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Here is some code that shows how to use itoa:

ctl-opt DftActGrp(*No) ActGrp(*new);

dcl-s str Char(33);
dcl-s rsp Char(10);


str = '899,999: ' + IToBase36(899999);
dsply str '*EXT' rsp;

str = '1,024: ' + IToBase36(1024);
dsply str '*EXT' rsp;

str = '999,999,999: ' + IToBase36(999999999);
dsply str '*EXT' rsp;

return;

dcl-proc IToBase36;
dcl-pi *n Char(33);
number Int(10) Const;
end-pi;

dcl-pr itoa Pointer ExtProc('__itoa');
num Int(10) Value;
str Char(33);
base Int(10) Value;
end-pr;

dcl-s string Char(33) Inz('');

string = %str(itoa(number: string: 36));

return string;
end-proc;

Mark Murphy
STAR BASE Consulting, Inc.
mmurphy@xxxxxxxxxxxxxxx


-----"Mark Murphy/STAR BASE Consulting Inc." <mmurphy@xxxxxxxxxxxxxxx> wrote: -----
To: Midrange Systems Technical Discussion <midrange-l@xxxxxxxxxxxx>
From: "Mark Murphy/STAR BASE Consulting Inc." <mmurphy@xxxxxxxxxxxxxxx>
Date: 06/18/2015 09:07AM
Subject: Re: The case of outnumbered numerator

I have done this using the C library function itoa() pass in an integer value and a base (base 36 will give 0-9 and A-Z). You can search google for usage documentation.

Mark Murphy
STAR BASE Consulting, Inc.
mmurphy@xxxxxxxxxxxxxxx


-----Gad Miron <gadmiron@xxxxxxxxx> wrote: -----
To: midrange-l@xxxxxxxxxxxx
From: Gad Miron <gadmiron@xxxxxxxxx>
Date: 06/17/2015 01:45AM
Subject: The case of outnumbered numerator

Hello Pundits

We have a six digits numerator that approaches 899999.

Since it happens to be a CHAR field I would like to start using the
letters A-Z
in lexical order when incrementing this field so that incrementing
899999 will result in 9AAAAA , 9AAAAB , ......., 9AAAAZ, 9AAAA0, 9AAAA1 etc.
(This way the numerator will sort the EBCDIC natural way)

Do you know an *elegant* way of doing it ?
(EBCDIC table having "holes" of unprintable chars does not expedite matters)

Thanks
Gad

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