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escaping ! in quoted string gives wrong result

From: adi
Subject: escaping ! in quoted string gives wrong result
Date: Mon, 14 Jun 2004 12:04:15 -0500 (CDT)

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' 
-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

        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.

        bash-2.05b$ echo hello
        bash-2.05b$ echo "\!echo"

        Expected output:
        % echo "\!echo"

        (Tested under zsh and Solaris csh.)

        [Description of how to fix the problem.  If you don't know a
        fix for the problem, don't include this section.]

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