ByteBuffer api giving wrong integer values

Question

I am using ByteBuffer APIs to convert an object into bytes. The object's class is as follows

public class Obj{
int a; // size 1 byte
int b; // size 4 bytes
int c; // size 4 bytes
}

Using ByteBuffer API, I have allocated an object

ByteBuffer bbf = ByteBuffer.allocate(9);
bbf.put((byte) this.getA());
bbf.putInt(this.getB());
bbf.putInt(this.getC());
byte[] msg = bbf.array();

I set the value of B as 100 but when I convert the byte array from offset 1 till length 4, I get a different integer value. Any idea where is the problem? thanks!

The offsets are a:0, b:1, c:5. Do you need the value to be big endian or little endian? — Peter Lawrey
– Peter Lawrey, Commented Nov 1, 2011 at 12:27
i set the byte order to Little Endian. But the problem is still there — hoshang.varshney
– hoshang.varshney, Commented Nov 1, 2011 at 12:29

Community · Accepted Answer · 2020-06-20 09:12:55Z

3

The code works as it should, and if you indeed select bytes with index 1,2,3,4 they will yield the value 100:

ByteBuffer bbf = ByteBuffer.allocate(9);
bbf.put((byte) 10);
bbf.putInt(100);
bbf.putInt(55);
byte[] msg = bbf.array();

byte[] from4to8 = Arrays.copyOfRange(msg, 1, 5);

ByteBuffer buf2 = ByteBuffer.wrap(from4to8);
System.out.println(buf2.getInt());             // Prints 100

Some notes:

Keep in mind that the endian may differ from system to system (so check the endian on both hosts if this is part of a protocol)
The array you get from the bbf.array() call is a backing array, i.e.:

Modifications to this buffer's content will cause the returned array's content to be modified, and vice versa.

edited Jun 20, 2020 at 9:12

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answered Nov 1, 2011 at 12:31

aioobe

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9 Comments

hoshang.varshney Over a year ago

tried your code, with byte order as LittleEndian. I get the printed value as 1677721600

aioobe Over a year ago

That's the value you get if you take bytes 0,1,2,3, instead of 1,2,3,4. (I.e. if you change from 1,5 to 0,4 in the example in my answer).

hoshang.varshney Over a year ago

By the second note do you mean I have to convert each instance variable separately using different ByteBuffer object? If not please explain in detail

Maurice Perry Over a year ago

set also the byte order of buf2 to little endian

aioobe Over a year ago

It simply says that you shouldn't use msg as an array containing a copy of the content of the bbf, it is the content of the bbf. If you have some lines in between, changing bbf the content of msg will change at the same time.

|

chown · Accepted Answer · 2011-11-18 01:58:36Z

1

As far as I can see, B is at offset 1

edited Nov 18, 2011 at 1:58

chown

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answered Nov 1, 2011 at 12:26

Maurice Perry

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1 Comment

hoshang.varshney Over a year ago

Sorry for the typo. Correction made.

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