Is the following difference intentional, a bug, or do I miss something?
unset a
set -- [EMAIL PROTECTED]
echo $#
Output: 0
typeset a
set -- [EMAIL PROTECTED]
echo $#
Output: 1
The man page says, If the word is double-quoted, ... [EMAIL PROTECTED] expands
each element of name to
Configuration Information [Automatically generated, do not change]:
Machine: i486
OS: linux-gnu
Compiler: gcc
Compilation CFLAGS: -DPROGRAM='bash' -DCONF_HOSTTYPE='i486'
-DCONF_OSTYPE='linux-gnu' -DCONF_MACHTYPE='i486-pc-linux-gnu'
-DCONF_VENDOR='pc' -DLOCALEDIR='/usr/share/locale'
Configuration Information [Automatically generated, do not change]:
Machine: i386
OS: linux-gnu
Compiler: gcc
Compilation CFLAGS: -DPROGRAM='bash' -DCONF_HOSTTYPE='i386'
-DCONF_OSTYPE='linux-gnu' -DCONF_MACHTYPE='i386-redhat-linux-gnu'
-DCONF_VENDOR='redhat' -DLOCALEDIR='/usr/share/locale'
[EMAIL PROTECTED] wrote:
Bash Version: 3.2
Patch Level: 39
Release Status: release
Description:
When nullglob option is enable (shopt -s nullglob), unset of an array
does not work.
You're right; it does. `unset' is a builtin, so all of the shell's word
expansions are
Björn Augustsson wrote:
Bash Version: 3.2
Patch Level: 33
Release Status: release
Description:
The test case below is pretty self-explanatory.
The assignment in fun_bad() doesn't exit the shell,
despite the set -e.
This is the version in Fedora 8.
Bernd Eggink wrote:
Is the following difference intentional, a bug, or do I miss something?
unset a
set -- [EMAIL PROTECTED]
echo $#
Output: 0
This is correct.
typeset a
set -- [EMAIL PROTECTED]
echo $#
Output: 1
The question is how to treat the variable created by
Chet Ramey wrote:
Björn Augustsson wrote:
fun_bad() { local bah=$( false ); }
fun_good() { local bah ; bah=$( false ); }
The `local' command returns success if the variable assignment succeeds,
which it does. The command substitution doesn't affect its exit status.
This is how all
