handling of variables

handling of variables

[Michael Prokop]

Hello,

I'm just wondering about:

% zsh -f
% FOO=BAR /bin/echo "$FOO"

%

Why doesn't this work? Am I running out of coffee? :)

Of course the following works:

% ( FOO=BAR ; echo "$FOO" )
BAR
% FOO=BAR && /bin/echo "$FOO"
BAR

And I'm wondering what's happening at:

% zsh -f
% FOO=BAR BAR=FOO echo $FOO $BAR

% FOO=BAR BAR=FOO echo $FOO $BAR
BAR
%

Can anyone please explain me what is happening here?

thx && regards,
-mika-
-- 
 ,'"`.         http://www.michael-prokop.at/
(  grml.org -» Linux for texttool-users and sysadmins
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Re: handling of variables

[Stephane Chazelas]

On Tue, Feb 01, 2005 at 12:57:37PM +0100, Michael Prokop wrote:
> Hello,
> 
> I'm just wondering about:
> 
> % zsh -f
> % FOO=BAR /bin/echo "$FOO"
> 
> %
> 
> Why doesn't this work? Am I running out of coffee? :)
[...]

Because that calls echo with FOO=BAR in its environment, it's
not echo that expands "$FOO", it's the shell.

You can do:

FOO=BAR eval '/bin/echo "$FOO"'

Here, it's eval that expands "$FOO", so it makes sense to put
FOO=BAR in its /environment/.

(note that with POSIX shells (and zsh in sh emulation), it's
FOO=BAR command eval '/bin/echo "$FOO"'
otherwise $FOO retains its value even after the eval).

-- 
Stéphane

Re: handling of variables

[Peter Stephenson]

Michael Prokop wrote:
> Hello,
> 
> I'm just wondering about:
> 
> % zsh -f
> % FOO=BAR /bin/echo "$FOO"
> 
> %
> 
> Why doesn't this work? Am I running out of coffee? :)

The sequence of interepretation is

line gets parsed to <environment-assignment> <command+args>
line gets expanded to /bin/echo "" (assuming FOO is currently empty)
shell forks
FOO=BAR is put in the environment
shell execs /bin/echo ""

Compare this with the result of

FOO=BAR printenv FOO

where the command itself examines the environment.

> And I'm wondering what's happening at:
> 
> % zsh -f
> % FOO=BAR BAR=FOO echo $FOO $BAR
> 
> % FOO=BAR BAR=FOO echo $FOO $BAR
> BAR
> %

This looks suspiciously like a bug.  The value of BAR is being restored
after the echo, but the value of FOO isn't.

-- 
Peter Stephenson <pws@csr.com>                  Software Engineer
CSR PLC, Churchill House, Cambridge Business Park, Cowley Road
Cambridge, CB4 0WZ, UK                          Tel: +44 (0)1223 692070


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Re: handling of variables

[Bart Schaefer]

On Feb 1,  2:10pm, Peter Stephenson wrote:
}
} > % zsh -f
} > % FOO=BAR BAR=FOO echo $FOO $BAR
} > 
} > % FOO=BAR BAR=FOO echo $FOO $BAR
} > BAR
} > %
} 
} This looks suspiciously like a bug.

I *think* (not yet sure) this is happening because during interpretation
of the assignments,

	BAR=FOO echo $FOO $BAR

examines "echo" and decides it looks like a simple builtin command and
therefore should have variables restored afterwards, but

	FOO=BAR BAR=FOO echo $FOO $BAR

examines "BAR=FOO" and fails to reach the same conclusion.

Re: handling of variables

[Bart Schaefer]

On Feb 1,  3:24pm, Bart Schaefer wrote:
} Subject: Re: handling of variables
}
} I *think* (not yet sure) this is happening because [...]

Nope, that's not it.

I now think that exec.c:save_params() needs to copy the value returned
by ecrawstr() before storing it in the restore_p linked list, but I'm
not sure what kind of memory to copy it into or what ramifications that
has for restore_params().

Re: handling of variables

[Michael Prokop]

* Peter Stephenson <pws@csr.com> [20050201 15:51]:
> Michael Prokop wrote:

> > I'm just wondering about:

> > % zsh -f
> > % FOO=BAR /bin/echo "$FOO"

> > %

> > Why doesn't this work? Am I running out of coffee? :)

> The sequence of interepretation is

> line gets parsed to <environment-assignment> <command+args>
> line gets expanded to /bin/echo "" (assuming FOO is currently empty)
> shell forks
> FOO=BAR is put in the environment
> shell execs /bin/echo ""

> Compare this with the result of

> FOO=BAR printenv FOO

> where the command itself examines the environment.

Thanks for explanation, Peter.

> > And I'm wondering what's happening at:

> > % zsh -f
> > % FOO=BAR BAR=FOO echo $FOO $BAR

> > % FOO=BAR BAR=FOO echo $FOO $BAR
> > BAR
> > %

> This looks suspiciously like a bug.  The value of BAR is being restored
> after the echo, but the value of FOO isn't.

Ok. ;-)

Thanks also to Stephane for his explanation!
thx && regards,
-mika-