If the data is not random you might find some advantage in removing the bit that is set the most first.
If the input data has some bits set more than 1/63 of the time you can get some advantage. You can see this in that if all of the data had a bit set then this algorithm would remove all the entries from the list on the first pass if it checked that bit first.
But I assume your data is random.
In reply to Re^2: Partition in to 63 parts
by gam3
in thread Need a faster way to find matches
by remzak
| For: | Use: | ||
| & | & | ||
| < | < | ||
| > | > | ||
| [ | [ | ||
| ] | ] |