I wrote the application about 15 years ago when we were then VMark
Software. I believe that would make it an internal IBM tool. I have
several other performance measuring tools that I use at customer sites when
analyzing system, application, and other performance issues.
The set of tests are lik
Mark,
What are these timing tests?
They kinda look like the tool PI has. I can almost remember the name of
it now, but it was so long ago.
Is this a tool we users can run or is this something internal at IBM?
If internal, do you have a complete set of comparative speeds for the
various operations
The original request was for a shortest code fragment. This likely meant
as line count, but could have been execution time.
I am using UniVerse 10.1.3 on a 1.7 GHz pentium. Here are some current
timings. Except for test 0001 which is an empty FOR NEXT loop, to the left
of the "|" is a hint of t
users@listserver.u2ug.org
Cc:
Subject: Re: [U2] Finding last day of month
Will,
This was exactly my method, except I was adding 35 days (it could have
been 30 as well) to the first day of the month to get to next month.
I
All,
The "Play Date" has been moved to the U2-Community.
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Will,
This was exactly my method, except I was adding 35 days (it could have been 30
as well) to the first day of the month to get to next month.
I also dislike the lookup table idea i.e. 31,28,30,31,etc, the leap year or
12th month logic.
Thanks for all the responses guys.
[EMAIL PROTECT
> But your's takes more CPU cycles so there :)
> Smaller in opcode? or smaller in milliseconds?
> Ah
Yeah, but now that you've challenged me, this one's smaller in size, op
codes and presumably cpu.
input oDate ; *format mmdd
lastDayOfMonth =
oconv(iconv(oconv(iconv(oDate,'ml#6':@
In a message dated 6/2/2005 8:17:25 PM Pacific Daylight Time,
[EMAIL PROTECTED] writes:
> That's an interesting approach, but it comes at the price of an extra
> ICONV, which is more expensive than a simple IF statement. Granted, the
> difference isn't significant for a small number of invoca
In a message dated 6/2/2005 9:40:08 PM Pacific Daylight Time,
[EMAIL PROTECTED] writes:
> [EMAIL PROTECTED] wrote on 06/02/2005 10:34:09 PM:
>
> > input oDate ; *format mmdd
> > iDate = iconv(oDate,'dymd')
> > loop until oconv(iDate+1,'dd) = 1 do iDate += 1 repeat
> > lastDay = oconv(iDate
Oh please, Kieran. That's overboard.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Kieran Clulow
Sent: Friday, 3 June 2005 11:43 AM
To: u2-users@listserver.u2ug.org
Subject: RE: [U2] Finding last day of month
... or as a one line coding horror
[EMAIL PROTECTED] wrote on 06/02/2005 10:34:09 PM:
> input oDate ; *format mmdd
> iDate = iconv(oDate,'dymd')
> loop until oconv(iDate+1,'dd) = 1 do iDate += 1 repeat
> lastDay = oconv(iDate,'dymd')
>
> Not sure if it's worth bragging about - but mine's smaller (146 v. 161
> bytes).
This w
L PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of [EMAIL PROTECTED]
Sent: Thursday, June 02, 2005 7:46 PM
To: u2-users@listserver.u2ug.org
Subject: Re: [U2] Finding last day of month
In a message dated 6/2/05 7:07:42 PM Pacific Daylight Time, [EMAIL PROTECTED]
writes:
<< Inpu
[EMAIL PROTECTED] wrote on 06/02/2005 09:54:59 PM:
> Input Date ; *format mmdd
> NextMonth = Oconv(Iconv(date[1,6]:'15','D') + 30,'D2')
> I.LastDayPreviousMonth = Iconv(NextMonth[1,2]:'01':NextMonth[5,2],'D') +
1
That's an interesting approach, but it comes at the price of an extra
ICONV, w
In a message dated 6/2/05 7:07:42 PM Pacific Daylight Time, [EMAIL PROTECTED]
writes:
<< Input Date ; * format mmdd
Middleofthismonth = date[1,6]:'15'
I.Middle = Iconv(Middleofthismonth,'D')
I.Next = I.Middle + 30
NextMonth = Oconv(I.NEXT,'D2-')
FirstDayNextMonth = NextMonth[1,2]:'01':Ne
[EMAIL PROTECTED] wrote on 03-06-2005 12.54.59:
>
> Now can we scrunch this down?
> Input Date ; *format mmdd
> NextMonth = Oconv(Iconv(date[1,6]:'15','D') + 30,'D2')
> I.LastDayPreviousMonth = Iconv(NextMonth[1,2]:'01':NextMonth[5,2],'D') +
1
input oDate ; *format mmdd
iDate = iconv(oD
Ok smartass you say let's see you do it
First of all, you can always get from the month you are in, to the next month
by simply add 30 days to the middle of the month. And you never need to
worry about leap years nor about 12th month logic
Input Date ; * format mmdd
Middleofthismonth
... or as a one line coding horror ;)
1 LAST.DAY =
REPLACE(CHANGE('31,28,31,30,31,30,31,31,30,31,30,31',',',@AM),2;28+(NOT(MOD((
THISDATE[1,4]),4)) AND (NOT(MOD((THISDATE[1,4]),400)) OR
MOD((THISDATE[1,4]),100<(THISDATE[5,2]+0)>
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People! use the system why are you doing all this math in your program? The
system knows what years are leap and what aren't. Let it do the work for you!
Will Johnson
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Add of you who are doing special logic on month 12 are FIRED!
Will Johnson
Fast Forward Technologies
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That incorrectly thinks 2100 is a leap.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Ross Morrissey
Sent: Friday, 3 June 2005 10:15 AM
To: u2-users@listserver.u2ug.org
Subject: RE: [U2] Finding last day of month
Aiming for SHORT (not clean, clear, or
YEAR = THISDATE[1,4] ; MONTH = THISDATE[5,2]+0
DAYS = CHANGE('31,28,31,30,31,30,31,31,30,31,30,31',',',@AM)
DAYS<2> += (NOT(MOD(YEAR,4)) AND (NOT(MOD(YEAR,400)) OR MOD(YEAR,100)))
LAST.DATE = DAYS
Should catch everything.
Kieran.
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To uns
On Behalf Of Marco Manyevere
Sent: Thursday, June 02, 2005 10:31 AM
To: u2-users@listserver.u2ug.org
Subject: [U2] Finding last day of month
Given a date like 20040203, I want to return the last valid date for that
month and year (20040229 in this case). What is the shortest code fragment
to
Oh dear, Oh dear!
What ever happend to :
DATE = ''
FOR I = 31 TO 28 STEP -1
IF OCONV(ICONV(I:rest of date,'D'),'D') = I:rest of date THEN
DATE = I:rest of date
EXIT
END
NEXT I
IF DATE = '' THEN Come to screeching HALT
Of course for the people with backward dates (!!!), the "I:rest of
Message-
From: Stevenson, Charles [mailto:[EMAIL PROTECTED]
Sent: Thursday, June 02, 2005 4:02 PM
To: u2-users@listserver.u2ug.org
Subject:RE: [U2] Finding last day of month
I use the method my dad taught me when I was a kid, cuz I couldn't
remember "Thiry days hath
I use the method my dad taught me when I was a kid, cuz I couldn't
remember "Thiry days hath September...".
Hold 2 fists side-by-side, forefingers of right & left hand touching.
Each knuckle and each valley between knuckles represents a month.
Start at the left pinky knuckle and count off months.
sible applications will be a zero wait state memory
device. Beam me up Scotty!
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of Richard A.
Wilson
Sent: Thursday, June 02, 2005 1:38 PM
To: u2-users@listserver.u2ug.org
Subject: Re: [U2] Finding last day of month
I'm
gettin' up there in my years too. Just turned 49 last March.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of Richard A.
Wilson
Sent: Thursday, June 02, 2005 12:29 PM
To: u2-users@listserver.u2ug.org
Subject: Re: [U2] Finding last day of month
y
This is documented back to at least UniData 5.2 and probably earlier than
that though I don't have easy access to the old manuals. It is in the
"Using UniData" manual in the "Creating Virtual Attributes" section. You
may wish to search for "Virtual Attribute Functions". HTH.
Regards,
Jim
[s
It's in my version 5 documentation - not sure when it was put in. There
is also ADD_MONTHS, LAST_DAY, MONTHS_BETWEEN and NEXT_DAY. See Virtual
Attribute Functions in chapter 5 Creating Virtual Attributes.
Colin Alfke
Calgary, Canada
Where it's Sunny, no pouring rain, no wait it's just cloudy, oops
PM
To: u2-users@listserver.u2ug.org
Subject: Re: [U2] Finding last day of month
you cant just use mod(x,4) logic
the following from some website (dont remember where/when I found it)
***
The rule for leap years is that all years divisable by 4 are leap years,
except
those years divisable by 100. The ex
[EMAIL PROTECTED]
Sent: Thursday, June 02, 2005 2:22 PM
To: u2-users@listserver.u2ug.org
Subject: RE: [U2] Finding last day of month
If it's for a UniData dictionary you can use the function LAST_DAY(date)
that will return the last day of the month. I'm not sure if UniVerse has
this functi
Here's yet another way...
END.OF.MONTH
0001 EDATES =
"31":@FM:"29":@FM:"31":@FM:"30":@FM:"31":@FM:"30":@FM:"31":@FM:"31":@FM:
"30":@FM:"31
":@FM:"30":@FM:"31"
0002 PRINT "INPUT DATE MMDD - ": ; INPUT ODATE
0003 EYEAR = ODATE[1,4]
0004 EMONTH = ODATE[5,2]
0005
unidata has a LAST_DAY function? what version is that?
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of
[EMAIL PROTECTED]
Sent: Thursday, June 02, 2005 2:22 PM
To: u2-users@listserver.u2ug.org
Subject: RE: [U2] Finding last day of month
If it's
: u2-users@listserver.u2ug.org
Subject: RE: [U2] Finding last day of month
If it's for a UniData dictionary you can use the function LAST_DAY(date)
that will return the last day of the month. I'm not sure if UniVerse has
this function as well.
Eg:
001: I
002: LAST_DAY(INVOICE.DATE)
003:
: Thursday, June 02, 2005 11:21 AM
To: u2-users@listserver.u2ug.org
Subject: Re: [U2] Finding last day of month
That method can actually backfire, for example, if your starting date is
20040130, you'll end up on Feb 29 instead of Jan 31. What I would do is
replace the day with 01 and add
If it's for a UniData dictionary you can use the function LAST_DAY(date)
that will return the last day of the month. I'm not sure if UniVerse has
this function as well.
Eg:
001: I
002: LAST_DAY(INVOICE.DATE)
003: D4
004: Last Day of Month
005: 15R
* User Supplied ISO Date
UserDate= '20040203'
UserDateYear= UserDate[1,4] + 0
UserDateMonth = UserDate[5,2] + 0
UserDateDay = UserDate[7,2] + 0
* First Day Of The Next Month
NextMonthYear = UserDateYear + (IF (UserDateMonth + 1) > 12 THEN 1 ELSE 0)
NextMonthMonth = (IF (UserDat
the calories out.
===
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] Behalf Of Allen Egerton
> Sent: Thursday, June 02, 2005 12:57 PM
> To: u2-users@listserver.u2ug.org
> Subject: Re: [U2] Finding last day of month
&g
quot;)-1
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Allen Egerton
Sent: Thursday, June 02, 2005 10:57 AM
To: u2-users@listserver.u2ug.org
Subject: Re: [U2] Finding last day of month
From: "Marco Manyevere" <[EMAIL PROTECTED]>
> Hi,
>
gt; From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of
> Marco Manyevere
> Sent: Thursday, June 02, 2005 10:31 AM
> To: u2-users@listserver.u2ug.org
> Subject: [U2] Finding last day of month
>
>
> Hi,
>
> Given a date like 20040203, I want to return t
in this, my allergies are truly messing with my
head today.and for the last weekGaaa
Allen
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of Dianne Ackerman
Sent: Thursday, June 02, 2005 11:21 AM
To: u2-users@listserver.u2ug.org
Subject: Re: [U2] Finding
:YR,"D") - 1
0012: CRT OCONV(NEW.DATE,"D")
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Marco Manyevere
Sent: Thursday, June 02, 2005 10:31 AM
To: u2-users@listserver.u2ug.org
Subject: [U2] Finding last day of month
Hi,
Given a date like 20040203,
[EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of Marco Manyevere
Sent: Thursday, June 02, 2005 01:31 PM
To: u2-users@listserver.u2ug.org
Subject: [U2] Finding last day of month
Hi,
Given a date like 20040203, I want to return the last valid date for that month
and year (20040229 in this
27;/':YR
Perry Taylor
-Original Message-
From: Marco Manyevere [mailto:[EMAIL PROTECTED]
Sent: Thu Jun 02 14:05:42 2005
To: u2-users@listserver.u2ug.org
Subject:[U2] Finding last day of month
Hi,
Given a date like 20040203, I want to return the last valid date for that
mont
That method can actually backfire, for example, if your starting date is
20040130, you'll end up on Feb 29 instead of Jan 31. What I would do is
replace the day with 01 and add 1 to the month, then iconv and subtract
1 day.
-Dianne
Marco Manyevere wrote:
Hi,
Given a date like 20040203, I w
Hi, Marco. I always ICONV the first day of the next month, then
subtract a day. For example:
ANY.DATE = '02-03-2004'
MO = ANY.DATE[1,2]
YR = ANY,DATE[7,4]
MO += 1
IF MO > 12 THEN
MO = 1
YR += 1
END
EOM.DATE = ICONV(MO:'-01-':YR, 'D') - 1
This lends itself to a subroutine or function as
Marco Manyevere <[EMAIL PROTECTED]> wrote on 06/02/2005 01:31:17 PM:
> Given a date like 20040203, I want to return the last valid date for
> that month and year (20040229 in this case). What is the shortest
> code fragment to achieve this?
Ooh! A good old-fashioned programming challenge. I t
I might offer a small modification to your code snippet as noted below.
BobW
>
> From: "Marco Manyevere" <[EMAIL PROTECTED]>
>
>
> > Hi,
> >
> > Given a date like 20040203, I want to return the last valid date for
> that
> month and year (20040229 in this case). What is the shortest code
fragme
You can try ICONV(date,code) and check STATUS(). If status() returns
other than 0 then date is valid
Marco Manyevere wrote:
Hi,
Given a date like 20040203, I want to return the last valid date for that month
and year (20040229 in this case). What is the shortest code fragment to achieve
thi
From: "Marco Manyevere" <[EMAIL PROTECTED]>
> Hi,
>
> Given a date like 20040203, I want to return the last valid date for that
month and year (20040229 in this case). What is the shortest code fragment
to achieve this?
Find the internal date of the first day of the next month, subtract 1, and
c
Hi,
Given a date like 20040203, I want to return the last valid date for that month
and year (20040229 in this case). What is the shortest code fragment to achieve
this?
At the moment I'm replacing the day with 01, then iconv, add 35 days to the
internal date and then oconv and replace the day
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