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? -- 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"
