How do shift operators work in Java

Question

I am trying to understand the shift operators and couldn t get much  When I tried to execute the below code  System out println Integer toBinaryString 2  lt  lt  11    System out println Integer toBinaryString 2  lt  lt  22    System out println Integer toBinaryString 2  lt  lt  33    System out println Integer toBinaryString 2  lt  lt  44    System out println Integer toBinaryString 2  lt  lt  55      I get the below  1000000000000     100000000000000000000000     100     10000000000000     1000000000000000000000000       Could somebody please explain

User · Answer

The typical usage of shifting a variable and assigning back to the variable can be rewritten with shorthand operators <<=, >>=, or >>>=, also known in the spec as Compound Assignment Operators.

For example,

i >>= 2

produces the same result as

i = i >> 2

User · Answer

Signed left shift Logically  Simple if 1 lt  lt 11 it will tends to 2048 and 2 lt  lt 11 will  give 4096   In java programming      int a   2  lt  lt  11      it will result in 4096  2 lt  lt 11   2  2 11    4096

User · Answer

The shift can be implement with data types  char  int and long int   The float and double data connot be shifted    value  value  gt  gt  steps     Right shift  signed data  value  value  lt  lt  steps     Left shift  signed data

User · Answer

System out println Integer toBinaryString 2  lt  lt  11       Shifts binary 2 10  by 11 times to the left  Hence  1000000000000  System out println Integer toBinaryString 2  lt  lt  22       Shifts binary 2 10  by 22 times to the left  Hence   100000000000000000000000  System out println Integer toBinaryString 2  lt  lt  33       Now  int is of 4 bytes hence 32 bits  So when you do shift by 33  it s equivalent to shift by 1  Hence   100

User · Answer

I believe this might Help       System out println Integer toBinaryString 2  lt  lt  0        System out println Integer toBinaryString 2  lt  lt  1        System out println Integer toBinaryString 2  lt  lt  2        System out println Integer toBinaryString 2  lt  lt  3        System out println Integer toBinaryString 2  lt  lt  4        System out println Integer toBinaryString 2  lt  lt  5      Result      10     100     1000     10000     100000     1000000   Edited   Must Read This  how-do-the-bitwise-shift-operators-work

User · Answer

It will shift the bits by padding that many 0 s   For ex     binary 10  which is digit 2 left shift by 2 is 1000 which is digit 8 binary 10  which is digit 2 left shift by 3 is 10000 which is digit 16

User · Answer

2 from decimal numbering system in binary is as follows  10   now if you do   2  lt  lt  11   it would be   11 zeros would be padded on the right side  1000000000000      The signed left shift operator   lt  lt   shifts a bit pattern to the left  and the signed right shift operator      shifts a bit pattern to the right  The bit pattern is given by the left-hand operand  and the number of positions to shift by the right-hand operand  The unsigned right shift operator       shifts a zero into the leftmost position  while the leftmost position after      depends on sign extension        left shifting results in multiplication by 2   2  in terms or arithmetic    For example   2 in binary 10  if you do  lt  lt 1 that would be 100 which is 4    4 in binary 100  if you do   lt  lt 1 that would be 1000 which is 8    Also See   absolute-beginners-guide-to-bit-shifting

User · Answer

I think it would be the following  for example    Signed left shift     2  lt  lt  1   is     10  binary of 2  add 1 zero at the end of the binary string  Hence 10 will be 100 which becomes 4   Signed left shift uses multiplication    So this could also be calculated as 2    2 1    4  Another example  2  lt  lt  11    2   2 11    4096    Signed right shift     4    1   is     100  binary of 4  remove 1 zero at the end of the binary string  Hence 100 will be 10 which becomes 2   Signed right shift uses division    So this could also be calculated as 4    2 1    2 Another example  4096    11    4096    2 11    2

User · Answer

Right and Left shift work on same way here is How Right Shift works  The Right Shift  The right shift operator      shifts all of the bits in a value to the right a specified number of times  Its general form   value  gt  gt  num   Here  num specifies the number of positions to right-shift the value in value  That is  the    moves all of the bits in the specified value to the right the number of bit positions specified by num  The following code fragment shifts the value 32 to the right by two positions  resulting in a being set to 8   int a   32  a   a  gt  gt  2     a now contains 8   When a value has bits that are    shifted off     those bits are lost  For example  the next code fragment shifts the value 35 to the right two positions  which causes the two low-order bits to be lost  resulting again in a being set to 8   int a   35  a   a  gt  gt  2     a still contains 8   Looking at the same operation in binary shows more clearly how this happens   00100011 35  gt  gt  2 00001000 8   Each time you shift a value to the right  it divides that value by two   and discards any remainder  You can take advantage of this for high-performance integer division by 2  Of course  you must be sure that you are not shifting any bits off the right end  When you are shifting right  the top  leftmost  bits exposed by the right shift are filled in with the previous contents of the top bit  This is called sign extension and serves to preserve the sign of negative numbers when you shift them right  For example     8  gt  gt  1 is    4  which  in binary  is  11111000    8  gt  gt 1 11111100    4   It is interesting to note that if you shift    1 right  the result always remains    1  since sign extension keeps bringing in more ones in the high-order bits  Sometimes it is not desirable to sign-extend values when you are shifting them to the right  For example  the following program converts a byte value to its hexadecimal string representation  Notice that the shifted value is masked by ANDing it with 0x0f to discard any sign-extended bits so that the value can be used as an index into the array of hexadecimal characters      Masking sign extension  class HexByte     static public void main String args          char hex              0    1    2    3    4    5    6    7          8    9    a    b    c    d    e    f           byte b    byte  0xf1   System out println  b   0x    hex  b  gt  gt  4   amp  0x0f    hex b  amp  0x0f          Here is the output of this program   b   0xf1

[java] How do shift operators work in Java?

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