datetime - How does the (Oracle) Java JVM know a leap second is occurring?

Question

A leap second will occur on June 30, 2015. Different Operating Systems seem to handle this situation differently. In my particular case, we are running a Red Hat 6.4 system with custom Java (JDK 1.7) software that is heavily time-dependent. According to some recent Red Hat released information I found, our system's NTP daemon will ensure the OS automatically handle the leap second by repeating 23:59:59 twice.

My question is: if I have a long running JDK 1.7 process, how does it know that a leap second is occurring? What I mean is, how does Java ultimately know the IERS people have decided to insert a leap second? The Date documentation seems to indicate is aware of leap seconds, but seems unhelpfully vague. Can I assume the JDK, when the appropriate Date object is constructed or Calendar.getInstance() is invoked, it is a pass through to the underlying OS's date-time handling to get the appropriate "real" time value? (Which, in my case, sounds like it would repeat second 23:59:59, because that's how the OS will handle it).

question from:https://stackoverflow.com/questions/30984599/how-does-the-oracle-java-jvm-know-a-leap-second-is-occurring

Answer 1

It depends on your jdk version. If for example you are running update 80, you can check the release notes:

JDK 7u80 contains IANA time zone data version 2015a. For more information, refer to Timezone Data Versions in the JRE Software.

Then follow the link to the timezone data versions and find 2015a. Then follow the link to TZ Updater version1.4.11:

New leap second 2015-06-30 23:59:60 UTC as per IERS Bulletin C 49. (Thanks to Tim Parenti.)

It does not seem to have been included before so if you are running an older version of JDK 7 you will probably not get the adjustment. More information about how it works internally is available here.

To be honest I have never tested to see how it works in practice.