bit shift - Java split the bits of a long in two parts using bit shifting -



bit shift - Java split the bits of a long in two parts using bit shifting -

i need split bits of number 9(1001) in 2 equal sized parts 10 , 01 in case.

my first thought shift right number dont expected result ,i suspect due sign(no unsigned in java :( ).

my current code following:

long num=9; system.out.println(long.tobinarystring(num)); long num1=num>>2; system.out.println(long.tobinarystring(num1)); long num2=num<<2; system.out.println(long.tobinarystring(num2));

output:

1001 10 100100

any workaround?

to lower part, need utilize bitwise and... if shift right 2 bits higher part, need , binary number 11 (two bits 1) lower part. here's code should shift:

long num = 9; int shift = 2 system.out.println(long.tobinarystring(num)); long num1 = num >> shift; system.out.println(long.tobinarystring(num1)); long num2 = num & ( (1<<shift) - 1); system.out.println(long.tobinarystring(num2));

explanation of calculating num2, shift 2, 0b means binary literal, in pseudocode:

( (1<<shift) - 1) == ( (1<<2) - 1) == ( 0b100 - 1) == 0b11 == 2 bits set

from should clear how work shift value.

java bit-shift

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