Re: how does escaping in "`...`" work?

[Matthew Woehlke]

Eric Blake wrote:
> On 06/07/2010 04:47 PM, Matthew Woehlke wrote:
> > But this does not:
> > $ echo "`echo \"you don\'t   say\"`"
> > you don\'t   say
>
> \' is not special inside ``.  If it is not special, then the \ is
> preserved on to the nested command.  Bash's behavior matches POSIX here.

Right; w.r.t. the ``'s, I didn't expect it was. The confusion was due to 
not understanding if the \' is or is not inside a ""'d string in the 
command substitution.

> > I expected:
> > "you don't say"
>
> Then don't use \'.  echo "`echo \"you don't  say\"`"

I think you meant:
echo "`echo \\\"you don\'t    say\\\"`"
:-)

(The above example indicated the quotes emitted in the output.)

> > That is, I expected that \" ->  " substitution would not happen inside
> > the ``'s outside of the context of the command substitution, since the
> > quotes do not need to be escaped in that context to be seen as quotes in
> > the substitution context.
>
> The quotes NEED to be escaped in the quoted command substitution
> context, since behavior is undefined if " quotes are not escaped inside
> "``".

Okay, thanks. (Actually your other reply was the most helpful of the two 
:-).)

--
Matthew
Please do not quote my e-mail address unobfuscated in message bodies.
--
Will your shell have salvation? Only if it's Bourne Again.