Efficiently shift an integer to the left and wrap around

Asked Dec 10 '18 at 13:39

Active Dec 10 '18 at 13:39

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How can I efficiently shift every bit in an integral type to the left, and wrap around so that the leftMost bit becomes the rightmost?

eg if the type was a byte, then:

1111,1110 becomes 1111,1101
0000,0001 becomes 0000,0010
1000,0000 becomes 0000,0001

At the moment I'm using this:

//num is an ulong
bool leftMostSet = (num & 0x8000000000000000) == 0x8000000000000000;
num = num << 1;
num = num | (leftMostSet ? (ulong)1 : 0);
return num;

I'm wondering if there is a more efficient method though

asked Dec 10 '18 at 13:39

Yair Halberstadt

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https://stackoverflow.com/questions/812022/c-sharp-bitwise-rotate-left-and-rotate-right – Dmitry Bychenko Dec 10 '18 at 13:45
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The dupe is for C, not C# but it looks like it'll work anyway. If not, come back and [edit] this so it can be reopened. – bommelding Dec 10 '18 at 13:46
Thanks, those answers work, although I don't know if the compiler optimizations will – Yair Halberstadt Dec 10 '18 at 13:46
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The jitter knows this pattern and will use rotate machine code. – Jon Hanna Dec 10 '18 at 13:47
Thanks @JonHanna. Do you have a source for that? – Yair Halberstadt Dec 10 '18 at 13:52
@YairHalberstadt I'm afraid not. I do recall it being noted as such by people in the know among the corefx developers. – Jon Hanna Dec 10 '18 at 14:24

Efficiently shift an integer to the left and wrap around

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