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.
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);