Open
Conversation
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.
Pr0methean
approved these changes
Aug 2, 2025
This file contains hidden or bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
Sign up for free
to join this conversation on GitHub.
Already have an account?
Sign in to comment
2 participants
Add this suggestion to a batch that can be applied as a single commit.This suggestion is invalid because no changes were made to the code.Suggestions cannot be applied while the pull request is closed.Suggestions cannot be applied while viewing a subset of changes.Only one suggestion per line can be applied in a batch.Add this suggestion to a batch that can be applied as a single commit.Applying suggestions on deleted lines is not supported.You must change the existing code in this line in order to create a valid suggestion.Outdated suggestions cannot be applied.This suggestion has been applied or marked resolved.Suggestions cannot be applied from pending reviews.Suggestions cannot be applied on multi-line comments.Suggestions cannot be applied while the pull request is queued to merge.Suggestion cannot be applied right now. Please check back later.
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.