Author: Sune Fischer
Date: 05:10:42 12/05/02
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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.
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