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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