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From: willcr@bud.sos.ivy.isc.com (Will Crowder)
Newsgroups: comp.lang.c
Subject: Re: Increment Operators vs. Precedence
Message-ID: <1991Mar07.174626.1688@ism.isc.com>
Date: 7 Mar 91 17:46:26 GMT
References: <THOMSON.91Mar5173421@zazen.macc.wisc.edu> <668294528.3716@mindcraft.com>
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In article <668294528.3716@mindcraft.com>, fred@mindcraft.com (Fred Zlotnick)
writes:

|> A related example is the conditional expression
|> 	x++ && y++ || z++
|> The precedence rules state that this means
|> 	((x++) && (y++)) || (z++)
|> As it happens, there is an order of evaluation rule for && and ||, and it
|> causes strange results here: x is always incremented, y is incremented
|> if x was not -1, and z is incremented only if x and y were both -1.  Thus
|> even though ++ is the highest precedence operator in this expression, not
|> all ++'s get evaluated.  Note that for most C operators (anything except &&,
|> ||, ?: and comma) there is no specified order of evaluation.

Actually, x is always incremented, y is incremented if x wasn't 0 (not -1)
and z is incremented if either x or y is 0.

Remember, the value of the expression (x++) is x *before* the increment, not
after.  Since all increment operators in this expression are postincrement,
the incremented values of x, y and z do not come into play in determining
which parts of the expression are evaluated.

Due to C's (rather handy) shortcircuiting of && and || operators, the right
side of the || is only evaluated if the left side is false.  For &&, the
right side is only evaluated if the left side is true.  Since the left side
of the || expression in this case is the && expression, the && expression
must be false (either x or y 0) for the right side to be evaluated.

Hope this helps,

Will

|> Hope this helps.
|> Fred Zlotnick                       |	#include <std.disclaimer>
|> fred@mindcraft.com                  |	#include <brilliant.quote>
|> ...!{decwrl,ames,hpda}!mindcrf!fred |