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List:       user-mode-linux-user
Subject:    Re: [uml-user] Question : how to get login prompt in the xterm that uml is kicked off from ?
From:       s-uml () rhythm ! cx
Date:       2004-02-25 19:09:20
Message-ID: 20040225190920.GC1995 () infinity ! rhythm ! cx
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> The command I use is
> 
> linux devfs=nomount mem=256M \
>  ubd0=redhat9.disk_uml_left_client1.cow,/usr/redhat9.image.workstation.disk 
> \
>  fakehd fake_ide con0=fd:0,fd:1 con=pty con1=xterm ubd=3 root=6201 \
>  eth0=daemon,10:00:01:02:00:00,,/tmp/left.ctl umid=uml_left_client_1
> 
> However, I do not understand why exactly the 'con,con0,con1' combo works 
> the way it does. Any pointers on how consoles,terminals etc link into the 
> boot process will be greately appreciated . 

This is what is going on in the the command line you have: 

con=pty: by default, system consoles in the uml will be given pty devices on
the host.

con0=fd:0,fd:1: overriding the above for con0, the uml's tty 0 (the main
system console) is sent to the uml's standard input and output. IE, any
activity on tye uml's tty0 will be sent to whatever tty/xterm on the host
you started uml in.

con1=xterm: overriding con=pty, the uml's tty 1 will appear in a new xterm
on the host.

> Also, I need to run a network of virtual machines and would like to reduce 
> the number of windows. I would like to have the login prompt show up in the 
> same xterm from which I kickoff the above command. 

Try these con= arguments instead of what you have: con=null con0=fd:0,fd:1.
This will send the uml's tty 0 to wherever you started uml from and
everything else will be canned. No xterms.

There are settings where ou can set defaults for these things if you compile
UML yourself.

> Plus, If I am to use a command line as above, how do I specify any 
> attributes for the xterm ?
> something like con0="xterm -fn 8x13" did not work

Not sure if that can be done with runtime arguments... might inovlve hacking
the uml source. Maybe someone else knows for sure.


-Steve


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