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aliased command always considered in $PATH (-> is tab-completed)
From: |
Jan . Stary |
Subject: |
aliased command always considered in $PATH (-> is tab-completed) |
Date: |
2 Apr 2004 08:41:55 -0000 |
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' -DSHELL -DHAVE_CONFIG_H -D_GNU_SOURCE
-D_FILE_OFFSET_BITS=64 -I. -I. -I./include -I./lib -I/usr/include -O2
-march=i386 -mcpu=i686
uname output: Linux praha1.ff.cuni.cz 2.4.9-e.38 #1 Thu Feb 19 05:40:22 CET
2004 i686 unknown
Machine Type: i386-redhat-linux-gnu
Bash Version: 2.05
Patch Level: 8
Release Status: release
Description:
Hello. If you setup an alias such as
antiword='antiword -s -w 79 -m 8859-2.txt'
then typing 'antiword' is gets tab-completed, even if there is not
really any 'antiword' command in $PATH. Similarly with other aliases.
Is this intentional? Are all aliased commands considered to 'be in
$PATH' and therefore subject to tab-completion? Is there a way to make
bash check that the 'actual' command (which starts the alias) is
really in $PATH?
Repeat-By:
alias aliacom='nonexistent arg arg arg'
aliac<TAB> # completes to aliacom
bash: nonexistent: command not found
Thank you
Jan Stary
- aliased command always considered in $PATH (-> is tab-completed),
Jan . Stary <=