Glen Barber <glen.j.bar...@gmail.com> writes: > Possibly off-topic... > > > 2009/7/19 Glen Barber <glen.j.bar...@gmail.com>: >> 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. >> >> > > I meant to say in the last line: "'#!/bin/sh' would override the 'zsh' shell." > > Can someone enlighten me if I am wrong about this?
The person to whom you were responding had it closer. The shell specified in the "#!" first line is only consulted if you run it as "./foo.sh". Otherwise, it's input to the shell that you started, and the line is only a comment. _______________________________________________ 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"
