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

P: n/a
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.

Nov 14 '05 #1
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"Stu" <be********@hotmail.com> wrote in message
news:11**********************@z14g2000cwz.googlegr oups.com...
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.
Nov 14 '05 #2

P: n/a
Stu wrote:

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;
}
Nov 14 '05 #3

P: n/a
On 22 Feb 2005 07:46:11 -0800, in comp.lang.c , "Stu"
<be********@hotmail.com> wrote:
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.

--
Mark McIntyre
CLC FAQ <http://www.eskimo.com/~scs/C-faq/top.html>
CLC readme: <http://www.ungerhu.com/jxh/clc.welcome.txt>

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Nov 14 '05 #4

P: n/a
"dandelion" <da*******@meadow.net> wrote:
"Stu" <be********@hotmail.com> wrote in message
news:11**********************@z14g2000cwz.googlegr oups.com...
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.


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

Richard
Nov 14 '05 #5

P: n/a

"Richard Bos" <rl*@hoekstra-uitgeverij.nl> wrote in message
news:42***************@news.individual.net...
"dandelion" <da*******@meadow.net> wrote:
"Stu" <be********@hotmail.com> wrote in message
news:11**********************@z14g2000cwz.googlegr oups.com...
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.


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.

Nov 14 '05 #6

P: n/a
dandelion wrote:
"Richard Bos" <rl*@hoekstra-uitgeverij.nl> wrote in message
"dandelion" <da*******@meadow.net> wrote:
"Stu" <be********@hotmail.com> wrote in message

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.
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

.... 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[i] := 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 *)

--
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the broken "Reply" link at the bottom of the article. Click on
"show options" at the top of the article, then click on the
"Reply" at the bottom of the article headers." - Keith Thompson
Nov 14 '05 #7

P: n/a
On Thu, 24 Feb 2005 12:18:24 +0100, "dandelion" <da*******@meadow.net>
wrote:
<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.

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
Nov 14 '05 #8

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