway, I have to wonder why CEETERM doesn't take a page from
the RETURN macro, and use LA from an offset into "R0", i.e.,
if I had CEETERM RC=12,MODIFIER=8, we'd get something like this:
LA 15,12(0)
LA 0,8(0)
L 13,4(13)
L 14,12(,13)
LM 1,12,24(13)
BR 14
No base register needed, correct? (This is one of those
cases where "register 0" is really just "zero", right, no
matter what value it actually holds?)
(Apologies to Russ Evans, if he is reading this, for taking
an unexpected side trip to LE land.)
Hope I've not shown too much ignorance.
Thanks!
Frank
