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PIPESTATUS not altered by $(())
From: |
Eric Blake |
Subject: |
PIPESTATUS not altered by $(()) |
Date: |
Sun, 16 Nov 2008 14:38:14 -0700 |
User-agent: |
Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.8.1.17) Gecko/20080914 Thunderbird/2.0.0.17 Mnenhy/0.7.5.666 |
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Hash: SHA1
Configuration Information [Automatically generated, do not change]:
Machine: i686
OS: cygwin
Compiler: gcc
Compilation CFLAGS: -DPROGRAM='bash.exe' -DCONF_HOSTTYPE='i686'
- -DCONF_OSTYPE='cygwin' -DCONF_MACHTYPE='i686-pc-cygwin' -DCONF_VENDOR='pc'
- -DLOCALEDIR='/usr/share/locale' -DPACKAGE='bash' -DSHELL -DHAVE_CONFIG_H
- -DRECYCLES_PIDS -I. -I/home/eblake/bash-3.2.39-20/src/bash-3.2
- -I/home/eblake/bash-3.2.39-20/src/bash-3.2/include
- -I/home/eblake/bash-3.2.39-20/src/bash-3.2/lib -O2 -pipe
uname output: CYGWIN_NT-5.1 LOUNGE 1.5.25(0.156/4/2) 2008-06-12 19:34 i686
Cygwin
Machine Type: i686-pc-cygwin
Bash Version: 3.2
Patch Level: 39
Release Status: release
Description:
$? and $PIPESTATUS[*] should always agree with each other, but
there are situations where the latter is not properly updated. In the
example below, a failed arithmetic substitution sets $? but not $PIPESTATUS.
Repeat-By:
$ PS1=' ${PIPESTATUS[*]} \$ '
0 $ echo $?
0
0 $ echo $((+))
bash: +: syntax error: operand expected (error token is "+")
0 $ echo $?
1
0 $
- --
Don't work too hard, make some time for fun as well!
Eric Blake ebb9@byu.net
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