date/time macros

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comrade

Joined: 16 Jun 2003
Posts: 1150
Location: Russian Federation

comrade 27 Feb 2004, 16:50

Privalov wrote these macros:

Code:

macro months [dayscount]
{
  forward
   if DAY > dayscount
    DAY = DAY-dayscount
    MONTH = MONTH+1
  forward
   end if
}
TIME = %T
DAY = TIME/(24*3600)
DAY = DAY - (DAY+365)/(3*365+366)
YEAR = 1970+DAY/365
DAY = DAY mod 365 + 1
MONTH = 1
months 31,28,31,30,31,30,31,31,30,31,30,31

TIME = TIME mod (24*3600)
HOUR = TIME/3600
MINUTE = (TIME mod 3600)/60
SECOND = (TIME mod 3600) mod 60
DATE equ (DAY / 10 + '0'),(DAY MOD 10 + '0'),".",(MONTH / 10 + '0'),(MONTH MOD 10 + '0'),".",(YEAR / 1000 + '0'),((YEAR / 100) MOD 10 + '0'),((YEAR / 10) MOD 10 + '0'),(YEAR MOD 10 + '0')
TIME equ (HOUR / 10 + '0'),(HOUR MOD 10 + '0'),":",(MINUTE / 10 + '0'),(MINUTE MOD 10 + '0'),":",(SECOND / 10 + '0'),(SECOND MOD 10 + '0')

It breaks down timestamp into individual fields. Today is 27th, but it says 25th. It worked before though, only now I notice there is problem. I am not too bright to figure out what is wrong. Does anyone have any idea?

_________________
comrade (comrade64@live.com; http://comrade.ownz.com/)

27 Feb 2004, 16:50

wolf

Joined: 28 Sep 2003
Posts: 11

wolf 23 Mar 2004, 07:34

hi,

I think there is a bug in the make_timestamp routine in fasm because there is a difference of up to two days in comparison to TimeDate-Functions calculated by GB32 for Windows depending on SystemTime.

In the calculation, shouldn't it be:

Code:

DAY = DAY - (DAY-730-60)/(3*365+366)-1

-730 subtracts the first two years 1970 and 1971
-1 takes into account the leap year 1972
- 60 takes into account that after 4 years plus 60 days there is a 29th of February

The macro has to be changed as well, as the 29th of February in a Leap Year will not be calculated correctly.

This is just a quick idea, maybe I'm wrong!

wolf

23 Mar 2004, 07:34

Tomasz Grysztar

Joined: 16 Jun 2003
Posts: 8466
Location: Kraków, Poland

Tomasz Grysztar 23 Mar 2004, 11:32

The macro is buggy, I have never dedicated enough time to fix it. However the make_timestamp gives me a good results (the same values as those given by Linux, which uses such time format internally)

23 Mar 2004, 11:32

wolf

Joined: 28 Sep 2003
Posts: 11

wolf 24 Mar 2004, 23:06

I wrote another macro:

Code:

macro calcdate 
      { 
         REPLOOP = DAY / 365 
         repeat REPLOOP 
            if YEAR mod 4 = 0 
               DAY = DAY - 366 
            else 
               DAY = DAY - 365 
            end if 
            YEAR = YEAR +1 
         end repeat 
         if YEAR mod 4 = 0 
            months 31,29,31,30,31,30,31,31,30,31,30,31 
         else 
            months 31,28,31,30,31,30,31,31,30,31,30,31 
         end if 
      } 

DAY = %T / 86400 +1 
YEAR = 1970 
MONTH = 1 
calcdate

The macro is probably buggy and you are right: the timestamp is ok.

But when I change my system time to the 29.02.2004 the macro shows a difference of two days.
And after I patched the timestamp routine with the above code, the macro gives correct results again.
When you subtract timestamp1 (29.02.) from timestamp2 (24.03.) you get a number of seconds which roughly represent 26 days, which resembles the difference in the macro results.
Sorry, but I still think there must be a bug in the timestamp routine.

wolf

24 Mar 2004, 23:06

comrade

Joined: 16 Jun 2003
Posts: 1150
Location: Russian Federation

comrade 24 Mar 2004, 23:10

timestamp routine works fine, because it is used for PE header and in resource macros. I checked PE header in a program which explodes stamp correctely with correct date.

_________________
comrade (comrade64@live.com; http://comrade.ownz.com/)

24 Mar 2004, 23:10

Tomasz Grysztar

Joined: 16 Jun 2003
Posts: 8466
Location: Kraków, Poland

Tomasz Grysztar 03 Apr 2004, 17:07

Here's the corrected macro:

Code:

macro months [dayscount]
{
  forward
   if DAY > dayscount
    DAY = DAY-dayscount
    MONTH = MONTH+1
  forward
   end if
}
TIME = %T
DAY = TIME/(24*3600)
DAY = DAY - (DAY+365)/(3*365+366)
YEAR = 1970+DAY/365
DAY = DAY mod 365 + 1
MONTH = 1
if YEAR mod 4 = 0
 FEBDAYS=29
else
 FEBDAYS=28
end if
months 31,FEBDAYS,31,30,31,30,31,31,30,31,30,31

It should give you correct values for dates up to the year 2100. Wink

03 Apr 2004, 17:07

comrade

Joined: 16 Jun 2003
Posts: 1150
Location: Russian Federation

comrade 04 Apr 2004, 03:10

Thanks!

_________________
comrade (comrade64@live.com; http://comrade.ownz.com/)

04 Apr 2004, 03:10

pelaillo
Missing in inaction

Joined: 19 Jun 2003
Posts: 878
Location: Colombia

pelaillo 04 Apr 2004, 15:27

This one even after 2100...

Code:

macro months [dayscount] 
{ 
  forward 
   if DAY > dayscount 
    DAY = DAY-dayscount 
    MONTH = MONTH+1 
  forward 
   end if 
} 
TIME = %T 
DAY = TIME/(24*3600) 
DAY = DAY - (DAY+365)/(3*365+366) 
YEAR = 1970+DAY/365 
DAY = DAY mod 365 + 1 
MONTH = 1 
if YEAR mod 4 = 0 
   if YEAR mod 100 = 0
         if YEAR mod 400 = 0
                 FEBDAYS=29
          else
                        FEBDAYS=28 
         end if
      else
                FEBDAYS=29 
 end if
else 
     FEBDAYS=28 
end if 
months 31,FEBDAYS,31,30,31,30,31,31,30,31,30,31

Gregorian's Calendar:
The day after Oct 4, 1582 was Oct 15, 1582.
Before reform, every 4 years were a leap year.
Leap year if divisible by 4 and not by 100
Leap year if divisible by 100 and by 400

04 Apr 2004, 15:27

Tomasz Grysztar

Joined: 16 Jun 2003
Posts: 8466
Location: Kraków, Poland

Tomasz Grysztar 04 Apr 2004, 16:19

But you are not taking into account the years divisible by 100 when they occur between 2000 and the year of timestamp.

04 Apr 2004, 16:19

pelaillo
Missing in inaction

Joined: 19 Jun 2003
Posts: 878
Location: Colombia

pelaillo 04 Apr 2004, 18:10

Because the years divisible by 4 and 100 and not by 400 are not leap.

04 Apr 2004, 18:10

Tomasz Grysztar

Joined: 16 Jun 2003
Posts: 8466
Location: Kraków, Poland

Tomasz Grysztar 04 Apr 2004, 20:30

So you should substract one form the count of days for each such year, that occurs between 1970 and the year you are calculating

04 Apr 2004, 20:30

Aster!x

Joined: 16 Jul 2004
Posts: 26

Aster!x 30 Jul 2004, 21:15

Maybe someone already tried to write such macro for fasm?

Code:

date MACRO
% FORC chr, @Date
db "&chr"
ENDM
ENDM

time MACRO
% FORC chr, @Time
db "&chr"
ENDM
ENDM

30 Jul 2004, 21:15

xanatose

Joined: 09 Jan 2004
Posts: 57

xanatose 19 Apr 2006, 21:16

Thank you for the macro

Here is a macro to get a string in the form YYY.MM.DD, this is usefull for version information.

Code:

macro datestring szName {
        common label szName
        db ((YEAR / 1000)+'0')
        db (((YEAR mod 1000) / 100) + '0')
        db (((YEAR mod 100) / 10) + '0')
        db ((YEAR mod 10) + '0')
        db '.'
        db ((MONTH / 10) + '0')
        db ((MONTH mod 10) + '0')
        db '.'
        db ((DAY / 10) + '0')
        db ((DAY mod 10) + '0')
        lenof.#szName = $ - szName
        db 0
}

I use this in the data section like this

Code:

; for example APR 19, 2006 would be '2006.04.19',0
datestring szVersionString

19 Apr 2006, 21:16

vid
Verbosity in development

Joined: 05 Sep 2003
Posts: 7103
Location: Slovakia

vid 12 Feb 2007, 15:31

could someone extend this to also get current time ?

12 Feb 2007, 15:31

comrade

Joined: 16 Jun 2003
Posts: 1150
Location: Russian Federation

comrade 04 Jul 2008, 04:51

What do you mean, "get current time" ? Isn't the timestamp %T itself, current time?

04 Jul 2008, 04:51

vid
Verbosity in development

Joined: 05 Sep 2003
Posts: 7103
Location: Slovakia

vid 04 Jul 2008, 08:43

I was talking about macro to parse timestamp into symbolic constants (hour / min / sec)

04 Jul 2008, 08:43

baldr

Joined: 19 Mar 2008
Posts: 1651

baldr 19 Sep 2008, 22:34

hour equ (%t / 3600) mod 24
min equ (%t / 60) mod 60
sec equ %t mod 60

All timestamps are UTC (GMT±0) Wink

19 Sep 2008, 22:34

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