freebsd stable - Jul 2009 - Shell execution ( [was] Re: Value of $? lost in the beginning of a function.)

Possibly off-topic...


2009/7/19 Glen Barber <glen.j.barber@gmail.com>:
> 2009/7/19 Romain Tarti?re <romain@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

In message <4ad871310907191717g1ed90be7y92250f2addc38d43@mail.gmail.com>,
Glen
Barber writes:
> Possibly off-topic...
> 
> 
> 2009/7/19 Glen Barber <glen.j.barber@gmail.com>:
> > 2009/7/19 Romain Tarti=E8re <romain@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. =A0sh 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' shel> l."
> 
> Can someone enlighten me if I am wrong about this?
> 
> --=20
> Glen Barber
> _______________________________________________
> freebsd-stable@freebsd.org mailing list
> http://lists.freebsd.org/mailman/listinfo/freebsd-stable
> To unsubscribe, send any mail to
"freebsd-stable-unsubscribe@freebsd.org"
"#!" is used to define the interpretor when the file is exec'd.

perl, AFAIK, is the only interpretor that will look at what is after
the "#!" and modify it's behaviour.  All other a interpretors
(shells)
treat "#!" as a comment.

Some shells used to examine the executable about to be called and
looked for "#!" and invoke the correct interpretor.  This was how
"#!" was supported before kernels has support for "#!".  It
was all
done in userland.

Mark
-- 
Mark Andrews, ISC
1 Seymour St., Dundas Valley, NSW 2117, Australia
PHONE: +61 2 9871 4742                 INTERNET: marka@isc.org

Glen Barber <glen.j.barber@gmail.com> writes:
> Possibly off-topic...
>
>
> 2009/7/19 Glen Barber <glen.j.barber@gmail.com>:
>> 2009/7/19 Romain Tarti?re <romain@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.

On Mon, Jul 20, 2009 at 3:12 PM, Doug Barton<dougb@freebsd.org>
wrote:
> Glen Barber wrote:
>> Possibly off-topic...
>
> It's very off topic for -stable. If you want to follow up on this,
> please do so in -hackers.
>
>> 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?
>
> It's trivial to create an example to test it yourself, don't ask
when
> you can do. :)
>
> #!/bin/foo
> echo yes
>
> $ ./foo
> bash: ./foo: /bin/foo: bad interpreter: No such file or directory
>
> $ sh foo
> yes
>
> $ bash foo
> yes
>
>
> Now it's your turn. :)
>
Hi, Doug.

The test I was going to run was:
---------------------
#!/bin/sh
echo $SHELL
---------------------

which I would execute from $PATH and from 'sh/csh/zsh foo.sh', but I
haven't had much time to play around with it.  I plan on having time
later today after ${JOB}.


-- 
Glen Barber

Glen Barber wrote:
> Possibly off-topic...
It's very off topic for -stable. If you want to follow up on this,
please do so in -hackers.
> 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?
It's trivial to create an example to test it yourself, don't ask when
you can do. :)

#!/bin/foo
echo yes

$ ./foo
bash: ./foo: /bin/foo: bad interpreter: No such file or directory

$ sh foo
yes

$ bash foo
yes


Now it's your turn. :)

Doug

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