ByteBuffer putLong() for long of 7-byte value

Question

I have a long. I know this long is not very big. It could be encoded as an unsigned integer into 7 bytes, actually.

I would like to write my long into byte[7] using ByteBuffer and the little endian.

This

long myLong = ... //some value
byte[] myLongBytes = ByteBuffer.allocate(7).order(ByteOrder.LITTLE_ENDIAN).putLong(myLong).array();

throws IndexOutOfBoundsException, of course. Can I even do this with ByteBuffer? Should I do it in a different way?

@Pooya But I need to encode it into 7 bytes only, because I know the value is not that big... I simply don't need 8 bytes. — vojta, Mar 02 '16 at 07:52
@vojta JVM doesn't know that and try to put 8 byte into 7 byte allocated buffer. you can simply use 8 bytes and then throw the most significant byte away (if it was positive ofcourse) — Pooya, Mar 02 '16 at 07:54
@Pooya Yes, of course, but I don't find it very efficient. I would have to allocate two arrays instead of one. — vojta, Mar 02 '16 at 07:56
@Pooya And the code above is just an example, I need to write many values, the long is just one of many in the row. — vojta, Mar 02 '16 at 07:56

score 6 · Accepted Answer · answered Mar 07 '16 at 20:43

(Assuming little endian byte order as indicated in the OP)

(Assuming you want 56 least significant bits from the long being stored)

long myLong = someLongValue & 0x00FFFFFFFFFFFFFFL;
ByteBuffer bb = ByteBuffer.allocate(xxx).order(ByteOrder.LITTLE_ENDIAN);

Option A: Write 64 bit long and than remove one byte

bb.putLong(myLong);
bb.position(bb.position()-1);

for tail write (i.e. this long is the last write to the ByteBuffer):

bb.putLong(myLong);
bb.limit(bb.position()-1);

Beware that this approach works only if you can ensure the ByteBuffer capacity is large enough to fit this temporary extra byte.

Some example cases where this approach should be OK:

where this 56bit value is always followed by another field (then you are sure there is some capacity reserve)
when you are strictly using the ByteBuffer.limit() feature and not (ab)using the ByteBuffer for comfortable byte array writes (which is implied by your post)

Option B: Write the long "per partes"

bb.putInt((int)myLong);
bb.putShort((short)(myLong>>>32));
bb.put((byte)(myLong>>>48));

This works every time.

Option C: Use byte[] instead of long

You might consider using byte array instead of long (depends on your use case).

Good luck!

Option A is fine provide you don't mind writing one extra bytes, Option B is better if this is a problem. Option C mean using the answer I gave. +1 for Option B — Peter Lawrey, Mar 08 '16 at 09:19

score 5 · Answer 2 · answered Mar 02 '16 at 09:28

You need to write a byte at a time such as (based on java.nio.Bits)

static private long makeLong(byte b6, byte b5, byte b4,
                             byte b3, byte b2, byte b1, byte b0)
{
    return ((((long)b6 & 0xff) << 48) |
            (((long)b5 & 0xff) << 40) |
            (((long)b4 & 0xff) << 32) |
            (((long)b3 & 0xff) << 24) |
            (((long)b2 & 0xff) << 16) |
            (((long)b1 & 0xff) <<  8) |
            (((long)b0 & 0xff)      ));
}

static long getLongL(ByteBuffer bb, int bi) {
    return makeLong(bb.get(bi + 6),
                    bb.get(bi + 5),
                    bb.get(bi + 4),
                    bb.get(bi + 3),
                    bb.get(bi + 2),
                    bb.get(bi + 1),
                    bb.get(bi    ));
}

private static byte long6(long x) { return (byte)(x >> 48); }
private static byte long5(long x) { return (byte)(x >> 40); }
private static byte long4(long x) { return (byte)(x >> 32); }
private static byte long3(long x) { return (byte)(x >> 24); }
private static byte long2(long x) { return (byte)(x >> 16); }
private static byte long1(long x) { return (byte)(x >>  8); }
private static byte long0(long x) { return (byte)(x      ); }

static void putLongL(ByteBuffer bb, int bi, long x) {
    bb.put(bi + 6, long6(x));
    bb.put(bi + 5, long5(x));
    bb.put(bi + 4, long4(x));
    bb.put(bi + 3, long3(x));
    bb.put(bi + 2, long2(x));
    bb.put(bi + 1, long1(x));
    bb.put(bi    , long0(x));
}

In general, I avoid making micro-optimisation like this as it adds complexity for little gain IMHO. If you want to save space I suggest using something like stop bit encoding which uses one byte for every 7 bits. i.e. a long might only use 1 byte for small values but could be much larger as required.

@h22 you are welcome to shorten it. ;) Reading/writing a 7 byte value isn't so trivial given there is no hardware support for it. — Peter Lawrey, Mar 02 '16 at 20:17

score -2 · Answer 3 · answered Mar 02 '16 at 07:50

-2

Do as below

byte[] myLongBytes = ByteBuffer.allocate(Long.SIZE / Byte.SIZE).putLong(myLong).array();

answered Mar 02 '16 at 07:50

sanky jain

873
1
6
14

But I need to encode it into 7 bytes only, because I know the value is not that big... I simply don't need 8 bytes. – vojta Mar 02 '16 at 07:52

score -2 · Answer 4 · answered Mar 02 '16 at 07:51

-2

Java long is the 64 bit so 64/8 = 8 byte value so you are not allocating enough space to it. Allocate 8 bytes, not seven.

answered Mar 02 '16 at 07:51

Audrius Meškauskas

20,936
12
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But I need to encode it into 7 bytes only, because I know the value is not that big... I simply don't need 8 bytes. – vojta Mar 02 '16 at 07:52
Then convert long into byte array and only write the last 7 bytes of that array to the buffer (highest byte goes first in Java). See http://stackoverflow.com/questions/4485128/how-do-i-convert-long-to-byte-and-back-in-java on how to convert. – Audrius Meškauskas Mar 02 '16 at 07:55

ByteBuffer putLong() for long of 7-byte value

4 Answers4

Option A: Write 64 bit long and than remove one byte

Option B: Write the long "per partes"

Option C: Use byte[] instead of long