UNIX for Dummies Questions & Answers

Is there a function call in std library or unit command that returns the number of current leap seconds?

GG

No. The standard for UNIX timekeeping does not include leap seconds.

NTP servers will report correct UTC values as leap seconds are inserted.
See:
NIST Time Scale Data Archive

Here is how I would do it - this assumes your system time is actually corrected by an ntp server.

1. calculate the Julian day for Jan 1, 1970 - call it day1

2. calculate the Julian day for today -- day2

There is code to do this on this forum - search for it. Julian days == number of days since Jan 1 4712 BCE

3. use the fact there are 86400 seconds in a day.

Code:

# this works at midnight 00:00:00 hh:mm:dd
time_t leapsec=time(NULL) - ( (day2 * 86400) - day1 * 86400) );

FYI as of this date - Jun 5 2009 - leapsec = 24

corrected by jmc

Jim,

Thanks! I will search for the sample code to convert Julian days to time.

My current legacy code (but just updated) uses the number of leap seconds as 15 seconds. Where do you get 25 seconds?

GG

I got 25 because I cannot count well at 4:00am my time. The correct number is 24.

Try this site -
NIST Time Scale Data Archive

-----Post Update-----

Try not to PM with questions.

Your "friend" got 15 from using GPS epoch seconds - 0000 (midnight) on Jan 6 1980.
Leap seconds started being added in 1972. So there are more leap seconds on the NIST calendar than on the GPS calendar, since GPS started with leap seconds under its belt.

TAI - UTC is now 35 seconds. So in another sense there have been 35.

Normally leap seconds are derived as changes to UTC - what computers are supposed to be synced to by the worldwide NTP service. 24 leap seconds.

It makes sense. We use leap seconds to convert GPS time to UTC time.

Again thanks!