Hi, Lowell On Mon, Jul 20, 2009 at 7:02 AM, Lowell Gilbert<freebsd-stable-lo...@be-well.ilk.org> wrote: > 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. >
I suppose that makes sense, but what if the script location is in the user's $PATH? -- 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"
