How to Convert Int to Unsigned Byte and Back

Question

I need to convert a number into an unsigned byte  The number is always less than or equal to 255  and so it will fit in one byte   I also need to convert that byte back into that number  How would I do that in Java  I ve tried several ways and none work  Here s what I m trying to do now   int size   5     Convert size int to binary String sizeStr   Integer toString size   byte binaryByte   Byte valueOf sizeStr     and now to convert that byte back into the number   Byte test   new Byte binaryByte   int msgSize   test intValue      Clearly  this does not work  For some reason  it always converts the number into 65  Any suggestions

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in java 7   public class Main       public static void main String   args            byte b    -2          int i   0           i     b  amp  0b1111 1111             System err println i             result   254

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A byte is always signed in Java  You may get its unsigned value by binary-anding it with 0xFF  though   int i   234  byte b    byte  i  System out println b      -22 int i2   b  amp  0xFF  System out println i2      234

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Java 8 provides Byte toUnsignedInt to convert byte to int by unsigned conversion   In Oracle s JDK this is simply implemented as return   int  x   amp  0xff  because HotSpot already understands how to optimize this pattern  but it could be intrinsified on other VMs   More importantly  no prior knowledge is needed to understand what a call to toUnsignedInt foo  does   In total  Java 8 provides methods to convert byte and short to unsigned int and long  and int to unsigned long   A method to convert byte to unsigned short was deliberately omitted because the JVM only provides arithmetic on int and long anyway   To convert an int back to a byte  just use a cast   byte someInt   The resulting narrowing primitive conversion will discard all but the last 8 bits

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If you want to use the primitive wrapper classes  this will work  but all java types are signed by default   public static void main String   args        Integer i 5      Byte b   Byte valueOf i        converts i to String and calls Byte valueOf       System out println b       System out println Integer valueOf b

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The solution works fine  thanks    but if you want to avoid casting and leave the low level work to the JDK  you can use a DataOutputStream to write your int s and a DataInputStream to read them back in  They are automatically treated as unsigned bytes then   For converting int s to binary bytes   ByteArrayOutputStream bos   new ByteArrayOutputStream    DataOutputStream dos   new DataOutputStream bos   int val   250  dos write byteVal       dos flush      Reading them back in      important to use a  non-Unicode   encoding like US ASCII or ISO-8859-1     i e   one that uses one byte per character ByteArrayInputStream bis   new ByteArrayInputStream     bos toString  ISO-8859-1   getBytes  ISO-8859-1     DataInputStream dis   new DataInputStream bis   int byteVal   dis readUnsignedByte      Esp  useful for handling binary data formats  e g  flat message formats  etc

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Even though it s too late  I d like to give my input on this as it might clarify why the solution given by JB Nizet works  I stumbled upon this little problem working on a byte parser and to string conversion myself  When you copy from a bigger size integral type to a smaller size integral type as this java doc says this happens  https   docs oracle com javase specs jls se7 html jls-5 html jls-5 1 3 A narrowing conversion of a signed integer to an integral type T simply discards all but the n lowest order bits  where n is the number of bits used to represent type T  In addition to a possible loss of information about the magnitude of the numeric value  this may cause the sign of the resulting value to differ from the sign of the input value  You can be sure that a byte is an integral type as this java doc says https   docs oracle com javase tutorial java nutsandbolts datatypes html byte  The byte data type is an 8-bit signed two s complement integer  So in the case of casting an integer 32 bits  to a byte 8 bits   you just copy the last  least significant 8 bits  of that integer to the given byte variable  int a   128  byte b    byte a     Last 8 bits gets copied System out println b       -128  Second part of the story involves how Java unary and binary operators promote operands  https   docs oracle com javase specs jls se7 html jls-5 html jls-5 6 2 Widening primitive conversion    5 1 2  is applied to convert either or both operands as specified by the following rules  If either operand is of type double  the other is converted to double  Otherwise  if either operand is of type float  the other is converted to float  Otherwise  if either operand is of type long  the other is converted to long  Otherwise  both operands are converted to type int  Rest assured  if you are working with integral type int and or lower it ll be promoted to an int     byte b 0x80  gets promoted to int  0xFF80  by the  amp  operator and then    0xFF80  amp  0xFF  0xFF translates to 0x00FF  bitwise operation yields     0x0080 a   b  amp  0xFF  System out println a      128  I scratched my head around this too     There is a good answer for this here by rgettman  Bitwise operators in java only for integer and long

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The Integer toString size  call converts into the char representation of your integer  i e  the char  5   The ASCII representation of that character is the value 65   You need to parse the string back to an integer value first  e g  by using Integer parseInt  to get back the original int value   As a bottom line  for a signed unsigned conversion  it is best to leave String out of the picture and use bit manipulation as  JB suggests

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Handling bytes and unsigned integers with BigInteger   byte   b                             your integer in big-endian BigInteger ui   new BigInteger b     let BigInteger do the work int i   ui intValue                  unsigned value assigned to i

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Except char  every other numerical data type in Java are signed   As said in a previous answer  you can get the unsigned value by performing an and operation with 0xFF  In this answer  I m going to explain how it happens    int i   234  byte b    byte  i  System out println b       -22  int i2   b  amp  0xFF           This is like casting b to int and perform and operation with 0xFF  System out println i2      234   If your machine is 32-bit  then the int data type needs 32-bits to store values  byte needs only 8-bits   The int variable i is represented in the memory as follows  as a 32-bit integer    0 24 11101010   Then the byte variable b is represented as   11101010   As bytes are unsigned  this value represent -22   Search for 2 s complement to learn more on how to represent negative integers in memory   Then if you cast is to int it will still be -22 because casting preserves the sign of a number   1 24 11101010   The the casted 32-bit value of b perform and operation with 0xFF    1 24 11101010  amp  0 24 11111111  0 24 11101010   Then you get 234 as the answer

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If you just need to convert an expected 8-bit value from a signed int to an unsigned value  you can use simple bit shifting   int signed   -119      11111111 11111111 11111111 10001001         Use unsigned right shift operator to drop unset bits in positions 8-31     int psuedoUnsigned    signed  lt  lt  24   gt  gt  gt  24      00000000 00000000 00000000 10001001 - gt  137 base 10          Convert back to signed by using the sign-extension properties of the right shift operator     int backToSigned    psuedoUnsigned  lt  lt  24   gt  gt  24     back to original bit pattern   http   docs oracle com javase tutorial java nutsandbolts op3 html  If using something other than int as the base type  you ll obviously need to adjust the shift amount  http   docs oracle com javase tutorial java nutsandbolts datatypes html  Also  bear in mind that you can t use byte type  doing so will result in a signed value as mentioned by other answerers   The smallest primitive type you could use to represent an 8-bit unsigned value would be a short

[java] How to Convert Int to Unsigned Byte and Back

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