Shell execution ( [was] Re: Value of $? lost in the beginning of a function.)

[Lowell Gilbert]

Glen Barber <glen.j.barber at gmail.com> writes:

> Possibly off-topic...
>
>
> 2009/7/19 Glen Barber <glen.j.barber at gmail.com>:
>> 2009/7/19 Romain Tartière <romain at 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.