Re: [Help-bash] Root access in subshell

[Bob Proulx]

Richard Taubo wrote:
> Greg Wooledge wrote:
> > Richard Taubo wrote:
> >> I am running the following command in the terminal logged in as root (not 
> >> via a script):
> >>    [#] $(find / -user myuser)      # This is a part of a little larger 
> >> script

Is that "[#]" your prompt?  It is hard to tell.  If so are you
actually running the subshell and then running the result of the
subshell?  That doesn't make any sense.

> I made the error of thinking that the subshell would return the
> result directly to the main shell, so I was confusing the lack of
> returned feedback with that the subshell was not run as root.  I saw
> that error messages came from the subshell, but not the result of
> the command itself, and was taken for a spin . . .

When you use $(...) it runs the command and replaces the output on the
command line and runs the result.  So the result is data dependent.
And could be dangerous!  If the first command found is "rm" then it
would run rm with the rest of the arguments.  But that would be
unlucky.  Probably it will run some other command and fail.  If the
first file is "." then it will try to source the first file.

  $ mkdir /tmp/test
  $ cd /tmp/test
  $ date > foo
  $ date > bar
  $ find . -print
  .
  ./bar
  ./foo
  $ cat foo
  Sun Feb  2 23:58:40 MST 2014
  $ $(find . -print)
  bash: Sun: command not found

But it might not say anything at all.

  $ echo echo foo > foo
  $ cat foo
  echo foo

And therefore:

  $ $(find . -print)
  foo

That is because 

  $ echo $(find . -print)
  . ./bar ./foo

And it will source ./bar which just contains "echo foo" which it does
and therefore no error.

Using $(...) is really just for assigning to variable or for placement
on the commandline.

> I see that if the subshell is set to a variable I can return 
> the feedback to the main shell — are there other ways to return such
> feedback to the main shell?

Huh?  What you are saying makes no sense to me.  What are you saying?
Returning feedback?  Simply run the command.  The output will be
printed.  Why are you using $(...) there?

Bob