bitwise shift in uint

Question

uint number = 0x418  in bits : 0000010000011000
uint number1 = 0x8041 in bits: 1000000001000001
uint number2 = 0x1804 in bits: 0001100000000100

I cannot get 0x8041 with

   number >> 4;

or

   (number >> 4) & 0xffff;

How I can get 0x8041 and 0x1804 from 0x418 with shift?

SOLUTION

(number >> nbits) | (number << (16 - nbits))

See http://stackoverflow.com/questions/776508/best-practices-for-circular-shift-rotate-operations-in-c for best practices for compiler-friendly rotates that guard against problems when the count is 0 or greater than the type width. Should work equally well in C#. — Peter Cordes, Aug 17 '15 at 22:07

score 4 · Accepted Answer · answered Feb 25 '13 at 03:18

4

C# does not have a bitwise rotate operator - bits shifted past the right end just fall off and vanish. What you can do to solve this is

(number >> nbits) | (number << (32 - nbits))

which will right-rotate a 32-bit unsigned integer by nbits bits.

answered Feb 25 '13 at 03:18

us2012

what has this question to do with rotation? – Aniket Inge Feb 25 '13 at 03:19
@Aniket Well, seeing the OP's edit to the question, this answer was exactly what he was looking for. – us2012 Feb 25 '13 at 03:31

score 2 · Answer 2 · answered Feb 25 '13 at 03:31

What you are describing is typically known as Rotation, not Shifting. In assembly (x86), this is exposed via ROR and ROL instructions.

I'm not aware of a bitwise operator available in C# to do this, but the algorithm is simple enough:

value = value & 0x1 ? (1 << Marshal.SizeOf(value) * 8 - 1) | (value >> 1) : ( value >> 1);

2 Answers2