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From: Matthias Andree <matthias.andree@gmx.de>
To: Michael Sierchio <ducatista@camber-thrust.net>
In-Reply-To: <20050126104355.GA3837@falcon.midgard.homeip.net> (Erik
	Trulsson's message of "Wed, 26 Jan 2005 11:43:56 +0100")
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Date: Wed, 26 Jan 2005 12:13:37 +0100
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cc: Michael Sierchio <kudzu@tenebras.com>
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Subject: Re: FreeBSD Port: qmail-1.03_3
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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