How does bitshifting work in Java

Question

I have this statement      Assume the bit value of byte x is 00101011  what is the result of x gt  gt 2    How can I program it and can someone explain me what is doing

User · Answer

You can use e g  this API if you would like to see bitString presentation of your numbers  Uncommons Math  Example  in jruby   bitString   org uncommons maths binary BitString new java math BigInteger new  12   toString 2   bitString setBit 1  true  bitString toNumber   gt  14   edit  Changed api link and add a little example

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These examples cover the three types of shifts applied to both a positive and a negative number      Signed left shift on 626348975 00100101010101010101001110101111 is   626348975 01001010101010101010011101011110 is  1252697950 after  lt  lt  1 10010101010101010100111010111100 is -1789571396 after  lt  lt  2 00101010101010101001110101111000 is   715824504 after  lt  lt  3     Signed left shift on -552270512 11011111000101010000010101010000 is  -552270512 10111110001010100000101010100000 is -1104541024 after  lt  lt  1 01111100010101000001010101000000 is  2085885248 after  lt  lt  2 11111000101010000010101010000000 is  -123196800 after  lt  lt  3      Signed right shift on 626348975 00100101010101010101001110101111 is   626348975 00010010101010101010100111010111 is   313174487 after  gt  gt  1 00001001010101010101010011101011 is   156587243 after  gt  gt  2 00000100101010101010101001110101 is    78293621 after  gt  gt  3     Signed right shift on -552270512 11011111000101010000010101010000 is  -552270512 11101111100010101000001010101000 is  -276135256 after  gt  gt  1 11110111110001010100000101010100 is  -138067628 after  gt  gt  2 11111011111000101010000010101010 is   -69033814 after  gt  gt  3      Unsigned right shift on 626348975 00100101010101010101001110101111 is   626348975 00010010101010101010100111010111 is   313174487 after  gt  gt  gt  1 00001001010101010101010011101011 is   156587243 after  gt  gt  gt  2 00000100101010101010101001110101 is    78293621 after  gt  gt  gt  3     Unsigned right shift on -552270512 11011111000101010000010101010000 is  -552270512 01101111100010101000001010101000 is  1871348392 after  gt  gt  gt  1 00110111110001010100000101010100 is   935674196 after  gt  gt  gt  2 00011011111000101010000010101010 is   467837098 after  gt  gt  gt  3

User · Answer

Firstly  you can not shift a byte in java  you can only shift an int or a long   So the byte will undergo promotion first  e g   00101011 -  00000000000000000000000000101011  or  11010100 -  11111111111111111111111111010100  Now  x  gt  gt  N means  if you view it as a string of binary digits     The rightmost N bits are discarded The leftmost bit is replicated as many times as necessary to pad the result to the original size  32 or 64 bits   e g    00000000000000000000000000101011  gt  gt  2 -  00000000000000000000000000001010  11111111111111111111111111010100  gt  gt  2 -  11111111111111111111111111110101

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The binary 32 bits for 00101011 is   00000000 00000000 00000000 00101011  and the result is     00000000 00000000 00000000 00101011    gt  gt  2 times                                           00000000 00000000 00000000 00001010   Shifts the bits of 43 to right by distance 2  fills with highest sign  bit on the left side   Result is 00001010 with decimal value 10   00001010     8 2   10

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You can t write binary literals like 00101011 in Java so you can write it in hexadecimal instead   byte x   0x2b    To calculate the result of x  gt  gt  2 you can then just write exactly that and print the result   System out println x  gt  gt  2

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public class Shift    public static void main String   args      Byte b   Byte parseByte  00101011  2     System out println b     byte val   b byteValue      Byte shifted   new Byte  byte   val  gt  gt  2      System out println shifted         often overloked  are the methods of Integer    int i   Integer parseInt  00101011  2     System out println  Integer toBinaryString i      i  gt  gt   2    System out println  Integer toBinaryString i           Output   43 10 101011 1010

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gt  gt  is the Arithmetic Right Shift operator  All of the bits in the first operand are shifted the number of places indicated by the second operand  The leftmost bits in the result are set to the same value as the leftmost bit in the original number   This is so that negative numbers remain negative    Here s your specific case   00101011   001010  lt -- Shifted twice to the right  rightmost bits dropped  00001010  lt -- Leftmost bits filled with 0s  to match leftmost bit in original number

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00101011   43 in decimal  class test           public static void main String   args          int a  43                String b  Integer toBinaryString a  gt  gt  2                  System out println b                Output   101011 becomes 1010

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byte x   51    00101011 byte y    byte   x  gt  gt  2     00001010 aka Base 10  10

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When you shift right 2 bits you drop the 2 least significant bits   So   x   00101011  x  gt  gt  2     now  notice the 2 new 0 s on the left of the byte  x   00001010   This is essentially the same thing as dividing an int by 2  2 times   In Java  byte b    byte  16  b   b  gt  gt  2     prints 4 System out println b

[java] How does bitshifting work in Java?

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