Date: Wed, 26 Jan 2005 12:13:37 +0100 From: Matthias Andree <matthias.andree@gmx.de> To: Michael Sierchio <ducatista@camber-thrust.net> Cc: lioux@freebsd.org Subject: Re: FreeBSD Port: qmail-1.03_3 Message-ID: <m3hdl4phni.fsf@merlin.emma.line.org> In-Reply-To: <20050126104355.GA3837@falcon.midgard.homeip.net> (Erik Trulsson's message of "Wed, 26 Jan 2005 11:43:56 %2B0100") References: <41F6F431.6060005@tenebras.com> <1106704507.16118.14.camel@shumai.marcuscom.com> <41F72C02.7060901@camber-thrust.net> <20050126104355.GA3837@falcon.midgard.homeip.net>
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Erik Trulsson <ertr1013@student.uu.se> writes:
> On Tue, Jan 25, 2005 at 09:34:58PM -0800, Michael Sierchio wrote:
>> ( 1 && 0 || 1 ) will ALWAYS evaluate to 1 on any ANSI C compiler.
>
> No, it will evaluate to 0.
It will evaluate to 1, as you're correctly stating...
> && has higher precedence than ||.
>
> (x && y || z) is equivalent to ((x && y) || z)
...here. 1 && 0 || 1
== 0 || 1
== 1
(this only holds as the numbers are side-effect free, if x and y were
function calls or macros with side effects, this simplification would
not take the side effects into account)
> which is different from (x && (y || z)).
Exactly.
--
Matthias Andree
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