It as elements of PL/I because of the DCL statement
-----Original Message-----
From: IBM Mainframe Discussion List [mailto:[email protected]] On Behalf
Of Paul Gilmartin
Sent: Thursday, August 3, 2017 3:35 PM
To: [email protected]
Subject: Re: Someone just too smart for his or her own good?
On Thu, 3 Aug 2017 19:11:07 +0000, Robert Prins wrote:
>Just came across the following, and please don't come back with
>pedantic remarks about undeclared variables, the code is just to show what's
>there:
>
>dcl sum fixed (7) init (-0.1);
>
I don't know what language this is, and I'm probsbly not skilled in it.
Regardless, "fixed" and "-0.1" seem contradictory; meriting a syntax error.
>for i = 1 to whatever;
> if a(i) >= 0 then
> sum = sum + a(i);
>
And there, at least I'd do:
> if a(i) > 0 then
> sum = sum + a(i);
>
as long as testing, why bother to add 0?
>if substr(unspec(sum), 25, 8) ^= '0d'bx then
> put data(sum);
>
What's this testing for; in what representation? Is a single non-numeric
comparison cheaper than a numeric comparison?
>In other words if all a(i) are negative, nothing is printed. A comment
>in the code suggests that this is faster code, on modern OoO z/OS
>systems, than the more logical:
>
>dcl sum fixed (7) init (-1);
>
>for i = 1 to whatever;
> if a(i) >= 0 then
> if sum ^= -1 then
> sum = sum + a(i);
> else
> sum = a(i);
>end;
>
Unless the preponderance of the items are non-positive, the test probably costs
more than it saves.
I like to structure my loops, so neither boundary, the first item nor tne last,
requires special treatment. This violates that.
And I studiously avoid the over-and-under read, such as:
read X # Get first item.
if not eof then do
process X
read X # Get next item.
if eof then break
done
>end;
>
>It probably is if the value of whatever is in the order of 42
>gazillion, but any other thoughts about this?
>
None suitable for publication, although events of the past week have eroded
that standard.
-- gil
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