Simplify perfect-square testing loop #2
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There's no need for the `bits` variable here. The number of loop iterations can be cut in half by repeatedly shifting right by 2 so long as the last 2 bits are zeroes. If there were in fact an odd number of trailing zero bits in `q0`, that will leave `tmp` with a last bit of 0, and so it won't pass the later `(tmp & 7) == 1` test. If there were an even number of trailing zero bits, `tmp` is left the same as before the change, and again the only real thing left to test is whether `(tmp & 7) == 1`. Caution: I didn't test or even compile this code! I was just browsing this on the web, and am using the web UI to edit. I've used the same trick in other projects, and am sure it's sound (give it a little thought - it's obvious ;-) ). Apologies in advance for any typos.
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There's no need for the
bitsvariable here. The number of loop iterations can be cut in half by repeatedly shifting right by 2 so long as the last 2 bits are zeroes. If there were in fact an odd number of trailing zero bits inq0, that will leavetmpwith a last bit of 0, and so it won't pass the later(tmp & 7) == 1test. If there were an even number of trailing zero bits,tmpis left the same as before the change, and again the only real thing left to test is whether(tmp & 7) == 1.Caution: I didn't test or even compile this code! I was just browsing this on the web, and am using the web UI to edit. I've used the same trick in other projects, and am sure it's sound (give it a little thought - it's obvious ;-) ). Apologies in advance for any typos.