Re: escaping ! in quoted string gives wrong result

[Chet Ramey]

adi@pirx.hexapodia.org wrote:

> Configuration Information [Automatically generated, do not change]:
> Machine: i386
> OS: openbsd3.4
> Compiler: cc
> Compilation CFLAGS:  -DPROGRAM='bash' -DCONF_HOSTTYPE='i386' 
> -DCONF_OSTYPE='openbsd3.4' -DCONF_MACHTYPE='i386-unknown-openbsd3.4' 
> -DCONF_VENDOR='unknown' -DSHELL  -DHAVE_CONFIG_H  -I.  
> -I/usr/ports/shells/bash2/w-bash-2.05b/bash-2.05b 
> -I/usr/ports/shells/bash2/w-bash-2.05b/bash-2.05b/include 
> -I/usr/ports/shells/bash2/w-bash-2.05b/bash-2.05b/lib  -O2
> uname output: OpenBSD pirx.hexapodia.org 3.4 GENERIC#18 i386
> Machine Type: i386-unknown-openbsd3.4
>
> Bash Version: 2.05b
> Patch Level: 0
> Release Status: release
>
> 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.

Chet
--
``The lyf so short, the craft so long to lerne.'' - Chaucer
( ``Discere est Dolere'' -- chet )
                                                Live...Laugh...Love
Chet Ramey, ITS, CWRU    chet@po.CWRU.Edu 
http://cnswww.cns.cwru.edu/~chet/