Java converting int to hex and back again

Question

I have the following code     int Val -32768  String Hex Integer toHexString Val     This equates to ffff8000  int FirstAttempt Integer parseInt Hex 16      Error  Invalid Int  int SecondAttempt Integer decode  0x  Hex       Error  Invalid Int    So  initially  it converts the value -32768 into a hex string ffff8000  but then it can t convert the hex string back into an Integer   In  Net it works as I d expect  and returns -32768   I know that I could write my own little method to convert this myself  but I m just wondering if I m missing something  or if this is genuinely a bug

User · Answer

Try using BigInteger class  it works   int Val -32768  String Hex Integer toHexString Val      int FirstAttempt Integer parseInt Hex 16      Error  Invalid Int    int SecondAttempt Integer decode  0x  Hex       Error  Invalid Int  BigInteger i   new BigInteger Hex 16   System out println i intValue

User · Answer

It s worth mentioning that Java 8 has the methods Integer parseUnsignedInt and Long parseUnsignedLong that does what you wanted  specifically   Integer parseUnsignedInt  ffff8000  16     -32768  The name is a bit confusing  as it parses a signed integer from a hex string  but it does the work

User · Answer

Java s parseInt method is actally a bunch of code eating  false  hex   if you want to translate -32768  you should convert the absolute value into hex  then prepend the string with  -    There is a sample of Integer java file    public static int parseInt String s  int radix    The description is quite explicit      Parses the string argument as a signed integer in the radix    specified by the second argument  The characters in the string            parseInt  0   10  returns 0   parseInt  473   10  returns 473   parseInt  -0   10  returns 0   parseInt  -FF   16  returns -255

User · Answer

Using Integer toHexString      is a good answer  But personally prefer to use String format        Try this sample as a test   byte   values   new byte 64   Arrays fill values   byte 8      Fills array with 8 just for test String valuesStr       for int i   0  i  lt  values length  i        valuesStr    String format  0x 02x   values i   amp  0xff         valuesStr trim

User · Answer

Hehe  curious  I think this is an  intentianal bug   so to speak    The underlying reason is how the Integer class is written  Basically  parseInt is  optimized  for positive numbers  When it parses the string  it builds the result cumulatively  but negated  Then it flips the sign of the end-result   Example   66   0x42  parsed like   4  -1    -4 -4   16   -64  hex 4 parsed   -64 - 2   -66  hex 2 parsed   return -66    -1    66   Now  let s look at your example FFFF8000  16  -1    -16  first F parsed  -16 16   -256   -256 - 16   -272  second F parsed  -272   16   -4352   -4352 - 16   -4368  third F parsed  -4352   16   -69888  -69888 - 16   -69904  forth F parsed  -69904   16   -1118464   -1118464 - 8   -1118472  8 parsed  -1118464   16   -17895552   -17895552 - 0   -17895552  first 0 parsed  Here it blows up since -17895552  lt  -Integer MAX VALUE   16  -134217728    Attempting to execute the next logical step in the chain  -17895552   16  would cause an integer overflow error    Edit  addition   in order for the parseInt   to work  consistently  for -Integer MAX VALUE  lt   n  lt   Integer MAX VALUE  they would have had to implement logic to  rotate  when reaching -Integer MAX VALUE in the cumulative result  starting over at the max-end of the integer range and continuing downwards from there  Why they did not do this  one would have to ask Josh Bloch or whoever implemented it in the first place  It might just be an optimization   However    Hex Integer toHexString Integer MAX VALUE   System out println Hex   System out println Integer parseInt Hex toUpperCase    16      works just fine  for just this reason  In the sourcee for Integer you can find this comment      Accumulating negatively avoids surprises near MAX VALUE

User · Answer

It overflows  because the number is negative   Try this and it will work   int n    int  Long parseLong  ffff8000   16

User · Answer

Below code would work   int a -32768  String a1 Integer toHexString a   int parsedResult  int Long parseLong a1 16   System out println  Parsed Value is    parsedResult

User · Answer

int val   -32768  String hex   Integer toHexString val    int parsedResult    int  Long parseLong hex  16   System out println parsedResult     That s how you can do it    The reason why it doesn t work your way  Integer parseInt takes a signed int  while toHexString produces an unsigned result  So if you insert something higher than 0x7FFFFFF  an error will be thrown automatically  If you parse it as long instead  it will still be signed  But when you cast it back to int  it will overflow to the correct value

User · Answer

int to Hex      Integer toHexString intValue    Hex to int    Integer valueOf hexString  16  intValue       You may also want to use long instead of int  if the value does not fit the int bounds     Hex to long   Long valueOf hexString  16  longValue    long to Hex  Long toHexString longValue

User · Answer

As Integer toHexString byte integer  is not working when you are trying to convert signed bytes like UTF-16 decoded characters you have to use   Integer toString byte integer  16     or  String format   02X   byte integer     reverse you can use  Integer parseInt hexString  16

[java] Java converting int to hex and back again

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