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 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 between them

User · Accepted Answer

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

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

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  amp  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   amp  0xFF  l      long b 1   amp  0xFF   lt  lt  8  l      long b 2   amp  0xFF   lt  lt  16  l      long b 3   amp  0xFF   lt  lt  24

User · Answer

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

User · Answer

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  lt  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   lt  lt  8     answer    sb 2   lt  lt  16     answer    sb 3   lt  lt  24     return answer

User · Answer

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  amp  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   amp  0xFF  l      long b 1   amp  0xFF   lt  lt  8  l      long b 2   amp  0xFF   lt  lt  16  l      long b 3   amp  0xFF   lt  lt  24

User · Answer

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

User · Answer

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  lt  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   lt  lt  8     answer    sb 2   lt  lt  16     answer    sb 3   lt  lt  24     return answer

User · Answer

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

User · Answer

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

User · Answer

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  amp  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   amp  0xFF  l      long b 1   amp  0xFF   lt  lt  8  l      long b 2   amp  0xFF   lt  lt  16  l      long b 3   amp  0xFF   lt  lt  24

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

java force indeed big endian   https   docs oracle com javase specs jvms se8 html jvms-2 html jvms-2 11

User · Answer

java force indeed big endian   https   docs oracle com javase specs jvms se8 html jvms-2 html jvms-2 11

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

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  amp  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   amp  0xFF  l      long b 1   amp  0xFF   lt  lt  8  l      long b 2   amp  0xFF   lt  lt  16  l      long b 3   amp  0xFF   lt  lt  24

[java] Does Java read integers in little endian or big endian?

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