Message-Id: <37B53B6E.7CF29EB1@intel.com>
Date: Sat, 14 Aug 1999 11:48:30 +0200
From: Kurt Alstrup <kurt DOT alstrup AT intel DOT com>
Organization: Intel Denmark Aps
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To: djgpp AT delorie DOT com
Subject: Re: Bit counting?
References: <37B45836 DOT EAD7C82D AT swipnet DOT se> <37B48ABE DOT 110D2C13 AT intel DOT com> <37B4D8BC DOT 6465 AT ns DOT sympatico DOT ca>
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Reply-To: djgpp AT delorie DOT com

Well ... Looking at it again then a modulo 077 (= 0x3f) could be
accompliced by using '&' instead.
Thus the line

    return (int) (((y + (y >> 3)) & 030707070707) & 077);

should do the same without using a potential slow mod operator.

Kurt Alstrup

Klaas wrote:

> Kurt Alstrup wrote:
> >
> > Try this little function, I guess it may gain if written in assembly
> > though ..
> >
> > /*
> >  *      Ones
> >  *
> >  * This magic counts the number of bits in one longword
> >  */
> >
> > int
> > Ones(unsigned long mask)
> > {
> >   register unsigned long y;
> >
> >   y = (mask >> 1) & 033333333333;
> >   y = mask - y - ((y >>1) & 033333333333);
> >   return (int) (((y + (y >> 3)) & 030707070707) % 077);
> > }
> >
> > Regards,
> > Kurt Alstrup
>
> Doesn't the modulo make it rather slow?
>
> -Mike