Why do I keep devoting thought to this?
If you desperately needed to optimize every last cycle out of the loop, and you
wanted to do this in a supported way, the trick would be two loops. Something
like
for i = 1 to whatever;
if a(i) >= 0 then do
sum = sum + a(i);
switch = 1;
break? /* what is the PL/I to exit from a loop? */ ;
end;
end;
for i = i to whatever;
if a(i) > 0 then sum = sum + a(i);
end;
I'm not a PL/I coder so not sure if you need a test for i already == whatever
when it gets to the second loop or whether it has already been incremented,
etc.
Probably faster than the original because avoids most of the a(i) == 0 cases.
Charles
-----Original Message-----
From: IBM Mainframe Discussion List [mailto:[email protected]] On Behalf
Of Robert Prins
Sent: Friday, August 4, 2017 10:48 AM
To: [email protected]
Subject: Re: Someone just too smart for his or her own good?
On 2017-08-04 13:41, Charles Mills wrote:
> OK, I get it. (Way too much time spent on this now.) The sum is also
> to be printed if there are any zero values in a[whatever].
>
> Is this a real business problem or an exercise in whatever?
The original code actually fills a set of buckets with values (or nothing), I
just simplified the code to a loop over an array, changing the values in the
array to do some testing, after first figuring out what was going on.
> I don't know the PL/I compiler but on a modern processor Store on
> Condition potentially makes
>
> if a(i) >= 0 then switch = 1;
It's more than likely that Enterprise PL/I V5.2 can already generate code that
utilizes new features of the z14 systems.
Robert
--
Robert AH Prins
robert.ah.prins(a)gmail.com
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