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From: tim@proton.amd.com (Tim Olson)
Newsgroups: comp.lang.c
Subject: Re: Compound Assignments (was Re: Another <sigh> error!)
Message-ID: <1991Apr9.145117.15100@dvorak.amd.com>
Date: 9 Apr 91 14:51:17 GMT
References: <1991Apr4.205257.15205@mccc.edu> <1991Apr6.195901.25255@dvorak.amd.com> <1991Apr7.185259.12709@druid.uucp>
Sender: usenet@dvorak.amd.com (Usenet News)
Reply-To: tim@amd.com (Tim Olson)
Organization: Advanced Micro Devices, Austin, TX
Lines: 35

In article <1991Apr7.185259.12709@druid.uucp> darcy@druid.uucp (D'Arcy J.M. Cain) writes:
| In article <1991Apr6.195901.25255@dvorak.amd.com> Tim Olson writes:
| >| 	x *= y;
| >| 	x = x * y;
| >In the second example, "x" is not evaluated twice -- it is evaluated
| >only once, just as in the first example.  The standard says just this
| >in 3.3.16.2 (Compound assignment):
| >	A compound assignment of the form E1 op= E2 differs from the
| >	simple assignment expression E1 = E1 op (E2) only in that the
| >	lvalue E1 is evaluated only once.
| 
| Huh?  Am I missing something or does that say that the two expressions
| *ARE* evaluated differently?

Yes, the two expressions you show are evaluated differently. I think
the confusion here is that the two expressions you show aren't the two
original expressions I was talking about -- they were (from the
original article):

|       x[i++] *= y;
|
| has the "x[i++]" part evaluated only once, while an expression such as
|
|       x *= y;
|
| has the "x" part evaluated twice, as in

Try substituting these two expressions into the discussion above, it
should make more sense.


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