Does Java read integers in little endian or big endian?

Question

I ask because I am sending a byte stream from a C process to Java. On the C side the 32 bit integer has the LSB is the first byte and MSB is the 4th byte.

So my ques

Answer 1

There are no unsigned integers in Java. All integers 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

Java is 'Big-endian' as noted above. That means that the MSB of an int is on the left if you examine memory (on an Intel CPU at least). The sign bit is also in the MSB for all Java integer types.
Reading a 4 byte unsigned integer from a binary file stored by a 'Little-endian' system takes a bit of adaptation in Java. DataInputStream's readInt() expects Big-endian format.
Here's an example that reads a four byte unsigned value (as displayed by HexEdit as 01 00 00 00) into an integer with a value of 1:

 // Declare an array of 4 shorts to hold the four unsigned bytes
 short[] tempShort = new short[4];
 for (int b = 0; b < 4; b++) {
    tempShort[b] = (short)dIStream.readUnsignedByte();           
 }
 int curVal = convToInt(tempShort);

 // Pass an array of four shorts which convert from LSB first 
 public int convToInt(short[] sb)
 {
   int answer = sb[0];
   answer += sb[1] << 8;
   answer += sb[2] << 16;
   answer += sb[3] << 24;
   return answer;        
 }

Answer 3

I stumbled here via Google and got my answer that Java is big endian.

Reading through the responses I'd like to point out that bytes do indeed have an endian order, although mercifully, if you've only dealt with “mainstream” microprocessors you are unlikely to have ever encountered it as Intel, Motorola, and Zilog all agreed on the shift direction of their UART chips and that MSB of a byte would be 2**7 and LSB would be 2**0 in their CPUs (I used the FORTRAN power notation to emphasize how old this stuff is :) ).

I ran into this issue with some Space Shuttle bit serial downlink data 20+ years ago when we replaced a $10K interface hardware with a Mac computer. There is a NASA Tech brief published about it long ago. I simply used a 256 element look up table with the bits reversed (table[0x01]=0x80 etc.) after each byte was shifted in from the bit stream.

Answer 4

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 5

java force indeed big endian : https://docs.oracle.com/javase/specs/jvms/se8/html/jvms-2.html#jvms-2.11

Answer 6

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