Re: escaping ! in quoted string gives wrong result

[Andy Isaacson]

On Fri, Jun 18, 2004 at 01:08:18AM -0400, Chet Ramey wrote:
> adi@pirx.hexapodia.org wrote:
> >Bash Version: 2.05b
> >
> >Description:
> >     Like some other shells, bash interprets ! inside double quotes.
> >     However, bash does not handle escaped ! like other shells.
> >
> >     "\!foo" should turn into `!foo' before being passed to the command.
> 
> It should not.  `!' is not one of the characters that is treated
> specially when in a double-quoted string.  This is one of the places
> where bash's sh heritage and the csh-derived history expansion features 
> collide.

I guess I don't understand your argument.  "!foo" is a history expansion
in bash, and I'm willing to accept that it should be ([cz]sh do so as
well).  Ergo, ! is treated specially in double-quotes.  The problem
arises when attempting to include a literal ! in a double-quoted string;
you can't say `echo "!foo"' (that's a history expansion) and you can't
say `echo "\!foo"' (that results in the string "\!foo").

I'm claiming that Bash should do the same thing that zsh and csh do, and
interpret `a.out "\!foo"' in the most useful way; a.out should receive
the C-style string "!foo" in argv[1], instead of the "\!foo" which it
currently receives.

(If I've misunderstood what you meant by "treated specially", my
apologies, and please feel free to educate me (or point me at a
definition, even source code.))

-andy