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No. Setll only matches the data type. It doesn't care about the number of keys.

Here is what I use:
setll (ctcom#:ctrctf:ctctr#:ctent#) icectlrr; (record format)
dou %eof(icect3); (filename)
reade (ctcom#:ctrctf:ctctr#:ctent#) icectlrr; (record format)
if not %eof(icect3); (filename)

endif;
enddo;


-----Original Message-----
From: RPG400-L [mailto:rpg400-l-bounces@xxxxxxxxxxxx] On Behalf Of Vinay Gavankar
Sent: Wednesday, August 15, 2018 9:55 AM
To: RPG programming on the IBM i (AS/400 and iSeries) <rpg400-l@xxxxxxxxxxxx>
Subject: Re: RPG Compiler Question

Since the fifth key is one character, wouldn't there be an error either in compilation or execution?

On Wed, Aug 15, 2018 at 9:37 AM, Michael Ryan <michaelrtr@xxxxxxxxx> wrote:

I’ll bet it concatenated the value of MyKey6 to the literal ‘R’.

Sent from my iPhone

On Aug 15, 2018, at 9:34 AM, Vinay Gavankar <vinaygav@xxxxxxxxx> wrote:

Hi,

We have a logical file with 6 keys.

We had following code running in Production for about 8 months:

Setll (MyKey1:MyKey2:MyKey3:MyKey4:'R' +
MyKey6) MyFile ;
If not %eof(MyFile) ;
ReadE (MyKey1:MyKey2:MyKey3:MyKey4:'R' +
MyKey6) MyFile ;

We recently discovered the missing colon between the 5th and 6th key
when trying to find the root cause of some missing data in the file.
It seemed that the 2 statements were being executed with just 5 keys.

I was wondering what the compiler did with the 6th parameter in the
statements. When I changed MyKey6 to an undefined variable, the
compiler gave an error.

TIA

Vinay
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