Path: utzoo!utgpu!jarvis.csri.toronto.edu!rutgers!mephisto!ncar!ames!amdcad!nucleus!tim
From: tim@nucleus.amd.com (Tim Olson)
Newsgroups: comp.graphics
Subject: Re: Integral Square Root
Message-ID: <28569@amdcad.AMD.COM>
Date: 31 Dec 89 22:31:57 GMT
References: <9170@cbmvax.commodore.com> <21550@mimsy.umd.edu> <9175@cbmvax.commodore.com>
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Reply-To: tim@amd.com (Tim Olson)
Organization: Advanced Micro Devices, Inc., Austin, Texas
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In article <9175@cbmvax.commodore.com> mitchell@cbmvax.commodore.com (Fred Mitchell - PA) writes:
| NOT WRONG- Let's analyze this line more closely (exploded IF, here we come!)...
| 
| 	if (
| 		( rb = r + 
| 			(b=1<<i) )
| 				* rb
| 					<= n)
| 
| Now the one stipulation of the C standard is that INNER PARENTHESIS
| GETS EVALUATED FIRST!!!!

No.  Parenthesis only change the grouping of an expression, not its
order of evaluation (this is true in K&R "classic" C, as well as the
proposed ANSI standard).  The standard explicitly states:

	 The grouping of an expression does not completely determine
	 its evaluation.  In the following fragment

	      #include <stdio.h>
	      int sum;
	      char *p;
	      /*...*/
	      sum = sum * 10 - '0' + ((*(p++)) = (getchar())));

	 the expression statement is groupsed as if it were written as

	      sum = (((sum * 10) - '0') + ((*(p++)) = (getchar())));

	 but the actual increment of p can occur at any time between
	 the previous sequence point and the next sequence point (the
	 ;), and the call to getchar can occur at any point prior to
	 the need of its returned value.

The key here is the concept of a "sequence point".  Any side-effects
(such as assignment) in an expression can occur at any time as long as
they are complete at the sequence point.  This leads to the
restriction in section 3.3:

	    Between the previous and next sequence point an object
	    shall have its stored value modified at most once by the
	    evaluation of an expression.  Furthermore, the prior value
	    shall be accessed only to determine the value to be
	    stored.


The expression

	(rb = r + (b=1<<i)) * rb <= n)

violates this restriction, because the second reference to rb is not
to determine the previous value of rb to use in the assignment.

Because the standard allows the assignment side-effect to occur at
anytime prior to the sequence point, the second reference to rb could
either be the old or new value of rb, depending upon the
implementation.

	-- Tim Olson
	Advanced Micro Devices
	(tim@amd.com)