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Bash interprets $! as an event


From: jeremy
Subject: Bash interprets $! as an event
Date: Tue, 27 Jul 2010 12:01:23 -0600

Configuration Information [Automatically generated, do not change]:
Machine: x86_64
OS: linux-gnu
Compiler: gcc
Compilation CFLAGS:  -DPROGRAM='bash' -DCONF_HOSTTYPE='x86_64' 
-DCONF_OSTYPE='linux-gnu' -DCONF_MACHTYPE='x86_64-pc-linux-gnu' 
-DCONF_VENDOR='pc' -DLOCALEDIR='/usr/share/locale' -DPACKAGE='bash' -DSHELL 
-DHAVE_CONFIG_H   -I.  -I../bash -I../bash/include -I../bash/lib   -g -O2 -Wall
uname output: Linux figment 2.6.32-24-server #38-Ubuntu SMP Mon Jul 5 10:29:32 
UTC 2010 x86_64 GNU/Linux
Machine Type: x86_64-pc-linux-gnu

Bash Version: 4.1
Patch Level: 5
Release Status: release

Description:

Event-catching seems to be catching cases it shouldn't. In particular, $! is 
caught as an event.

Repeat-By:

This command fails as follows:

> nickurak@woking:~$ sleep 5 & PID=$!; echo "pid is $PID"  
> bash: !: event not found

Adding quotation marks doesn't help:

> nickurak@woking:~$ sleep 5 & PID="$!"; echo "pid is $PID" 
> bash: !": event not found

As a workaround, adding whitespace after the ! fixes it:

> nickurak@woking:~$ sleep 5 & PID=$! ; echo "pid is $PID"
> [1] 1614
> pid is 1614

or

> nickurak@woking:~$ sleep 5 & PID="$! "; echo "pid is $PID"
> [3] 2048
> pid is 2048 



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