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); 

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@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.