Author: Dan Newman
Date: 05:27:54 12/05/02
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On December 05, 2002 at 08:10:42, Sune Fischer wrote: >On December 05, 2002 at 07:34:37, Matt Taylor wrote: > >><snip> >>>> I went on and did some testing with the b & -b >>> >>>Okay, I'm going to ask now. Can someone explain to me the meaning of b& -b? My >>>compiler generates a warning that changing a sign on an unsigned accomplishes >>>nothing, so the expression reduces to b & b which is b? >><snip> >> >>You may have to cast if your compiler is too "smart" for you. The b & -b gives a >>mask such that the only bit set in the mask is the first bit set in b. (e.g. >>1111b -> mask of 0001b, 1000b -> mask of 1000b, 1010b -> mask of 0010b) >> >>-Matt > > >I still don't understand how it works, "unary minus operator applied to unsigned >type", how is that even defined, what is -b? > >But basicly the result is the same as b^(b&(b-1)) ? > >-S. The b & -b trick relies on having twos-complement representation for negative numbers. But it only works if b is signed (or perhaps with a cast to make it so). It can be done on unsigned b with an extra operation: b & (~b + 1). Here's an example of how it works (on 8-bit numbers): Let b = 11100100 then ~b = 00011011 and ~b + 1 = 00011100 (same as -b if b is signed) and finally b & (~b + 1) = 00000100 which has only the first one-bit of b turned on. -Dan.
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