Re: escaping ! in quoted string gives wrong result

[Paul Jarc]

Andy Isaacson <adi@hexapodia.org> wrote:
> Ergo, ! is treated specially in double-quotes.

By the interactive history substitutor, yes, but not by the command
language parser, which is what Chet was saying, I think, and which is
exactly your problem: "\!" remains "\!", just as "\x" remains "\x".
"\$" is treated specially, and becomes "$".

> The problem arises when attempting to include a literal ! in a
> double-quoted string;

Right.  So don't do that when history is enabled.  This works:
echo "foo"\!"bar"

> 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

That would be incompatible with other sh implementations, including
FreeBSD, NetBSD, OpenBSD, pdksh, and Solaris.


paul