C# bitwise rotate left and rotate right

Question

Hi

I am looking for C# equivalent (.Net 2) of _rotl and _rotr from c++.

Any ideas ?

Thanks,

Answer 1

There's no built-in language feature for bit rotation in C#, but these extension methods should do the job:

public static uint RotateLeft(this uint value, int count)
{
    return (value << count) | (value >> (32 - count))
}

public static uint RotateLeft(this uint value, int count)
{
    return (value >> count) | (value << (32 - count))
}

Note: As Mehrdad points out, right-shift (>>) for signed integers is a peculiarity: it fills the MSBs with sign bit rather than 0 as it does for unsigned numbers. I've now changed the methods to take and return uint (unsigned 32-bit integer) instead - this is also in greater accordance with the C++ rotl and rotr functions. If you want to rotate integers, just case them before passing, and again cast the return value, of course.

Example usage:

int foo1 = 8.RotateRight(3); // foo1 = 1
int foo2 = int.MinValue.RotateLeft(3); // foo2 = 4

(Note that int.MinValue is 111111111111111111111111 - 32 1s in binary.)

Answer 2

Is this what you are trying to do?

Jon Skeet answered this in another site

Basically what you want is

(for left)

(original << bits) | (original >> (32 -bits))

or

(for right)

(original >> bits) | (original << (32 -bits))

Also, as Mehrdad has already suggested, this only works for uint, which is the example that Jon gives as well.

Answer 3

The naive version of shifting won't work. The reason is, right shifting signed numbers will fill the left bits with sign bit, not 0:

You can verify this fact with:

Console.WriteLine(-1 >> 1);

The correct way is:

public static int RotateLeft(this int value, int count)
{
    uint val = (uint)value;
    return (int)((val << count) | (val >> (32 - count)));
}

public static int RotateRight(this int value, int count)
{
    uint val = (uint)value;
    return (int)((value >> count) | (value << (32 - count)));
}