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From: kdq@demott.COM (Kevin D. Quitt)
Newsgroups: comp.lang.c
Subject: Re: count of bits set in a long
Keywords: bits set, count of, long
Message-ID: <644@demott.COM>
Date: 2 Oct 90 21:38:16 GMT
References: <12330@crdgw1.crd.ge.com>
Reply-To: kdq@demott.COM (Kevin D. Quitt)
Organization: DeMott Electronics Co., Van Nuys CA
Lines: 27

In article <12330@crdgw1.crd.ge.com> davidsen@antarctica.crd.GE.COM (william E Davidsen) writes:
>
>  There have been lots of fast but non-portable solutions posted.
>
>  I like the one which said (paraphrase)
>
>	while (n) {
>	  count += (n & 1);
>	  n >>= 1;
>	}

    This one, of course, assumes that the shift is logical, not arithmetic,
or your first negative number will keep you occupied for quite a while.  I
think the best solution is:

	char	*ptr	= (char *)&lvar;
	int	 count	= 0;

	for ( i = 0; i < sizeof( long ); i++ )
	    count	+= bits_in_byte[ *p++ ];
-- 
 _
Kevin D. Quitt         demott!kdq   kdq@demott.com
DeMott Electronics Co. 14707 Keswick St.   Van Nuys, CA 91405-1266
VOICE (818) 988-4975   FAX (818) 997-1190  MODEM (818) 997-4496 PEP last

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