Subject | (Addendum)Re: [firebird-support] Problem Using MOD |
---|---|
Author | Mahesh Ishwar |
Post date | 2003-07-19T06:16:57Z |
Sorry I forgot to write my firebird version
ISQL Version: WI-T1.5.0.3481 Firebird 1.5 Release Candidate 3
Firebird/x86/Windows NT (access method), version "WI-T1.5.0.3481 Firebird 1.5 Release Candidate 3"
Thanx.
Mahesh.
Nairo Granados <nairox@...> wrote:
Hi All,
I'm using the function MOD in a stored procedure in Firebird 1.5 to
get the remainder of a division but the operation get the wrong
result.
For Example
A = MOD(2000, 19);
The result is 105 instead of 5.
I don't know why is this happen. Somebody could help me???
This is the UDF function
DECLARE EXTERNAL FUNCTION MOD
INTEGER,
INTEGER
RETURNS INTEGER
ENTRY_POINT 'IB_UDF_mod' MODULE_NAME 'ib_udf';
Thanks a lot,
Nairo Granados
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ISQL Version: WI-T1.5.0.3481 Firebird 1.5 Release Candidate 3
Firebird/x86/Windows NT (access method), version "WI-T1.5.0.3481 Firebird 1.5 Release Candidate 3"
Thanx.
Mahesh.
Nairo Granados <nairox@...> wrote:
Hi All,
I'm using the function MOD in a stored procedure in Firebird 1.5 to
get the remainder of a division but the operation get the wrong
result.
For Example
A = MOD(2000, 19);
The result is 105 instead of 5.
I don't know why is this happen. Somebody could help me???
This is the UDF function
DECLARE EXTERNAL FUNCTION MOD
INTEGER,
INTEGER
RETURNS INTEGER
ENTRY_POINT 'IB_UDF_mod' MODULE_NAME 'ib_udf';
Thanks a lot,
Nairo Granados
Yahoo! Groups SponsorADVERTISEMENT
To unsubscribe from this group, send an email to:
firebird-support-unsubscribe@yahoogroups.com
Your use of Yahoo! Groups is subject to the Yahoo! Terms of Service.
---------------------------------
Yahoo! Plus - For a better Internet experience
[Non-text portions of this message have been removed]