Author: Ricardo Gibert
Date: 08:39:05 07/27/03
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On July 27, 2003 at 11:10:07, Ratko V Tomic wrote: >>To compute b = a for mask m: b = (b&~m)|(a&m) [untested] >> >>That's 4 operations. 3 if m is a constant when the "~" will optimized away. > >A quicker way (max 3 ops): > > b^=(a^b)&m; Neat. My way may gain from parallelizing operations, while yours is sequential. However, I suspect mine uses an extra register and this should cause it to be slower than yours depending on the context in which it is used. It would be interesting to benchmark the two.
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