Linmodems Mailing list Archives

	Web archives.linmodems.org

Return-Path: <neilharwani@hotmail.com>
Mailing-List: contact discuss-help@linmodems.org; run by ezmlm
Delivered-To: mailing list discuss@linmodems.org
Received: (qmail 32661 invoked from network); 7 Oct 2002 12:40:25 -0000
Received: from dav58.sea2.hotmail.com (HELO hotmail.com) (207.68.164.193)
  by www.linmodems.org with SMTP; 7 Oct 2002 12:40:25 -0000
Received: from mail pickup service by hotmail.com with Microsoft SMTPSVC;
	 Mon, 7 Oct 2002 05:39:57 -0700
X-Originating-IP: [203.88.129.158]
From: "Neil Harwani" <neilharwani@hotmail.com>
To: <discuss@linmodems.org>
Subject: Kernel architecture modem etc.
Date: Mon, 7 Oct 2002 18:05:15 +0530
MIME-Version: 1.0
Content-Type: text/plain;
	charset="iso-8859-1"
Content-Transfer-Encoding: 7bit
X-Priority: 3
X-MSMail-Priority: Normal
X-Mailer: Microsoft Outlook Express 5.50.4522.1200
X-MimeOLE: Produced By Microsoft MimeOLE V5.50.4522.1200
Message-ID: <DAV58zXfhCuIKe4YSeE000060b7@hotmail.com>
X-OriginalArrivalTime: 07 Oct 2002 12:39:57.0578 (UTC) FILETIME=[A4EAFEA0:01C26DFE]

Hi friends,
    I am a bit new to linux. I know a little bit basics as such but I am a
big linux enthusiast. I have a Conexant internal modem for which I went to
http://www.mbsi.ca/cnxtlindrv/ Now how to do I determine my chipset
architecture whether it is I586 or I686 ? I have a Pentium 4 1.7 GHZ
processor and an Intel D845WN motherboard, I have assembled the system just
a month back. Once i know that I also need to decide which kernel version I
have, please guide me as to what exact command I should use for finding both
things. Main thing is my architecture, I can find kernel version with uname
but I am confused about architecture cos when i had downloaded the Java
Virtual Machine for windows it said I have I586 architecture but in linux it
tells me I have I686 architecture by uname -m. What exactly to type for this
command: rpm -q --qf '%{ARCH}\n' kernel
The site tells me this will give me exact architecture.
Regards and respect to everyone on the list,
Neil.