Re: Newbie Q: How to program in UTC (time/calendar) ?

[reinkor <reinkor@gmail.com>]

Sorry, I now realized that this was not so intuitive for me.
Given the following (modified) code:

------------------------------------------------
with Text_IO;
use  Text_IO;

with Ada.Calendar.Formatting,Ada.Calendar.Time_Zones;
use  Ada.Calendar,Ada.Calendar.Formatting,Ada.Calendar.Time_Zones;

procedure t1 is

   package Int_Io  is new Text_IO.Integer_Io (Integer);
   use Int_Io;

   Time1,Time2 : Time;

begin

   Time1 := Value("1970-01-01 00:00:00");
   Time2 := Ada.Calendar.Time_Of(1970,1,1,0.0);

   Put("Time1: ");Put(Image(Time1));New_Line;
   Put("Time2: ");Put(Image(Time2));New_Line;

   Put("UTC_Time_Offset: ");Put(Integer(UTC_Time_Offset),8);

end t1;
------------------------------------

This gives the following output on my computer:

Time1: 1970-01-01 00:00:00
Time2: 1969-12-31 23:00:00
UTC_Time_Offset:      120



The first line I now understand reflects that both "Value"
and "Image" has parameter Time_Zone = 0 (UTC). I.e. "Time1"
can be understood to have value 1970-01-01 00:00:00 (UTC).
"But "Time2" is 1970-01-01 00:00:00 *minus* one hour (only), and
UTC_Time_Offset = 120 (2 hours).
I would expect that it should be UTC - 2 hours
(i.e. 1969-12-31 22:00:00 and *not* 1969-12-31 23:00:00) ?

I here fight with my intuition :-)

reinert





On 27 Mar, 18:58, "Jeffrey R. Carter"
<spam.jrcarter....@nospam.acm.org> wrote:
> reinkor wrote:
> >    Time1 := Value("1970-01-01 00:00:00");
> >    Put(Image(Time1));New_Line;
> >    Put(Image(Ada.Calendar.Time_Of(1970,1,1,Duration(0))));
>
> > 1970-01-01 00:00:00
> > 1969-12-31 23:00:00
>
> > The last output line here reflects the fact that Ada takes
> > "local time" from the computer - and the result would be
> > different if I wait to run my program till next Monday
> > (this weekend we change to summer daylight saving time).
>
> No. It reflects the fact that Image takes an Ada.Calendar.Time_Zones.Time_Offset
> parameter (Time_Zone) that defaults to 0 (UTC). Image is therefore converting
> the supplied time from your computer's offset
> (Ada.Calendar.Time_Zones.UTC_Time_Offset) to the default offset of 0. Your
> offset seems to be 60 (GMT + 1 hr). To get the image of what you passed to it,
> you must specify an offset of UTC_Time_Offset.
>
> The 1st value comes out as expected because Value also takes an
> Ada.Calendar.Time_Zones.Time_Offset parameter (Time_Zone) that defaults to 0
> (UTC). It is therefore taking the supplied string, interpreting it as a time in
> UTC, and converting it to the corresponding time for your computer's offset
> (UTC_Time_Offset). When you pass this time to Image with the default for
> Time_Zone, the conversion is then reversed as described above.
>
> You can compare the representations provided by Image with the values obtained
> by Ada.Calendar.Split to observe the conversions that the operations in
> Ada.Calendar.Formatting perform.
>
> Two comments on your code: "with Ada.Calendar.Formatting;" makes Ada,
> Ada.Calendar, and Ada.Calendar.Formatting visible; there is no need to also with
> Ada.Calendar.
>
> I cannot understand why you would say "Duration (0)" when "0.0" is clearer and
> shorter.
>
> --
> Jeff Carter
> "My mind is a raging torrent, flooded with rivulets of
> thought, cascading into a waterfall of creative alternatives."
> Blazing Saddles
> 89