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From: ok@quintus.uucp (Richard A. O'Keefe)
Newsgroups: comp.lang.c
Subject: Re: Shifting question
Message-ID: <178@quintus.UUCP>
Date: 20 Jul 88 04:35:17 GMT
References: <705@bnr-rsc.UUCP> <11556@steinmetz.ge.com> <60290@sun.uucp> <1818@spar.SPAR.SLB.COM>
Sender: news@quintus.UUCP
Reply-To: ok@quintus.UUCP (Richard A. O'Keefe)
Organization: Quintus Computer Systems, Inc.
Lines: 50

In article <1818@spar.SPAR.SLB.COM> hunt@spar.UUCP (Neil Hunt) writes:
>shift_pixel(image, count)
>    struct image *image;
>    int count;
>    {
>	int i, j;
>
>	for(j = 0; j < image->rows; j++)
>		for(i = 0; i < image->cols; i++)
>			image->pixels[i][j] >>= count;
>    }
>... the last line has to be:
>
>			if(count > 0)
>				image->pixels[i][j] >>= count;
>			else
>				image->pixels[i][j] <<= -count;
>
It is better to move the "if" outside the loop.
C arrays using the Algol convention rather than the Fortran one,
it may be better to vary j faster than i (and it might be faster
still to use pointers).

void shift_pixel(image, count)
    struct image *image;
    int count;
    {
	int i, j;

	if (count > 0) {
	    for (i = image->cols; --i >= 0; )
		for (j = image->rows; --j >= 0; )
		    image->pixels[i][j] >>= count;
	} else
	if (count < 0) {
	    count = -count;
	    for (i = image->cols; --i >= 0; )
		for (j = image->rows; --j >= 0; )
		    image->pixels[i][j] <<= count;
	}
    }

The bottom line is that because the C standard specifies that shifting
is undefined for negative or overlarge shift counts, it can generate
**faster** code for (this version of) the operation you want than it
could if shifting were more tightly specified.  Since several machines
take the bottom N bits of the shift count and assume it is negative,
if you didn't write the if somewhere, the compiler would have to, and
putting that if (count > 0) shift right; else shift left; in the inner
loop would really hurt pipelined and vectored machines.  (Unless the
compiler was smart enough to move the test outside the loop, of course.)