2009/7/19 Romain Tartière <rom...@blogreen.org>: > Hi Glen, > > On Sun, Jul 19, 2009 at 04:32:28PM -0400, Glen Barber wrote: >> > % sh foo.sh >> > % zsh foo.sh >> > % bash foo.sh >> What happens if you replace '#!/bin/sh' with '#!/usr/local/bin/zsh' ? > > This is not related to my problem since I am not running the script > using ./foo.sh but directly using the proper shell. sh just behaves > differently, that looks odd so I would like to know if it is a bug in sh > or if there is no specification for this and the behaviour depends of > the implementation of each shell, in which case I have to tweak the > script I am porting to avoid this construct (passing $? as an argument > for example). > > Romain >
My understanding was this: If you specify 'sh foo.sh' at the shell, the script will be run in a /bin/sh shell, _unless_ you override the shell _in_ the script. Ie, 'sh foo.sh' containing '#!/bin/sh' being redundant, but 'zsh foo.sh' containing '#!/bin/sh' would execute using zsh. -- Glen Barber _______________________________________________ freebsd-stable@freebsd.org mailing list http://lists.freebsd.org/mailman/listinfo/freebsd-stable To unsubscribe, send any mail to "freebsd-stable-unsubscr...@freebsd.org"
