Are Integers in Java little endian or big endian?

Question

I ask because I am sending a byte stream from a C processto Java (through a priority middle ware). On the C side the 32 bit integer has the LSB is the first byte and MSB is the 4th byte.

So my question is on the Java side when we read the byte as it was sent from the C process, what is endian on the Java side?

A follow up question if the endian on the Java side is not the same as the one sent how can I convert betwen them?

thanks:)

Answer 1

There are no unsigned integers in Java (all are signed and in Big Endian).

On the C side the each byte has tne LSB at the start is on the left and the MSB at the end.

It sounds like you are using LSB as Least significant bit, are you? LSB usually stands for least significant byte. Endianness is not bit based but byte based.

To convert from unsigned byte to a Java integer:

int i = (int) b & 0xFF;

To convert from unsigned 32-bit little-endian in byte[] to Java long (from the top of my head, not tested):

long l = (long)b[0] & 0xFF;
l += ((long)b[1] & 0xFF;) << 8;
l += ((long)b[2] & 0xFF;) << 16;
l += ((long)b[3] & 0xFF;) << 24;

Answer 2

I would read the bytes one by one, and combine them into a long value. That way you control the endianness, and the communication process is transparent.

Answer 3

Use the network byte order (big endian), which is the same as Java uses anyway. See man htons for the different translators in C.

Answer 4

If it fits the protocoll you use, use a DataInputStream, where the behavior is very well defined.

Answer 5

There's no way this could influence anything in Java, since there's no (direct non-API) way to map some bytes directly into an int in Java.

Every API that does this or something similar defines the behaviour pretty precisely, so you should look up the documentation of that API.

Answer 6

It doesn't have to do with Java or C Little Endian and Big Endian are the property of the processor.