Re: hurd does NOT need /hurd

[Diego Roversi <diegor@tiscalinet.it>]

On Mon, May 20, 2002 at 04:50:43PM +1000, Anthony Towns wrote:
> On Sun, May 19, 2002 at 11:24:07PM -0700, Thomas Bushnell, BSG wrote:
> > > They're referred to by the user, but they're not invoked by the user.
> > In one sense, no program on Linux is invoked by the user--only the
> > shell actually invokes the program, at the user's direction.
> 
> That's not a very helpful sense for this discussion.
> 
> If you want to indirectly invoke a command via the shell, you type:
> 
> 	$ /bin/sh -c 'ls'
> 

Perhaps we have to change the point of view. When you write 'ls' you are
telling to the shell to:
a) fork a new process
b) (optionaly) set some file handler (ie: stdin & stdout)
c) exec the indicated executables 

So you are not really directly invoking the executable, but telling to the
shell to do it, using a implicit protocol. Under the Hurd exist a different
protocol to run process that is used only for translators, because it
requires a more complex dialog. At the moment the shell doesn't speak this
protocol, so you need a helper process: settrans. 

So in this case you have a double indirectly invoked process, but only
because your shell is not hurd-speaking ;). Btw even if bash was extended to
support hurd-process, it should look for binary in a different path, just
because of the different way they work, regardless that is invoked directly
by user or not.

I hope I was clear... :)

-- 
Saluti / Regards

Diego Roversi | 
              | diegor at tiscalinet.it 


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