Byte Array and Int conversion in Java

Question

I am having some difficulty with these two functions  byteArrayToInt and intToByteArray   The problem is that if I use one to get to another and that result to get to the former  the results are different  as you can see from my examples below   I cannot find the bug in the code  Any ideas are very welcome  Thanks   public static void main String   args        int a   123      byte   aBytes   intToByteArray a       int a2   byteArrayToInt aBytes        System out println a              prints  123      System out println aBytes         prints   B 459189e1      System out println a2             prints  2063597568             System out println intToByteArray a2        prints   B 459189e1     public static int byteArrayToInt byte   b         int value   0      for  int i   0  i  lt  4  i              int shift    4 - 1 - i    8          value     b i   amp  0x000000FF   lt  lt  shift            return value     public static byte   intToByteArray int a        byte   ret   new byte 4       ret 0     byte   a  amp  0xFF          ret 1     byte    a  gt  gt  8   amp  0xFF          ret 2     byte    a  gt  gt  16   amp  0xFF          ret 3     byte    a  gt  gt  24   amp  0xFF       return ret

User · Answer

Instead of allocating space  et al  an approach using ByteBuffer from java nio      byte   arr     0x01  0x00  0x00  0x00  0x48  0x01       say we want to consider indices 1  2  3  4  0x00  0x00  0x00  0x48   ByteBuffer bf   ByteBuffer wrap arr  1  4      big endian by default int num   bf getInt          72   Now  to go the other way    ByteBuffer newBuf   ByteBuffer allocate 4   newBuf putInt num   byte   bytes   newBuf array         0  0  0  72   0x48   72

User · Answer

I took a long look at many questions like this  and found this post    I didn t like the fact that the conversion code is duplicated for each type  so I ve made a generic method to perform the task   public static byte   toByteArray long value  int n        byte   ret   new byte n       ret n-1     byte    value  gt  gt   0 8   amp  0xFF          ret n-2     byte    value  gt  gt   1 8   amp  0xFF                  ret 1     byte    value  gt  gt    n-2  8   amp  0xFF          ret 0     byte    value  gt  gt    n-1  8   amp  0xFF          return ret      See full post

User · Answer

You re swapping endianness between your two methods   You have intToByteArray int a  assigning the low-order bits into ret 0   but then byteArrayToInt byte   b  assigns b 0  to the high-order bits of the result   You need to invert one or the other  like   public static byte   intToByteArray int a        byte   ret   new byte 4       ret 3     byte   a  amp  0xFF          ret 2     byte    a  gt  gt  8   amp  0xFF          ret 1     byte    a  gt  gt  16   amp  0xFF          ret 0     byte    a  gt  gt  24   amp  0xFF       return ret

User · Answer

You can also use BigInteger for variable length bytes  You can convert it to Long  Integer or Short  whichever suits your needs   new BigInteger bytes  intValue      or to denote polarity   new BigInteger 1  bytes  intValue      To get bytes back just   new BigInteger bytes  toByteArray

User · Answer

here is my implementation   public static byte   intToByteArray int a        return BigInteger valueOf a  toByteArray       public static int byteArrayToInt byte   b        return new BigInteger b  intValue

User · Answer

I found a simple way in com google common primitives which is in the  Maven com google guava guava 12 0 1   long newLong   Longs fromByteArray oldLongByteArray   int newInt   Ints fromByteArray oldIntByteArray     Have a nice try

User · Answer

Your methods should be  something like    public static int byteArrayToInt byte   b         return   b 3   amp  0xFF                b 2   amp  0xFF   lt  lt  8                b 1   amp  0xFF   lt  lt  16                b 0   amp  0xFF   lt  lt  24     public static byte   intToByteArray int a        return new byte              byte    a  gt  gt  24   amp  0xFF            byte    a  gt  gt  16   amp  0xFF               byte    a  gt  gt  8   amp  0xFF               byte   a  amp  0xFF             These methods were tested with the following code    Random rand   new Random System currentTimeMillis     byte   b  int a  v  for  int i 0  i lt 10000000  i          a   rand nextInt        b   intToByteArray a       v   byteArrayToInt b       if  a    v            System out println  ERR      a            Arrays toString b             v           System out println  Done

User · Answer

I like owlstead s original answer  and if you don t like the idea of creating a ByteBuffer on every method call then you can reuse the ByteBuffer by calling it s  clear   and  flip   methods   ByteBuffer  intShifter   ByteBuffer allocate Integer SIZE   Byte SIZE                                      order ByteOrder LITTLE ENDIAN    public byte   intToByte int value         intShifter clear         intShifter putInt value             return  intShifter array       public int byteToInt byte   data         intShifter clear         intShifter put data  0  Integer SIZE   Byte SIZE        intShifter flip        return  intShifter getInt

User · Answer

That s a lot of work for   public static int byteArrayToLeInt byte   b        final ByteBuffer bb   ByteBuffer wrap b       bb order ByteOrder LITTLE ENDIAN       return bb getInt       public static byte   leIntToByteArray int i        final ByteBuffer bb   ByteBuffer allocate Integer SIZE   Byte SIZE       bb order ByteOrder LITTLE ENDIAN       bb putInt i       return bb array        This method uses the Java ByteBuffer and ByteOrder functionality in the java nio package  This code should be preferred where readability is required  It should also be very easy to remember   I ve shown Little Endian byte order here  To create a Big Endian version you can simply leave out the call to order ByteOrder      In code where performance is higher priority than readability  about 10x as fast    public static int byteArrayToLeInt byte   encodedValue        int value    encodedValue 3   lt  lt   Byte SIZE   3        value     encodedValue 2   amp  0xFF   lt  lt   Byte SIZE   2       value     encodedValue 1   amp  0xFF   lt  lt   Byte SIZE   1       value     encodedValue 0   amp  0xFF       return value     public static byte   leIntToByteArray int value        byte   encodedValue   new byte Integer SIZE   Byte SIZE       encodedValue 3     byte   value  gt  gt  Byte SIZE   3       encodedValue 2     byte   value  gt  gt  Byte SIZE   2          encodedValue 1     byte   value  gt  gt  Byte SIZE          encodedValue 0     byte  value      return encodedValue      Just reverse the byte array index to count from zero to three to create a Big Endian version of this code     Notes    In Java 8 you can also make use of the Integer BYTES constant  which is more succinct than Integer SIZE   Byte SIZE

User · Answer

sorry this is the correct          public byte   IntArrayToByteArray int   iarray   int sizeofintarray                final ByteBuffer bb          bb   ByteBuffer allocate  sizeofintarray   4          for int k   0  k  lt  sizeofintarray   k           bb putInt k   4  iar k           return bb array

[java] Byte Array and Int conversion in Java

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