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Change
myArray(11-i) = myArray(10-i);
to
myArray(%Elem(myArray) +1 -i) = myArray(%Elem(myArray) -i);

-----Original Message-----
From: rpg400-l-bounces@xxxxxxxxxxxx [mailto:rpg400-l-bounces@xxxxxxxxxxxx] On Behalf Of Jeff Crosby
Sent: Friday, April 09, 2010 6:52 AM
To: RPG programming on the IBM i / System i
Subject: Re: Redefining a DS

I tweaked it a bit to use the %Elem BIF. Will this work?

d myArray s 9p 2 dim(10)
d i s 10i 0

/FREE

// Move One Right
for i = 1 to %Elem(myArray)-1;
myArray(11-i) = myArray(10-i);
endfor;
myArray(1) = 0;

// Move One Left
for i = 1 to %Elem(myArray)-1;
myArray(i) = myArray(i+1);
endfor;
myArray(10) =

*INLR = *On;
Return;

/END-FREE

Hey! It compiles so there's no need to test!

:)


On Fri, Apr 9, 2010 at 7:48 AM, <hr@xxxxxxxxxxxx> wrote:
Here is a NON-DS solution:

d myArray s 9p 2 dim(10)
d i s 10i 0

// Move One Right
for i = 1 to 9;
myArray(11-i) = myArray(10-i);
endfor;
myArray(1) = 0;

// Move One Left
for i = 1 to 9;
myArray(i) = myArray(i+1);
endfor;
myArray(10) = 0;

/Henrik





"James Newman, CDP" <newmanas400@xxxxxxxxx>
Sent by: rpg400-l-bounces@xxxxxxxxxxxx
09-04-2010 03:16
Please respond to
RPG programming on the IBM i / System i <rpg400-l@xxxxxxxxxxxx>


To
rpg400-l@xxxxxxxxxxxx
cc

Subject
Redefining a DS






Okay, chalk this up to I'm an old fart and can't remember how to do this,
although I know I've done it in the past.

I have an array of 10 elements. I want to move #9 to #10, #8 to #9, etc.
I
can write code that does that. In talking over this with a friend today
he
suggested moving the array to a data structure, redefining it, and moving
all but 1 element back to the array. "Pretty smart", I said. Now I wish
I'd said "hey, can you send me an example of how you do that?"

Could one of you young whipper-snappers post the way the DS should be
coded? TIA.


Jim Newman
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