Re: Running commands as "$@"

[Cristian Zoicas]

Davide Brini wrote:
> On Thu, 2 Feb 2023 12:11:32 +0100, Cristian Zoicas
> <zoicas@medialab.sissa.it> wrote:
>
> > Now, if a call the script with the commands
> >
> >       $ sh runner.sh A=B
> >
> > we get the following errors:
> >
> >       runner.sh: 10: runner.sh: A=B: not found
> >
> > Why the shell does not treat the first string in "$@" ('A=B'
> > in these cases) as the first possibile token of command (which
> > is an assignemnt)?
>
> It works for me ("sh" points to bash).
>
> > In addition, the problem is NOT solved by using 'eval "$@"'.
> > Apparently it works. For example the command
> >
> >        $ sh runner.sh A=BCD
> >
> >    works, but
> >
> >        $ sh runner.sh A="B    CD"
> >
> > gives an error. In this case the error (I think) comes
> > from the way the expansion of "$@" is performed: eval has to evaluate the
> > string "A=B   CD", so it treats A=B as an assignemnt and 'CD' as
> > a command.
>
> This happens because the quotes are removed by the calling shell before eval
> has a chance to see them. This works:
>
> $ sh runner.sh 'A="B    CD"'
>
> or even, to check
>
> $ sh runner.sh 'A="B    CD"; echo "$A"'
> B    CD
>
> > I am curios what is going on and if there is any possibility of
> > running a command (any command, including an assignment) by using
> > a very simple invocation (someting like eval "$@").
>
> The short answer is: yes, but not where quoting is involved unless you take
> special measures to ensure that the relevant quotes are seen by eval.
Correct, but I want to avoid taking these special measures. It is (and it was) 
clear for
me why the script does not work with eval.

There was a mistake in my original question and I am extremely sorry for that.

I wanted to ask the question for this script

# ---- begin script runner.sh ----
"$@"
# ---- end script runner.sh ----

and not for the one with "eval"

Here is what I want to obtain:

    A command is a (parsed) sequence of strings and also "$@" is a sequence
    of strings.

    So expected that "$@" would allow me to run anything that was a valid 
command.

    In fact the script above lets me to run "ls -l" but fails to run "A=B"

Hope that clarifies my issue.