Converting seconds to (Days, Hours, Minutes, seconds)

[Stu]

Is there a simple function call within "C" that I can use to convert
number of seconds (keep in mind this may be a type longlong and has to
work on UNIX and NT) into Days, months, Hours, Minutes and seconds.

If anybody can provide me with an example I would be very grateful.

Thanks in advance for all that answer this post.

 

[dandelion]

Is there a simple function call within "C" that I can use to convert
number of seconds (keep in mind this may be a type longlong and has to
work on UNIX and NT) into Days, months, Hours, Minutes and seconds.
No.

If anybody can provide me with an example I would be very grateful.

Calculating a number of months, days, hours and minutes from a number
of seconds is rather trivial, so I won't bother.

If you intend to calculate a date/time pair from a number os seconds since
some predefined point (the famous UNIX EPOCH for instance) things will be a
lot trickier, since you will have to take leapyears into account too. In
that case i'd suggest using the systems functions for that and write a
wrapper that will (depending on some compiler definition) call the
appropriate function out of your system libs.

 

[infobahn]

Is there a simple function call within "C" that I can use to convert
number of seconds (keep in mind this may be a type longlong and has to
work on UNIX and NT) into Days, months, Hours, Minutes and seconds.

If anybody can provide me with an example I would be very grateful.

Thanks in advance for all that answer this post.

The following code ignores months, pending your definition thereof.
(30 days? 365/12.0 days? 28 days? 27.whatever days?)

typedef unsigned long TIMEINT; /* change to long long if you wish */

void ConvSeconds(TIMEINT *d,
TIMEINT *hr,
TIMEINT *min,
TIMEINT *sec,
TIMEINT s)
{
*d = s / 86400;
s %= 86400;
*hr = s / 3600;
s /= 3600;
*min = s / 60;
*sec = s % 60;
}

 

[Mark McIntyre]

Is there a simple function call within "C" that I can use to convert
number of seconds (keep in mind this may be a type longlong and has to
work on UNIX and NT) into Days, months, Hours, Minutes and seconds.

I have a feeling that your "time in seconds" is a time_t object maybe....
Note that this is not guaranteed to be a measure of seconds.

If anybody can provide me with an example I would be very grateful.

localtime() and gmtime() will do this for time_t objects.

 

[Richard Bos]

Calculating a number of months, days, hours and minutes from a number
of seconds is rather trivial, so I won't bother.

Actually, calculating months is not that trivial. The problem is, what
months? 30 days, 31, 28, 30.5, 365.2425/12?

Richard

 

[dandelion]

Actually, calculating months is not that trivial. The problem is, what
months? 30 days, 31, 28, 30.5, 365.2425/12?

I'd just create a table holding the days_in_month. Something like

days_in_month[] =
{
31, /* January */
28,
31,
30,

/* etc. */

31 /* December */
};

Then, you need 'leapyear' info for February, which isn't that hard to
calculate. IIRC it's a leapyear if year % 4 == 0 && year % 100 != 0.

But ideas on what's trivial and what's not can be very different. I agree to
that.

 

[CBFalconer]

I'd just create a table holding the days_in_month. Something like
.... snip ...

Then, you need 'leapyear' info for February, which isn't that
hard to calculate. IIRC it's a leapyear if year % 4 == 0 &&
year % 100 != 0.

Here is a routine I used to use to compute yr, mo, day from the
CP/M date stamp, which corresponded to days since 1 Jan 1978. To
use it you need to make a preliminary conversion of seconds since
something into days since something. Dealing with quad years of
1461 days avoids a plethora of silly errors. No need to worry
further about leap year until 2100.

PROCEDURE drtodate(thedate : integer; VAR yr, mo, day : integer);
(* 1 Jan 1978 corresponds to Digital Research date = 1 *)
(* BUG - cannot handle negative values for dates > 2067 *)

VAR
i, y1 : integer;
dayspermonth : ARRAY[1..12] OF 1..31;

BEGIN (* drtodate *)
FOR i := 1 TO 12 DO dayspermonth := 31;
dayspermonth[4] := 30; dayspermonth[6] := 30;
dayspermonth[9] := 30; dayspermonth[11] := 30;
IF thedate > 731 THEN BEGIN (* avoid overflows *)
yr := 1980; thedate := thedate - 731; END
ELSE BEGIN
thedate := thedate + 730; yr := 1976; END;
(* 0..365=y0; 366..730=y1; 731..1095=y2; 1096..1460=y3 *)
i := thedate DIV 1461; thedate := thedate MOD 1461;
y1 := (thedate-1) DIV 365; yr := yr + y1 + 4*i;
IF y1 = 0 THEN (* leap year *) dayspermonth[2] := 29
ELSE BEGIN
thedate := thedate - 1; (* 366 -> 365 -> 1 Jan *)
dayspermonth[2] := 28; END;
day := thedate - 365*y1 + 1; mo := 1;
WHILE day > dayspermonth[mo] DO BEGIN
day := day - dayspermonth[mo];
mo := succ(mo); END;
END; (* drtodate *)

 

[Dave Thompson]

Then, you need 'leapyear' info for February, which isn't that hard to
calculate. IIRC it's a leapyear if year % 4 == 0 && year % 100 != 0.

Officially || year % 400 == 0. Or equivalently but more symmetrically
year % 4 == 0 && !(year % 100 == 0 && !(year % 400 == 0 )) .

2000 was a leap year. The next effect of the 400 term is 2400, and
it's not unthinkable that we will have a very different calendar by
then. In fact if you want to repeat the Y2K type of mistake you can
just do year % 4 and assume (or try to require) your application won't
be used for dates beyond 2099 or before 1901.
- David.Thompson1 at worldnet.att.net