/bin/sh's builtin 'rm' busted: 'rm -f' without arguments return error
|Category:||cmd - userland programs|
- How to reproduce:
Check that rm is actually a built-in:
$ /bin/sh 'type rm'
rm is a shell builtin version of /usr/xpg4/bin/rm
Look it misbehave:
$ /bin/sh 'rm -f'; echo st = $?
Usage: rm [-cFdfirRuv] file ...
st = 2
- Good news:
The issue is not shared by the 'rm' executables on the system:
$ /bin/rm -f; echo st = $?
st = 0
$ /usr/xpg4/bin/rm -f; echo st = $?
st = 0
- Version details:
$ cat /etc/release
OpenIndiana Development oi_151.1.7 X86 (powered by illumos)
Copyright 2011 Oracle and/or its affiliates. All rights reserved.
Use is subject to license terms.
Assembled 03 October 2012
- Why high priority:
It's worth noting that future version of Automake will start assuming
that "rm -f" with further arguments works OK; so, packages
bootstrapped with those future Automakes will not be able to build
properly when the Illumos shell is in use, unless this bug is fixed.
#1 Updated by Stefano Lattarini about 5 years ago
- Future versions of the POSIX standard will mandate that "rm -f" without
further arguments work correctly, that is, as a no-op:
- More information about the planned Automake change to have it assume
"rm -f" without further arguments work:
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