Getting Date Time in Unix Time as Byte Array which size is 4 bytes with Java

Question

How Can I get the date time in unix time as byte array which should fill 4 bytes space in Java?

Something like that:

byte[] productionDate = new byte[] { (byte) 0xC8, (byte) 0x34,
                    (byte) 0x94, 0x54 };

Solution 1

First: Unix time is a number of seconds since 01-01-1970 00:00:00 UTC. Java's System.currentTimeMillis() returns milliseconds since 01-01-1970 00:00:00 UTC. So you will have to divide by 1000 to get Unix time:

int unixTime = (int)(System.currentTimeMillis() / 1000);

Then you'll have to get the four bytes in the int out. You can do that with the bit shift operator >> (shift right). I'll assume you want them in big endian order:

byte[] productionDate = new byte[]{
        (byte) (unixTime >> 24),
        (byte) (unixTime >> 16),
        (byte) (unixTime >> 8),
        (byte) unixTime

};

Solution 2

You can use ByteBuffer to do the byte manipulation.

int dateInSec = (int) (System.currentTimeMillis() / 1000);
byte[] bytes = ByteBuffer.allocate(4).putInt(dateInSec).array();

You may wish to set the byte order to little endian as the default is big endian.

To decode it you can do

int dateInSec = ByteBuffer.wrap(bytes).getInt();

Comments

[Joop Eggen]

Tip, though not more efficient: ... = ByteBuffer.allocate(4).putInt(unixTime).array();.

[Jesper]

It's shorter, but a lot more overhead than my solution, to allocate a ByteBuffer just for this...

[Vishy]

@Jesper true, although with escape analysis in Java 8 the ByteBuffer can be allocated on the stack which is not so expensive.