From mboxrd@z Thu Jan 1 00:00:00 1970 Received: from mailman by lists.gnu.org with tmda-scanned (Exim 4.43) id 1EXz3c-0001f4-D1 for qemu-devel@nongnu.org; Fri, 04 Nov 2005 05:44:29 -0500 Received: from exim by lists.gnu.org with spam-scanned (Exim 4.43) id 1EXz2F-0001XN-LC for qemu-devel@nongnu.org; Fri, 04 Nov 2005 05:44:24 -0500 Received: from [199.232.76.173] (helo=monty-python.gnu.org) by lists.gnu.org with esmtp (Exim 4.43) id 1EXyja-0000Wl-06 for qemu-devel@nongnu.org; Fri, 04 Nov 2005 05:23:47 -0500 Received: from [64.233.184.207] (helo=wproxy.gmail.com) by monty-python.gnu.org with esmtp (Exim 4.34) id 1EXyja-0001sR-2A for qemu-devel@nongnu.org; Fri, 04 Nov 2005 05:23:46 -0500 Received: by wproxy.gmail.com with SMTP id i28so357561wra for ; Fri, 04 Nov 2005 02:23:41 -0800 (PST) Message-ID: Date: Fri, 4 Nov 2005 18:23:41 +0800 From: zheng sw MIME-Version: 1.0 Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: quoted-printable Content-Disposition: inline Subject: [Qemu-devel] How to get a guest IP Reply-To: qemu-devel@nongnu.org List-Id: qemu-devel.nongnu.org List-Unsubscribe: , List-Archive: List-Post: List-Help: List-Subscribe: , To: qemu-devel@nongnu.org Host want to get the guest's IP when the host start the guest? My setting is: debian(2.6.8), qemu, bridge-utils, TUN/TAP. The guest get a IP from a DHCP server which is in the same subnet whit host. My purpose is use bash to run ssh connect to the host, then run qemu, After the guest is started, the host want to get the guest's IP and send to me, but nobody will control the guest. Now I am puzzle in how the host get the guest IP. I have try 1. redirect the guest stdio to a terminal. but false, 2. Use DHCP server bound MAC address with IP, and use -macaddr set the guest MAC. but I can't control the DHCP server now. 3, mount the qemu image, but document say it can't mounted. 4. guest use static IP, but I may start several guest with a image. 5. use -seral pty, but I can't read the pts. Thank you! My english is poor, sorry.