2 byte array in Little Endian to int without java.nio.*

Question

I want to convert 2byte array in Little Endian to Int without using java.nio.*. How can I accomplish this?

With regards

Also you may take a look at http://stackoverflow.com/questions/7619058/convert-a-byte-array-to-integer-in-java-and-vise-versa — Jk1, Jan 28 '14 at 03:05
@George: Problem is in converting 2 byte array to Int, it gives negative value — nurgasemetey, Jan 28 '14 at 03:07
@George: I read just now that I must use 0xff to give only unsigned values. Is that true? — nurgasemetey, Jan 28 '14 at 03:10
Doing `byte & 0xff` is a trick to get the unsigned value of a byte, yes. — George, Jan 28 '14 at 03:11
`&` and `<<` is pretty much all you need. (Well, maybe the occasional `+` or `|`.) — Hot Licks, Jan 28 '14 at 03:30

score 1 · Accepted Answer · answered Jan 28 '14 at 03:16

1

This should do the trick int val = (anArray[1] & 0xff) << 8 + (anArray[0] & 0xff);

answered Jan 28 '14 at 03:16

George

score 1 · Answer 2 · answered Feb 19 '18 at 18:00

1

Just came across this post and realised that the accepted answer will not work correctly because + has a higher precedence than <<.

Therefore it should be int val = ((anArray[1] & 0xff) << 8) + (anArray[0] & 0xff); instead.

answered Feb 19 '18 at 18:00

benjay

score 0 · Answer 3 · answered Jan 28 '14 at 03:14

0

you have 2 Byte means 16 bit because in Little indian The least significant 16-bit unit stores the value you can use bitvise operations in java

answered Jan 28 '14 at 03:14

Rohit Sachan

3 Answers3