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can't avoid double message in trap 'echo; exit' with backquoted code
From: |
ahuxley |
Subject: |
can't avoid double message in trap 'echo; exit' with backquoted code |
Date: |
Fri, 30 Nov 2001 10:53:19 +0100 (MET) |
Configuration Information [Automatically generated, do not change]:
Machine: hppa2.0
OS: hpux11.00
Compiler: gcc
Compilation CFLAGS: -DPROGRAM='bash' -DCONF_HOSTTYPE='hppa2.0'
-DCONF_OSTYPE='hpux11.00' -DCONF_MACHTYPE='hppa2.0-hp-hpux11.00'
-DCONF_VENDOR='hp' -DSHELL -DHAVE_CONFIG_H -DHPUX -I. -I. -I./lib
-I/opt/bash/include -g -O2
uname output: HP-UX te35 B.11.00 A 9000/785 2008900992 two-user license
Machine Type: hppa2.0-hp-hpux11.00
Bash Version: 2.03
Patch Level: 0
Release Status: release
Description:
Setting up a trap to display a message and exit, and then
invoking the signal handler while in a backquoted section of
code results in the trap's message appearing twice.
All other OS's and Bourne-like shells produce the message only
once. But despite the subshell, the trap message comes from the
same pid each time.
No amount of '|| exit' added to each read/function call prevent
the second message appearing.
(I tried to verify whether this bug was present in latest bash
(2.05a) but I couldn't compile it for HP-UX 10.20 or 11.00.
A separate bug report has been files for that.)
Repeat-By:
Run the following script:
#!/bin/bash
get_x()
{
typeset LOCAL_X
echo "Press CTRL-C now: " > /dev/tty
read LOCAL_X < /dev/tty
echo "$LOCAL_X"
}
main()
{
trap 'echo "got CTRL-C, this message should appear only once, pid
$$ exiting ..." > /dev/tty; exit 1' 2
X=`get_x`
}
main "$@"
--
Alexis Huxley
ahuxley@eso.org
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