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Date:      Tue, 6 May 2003 10:25:42 -0500
From:      "Jacques A. Vidrine" <nectar@FreeBSD.org>
To:        Harti Brandt <brandt@fokus.fraunhofer.de>
Cc:        freebsd-arch@freebsd.org
Subject:   Re: `Hiding' libc symbols
Message-ID:  <20030506152542.GC77708@madman.celabo.org>
In-Reply-To: <20030506095424.G838@beagle.fokus.fraunhofer.de>
References:  <20030505225021.GA43345@nagual.pp.ru> <Pine.GSO.4.10.10305051855570.10283-100000@pcnet1.pcnet.com> <20030505232012.GC21953@madman.celabo.org> <20030506095424.G838@beagle.fokus.fraunhofer.de>

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On Tue, May 06, 2003 at 09:56:06AM +0200, Harti Brandt wrote:
> There is no guarantee that you 'fix' the port by hiding the symbol.  You
> may as well break it. This depends on the function itself and on the
> internal relationships in libc. You have to go through each individual
> port and see what happens anyway.

Please explain.  I _am_ guaranteed that keeping the port from hijacking
strlcpy calls in libc will not break the port.  How could the port rely
on the internals of libc?

Cheers,
-- 
Jacques Vidrine   . NTT/Verio SME      . FreeBSD UNIX       . Heimdal
nectar@celabo.org . jvidrine@verio.net . nectar@freebsd.org . nectar@kth.se



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