Is there a way to perform a circular bit shift in C#?

Question

I know that the following is true

int i = 17; //binary 10001
int j = i << 1; //decimal 34, binary 100010

But, if you shift too far, the bits fall off the end. Where this happens is a matter of the size of integer you are working with.

Is there a way to perform a shift so that the bits rotate around to the other side? I'm looking for a single operation, not a for loop.

Answer 1

If you know the size of type, you could do something like:

uint i = 17;
uint j = i << 1 | i >> 31;

... which would perform a circular shift of a 32 bit value.

As a generalization to circular shift left n bits, on a b bit variable:

/*some unsigned numeric type*/ input = 17;
var result = input  << n | input  >> (b - n);

---

@The comment, it appears that C# does treat the high bit of signed values differently. I found some info on this here. I also changed the example to use a uint.

Answer 2

One year ago I've to implement MD4 for my undergraduate thesis. Here it is my implementation of circular bit shift using a UInt32.

private UInt32 RotateLeft(UInt32 x, Byte n)
{
      return UInt32((x << n) | (x >> (32 - n)));
}

Answer 3

Sincce .NET Core 3.0 and up there's BitOperations.RotateLeft() and BitOperations.RotateRight() so you can just use something like

BitOperations.RotateRight(12, 3);
BitOperations.RotateLeft(34L, 5);

In previous versions you can use BitRotator.RotateLeft() and BitRotator.RotateRight() in Microsoft.VisualStudio.Utilities