Re: problem piping output of shell builtin

[Vincent Stemen <zsh@hightek.org>]

On Mon, Jan 05, 2004 at 03:36:38PM -0500, Pavol Juhas wrote:
> On Mon, Jan 05, 2004 at 07:26:15PM +0000, gj@sdf.lonestar.org wrote:
> > Hi all,
> > 
> > I'm migrating from bash to zsh. It hasn't been so bad because I'm sort of new
> > to shell programming anyways ( though I did have "fun" figuring out that zsh
> > arrays start incrementing from 1 as opposed to bash's 0 :). I thought I'd
> > share the latest hiccup...
> > 
> > Why can't I pipe the output of 'jobs' thusly?
> 
> AFAIK, all the shells run one side of the pipe in a subshell.  bash
> executes subshell for the right side of the pipe, however zsh does
> so for the left side.  Therefore the `jobs' command in 
> `jobs|read line' is evaluated in the subshell of zsh, which has no
> knowledge about processes in the parent shell - and produces no
> output.  Left side subshell is however advantageous in other
> situations, just compare
> 
>   zsh -c 'echo 10|read a; echo .$a'
>   .10
>   bash -c 'echo 10|read a; echo .$a' 
>   .
> 
> To access information in the zsh job table, you need to use the
> builtin associate arrays jobtexts, jobstates and jobdirs, for example:
> 
>   for j in ${(k)jobstates}; do
>     print -- "[$j] ${jobstates[$j]} ${jobtexts[$j]} in ${jobdirs[$j]}"
>   done
> 
> HTH,
> 
> Pavol

Under bash, at least, the semi-colon is ending the pipe command and
then executing "echo .$a" as new command in the original shell.  So
you need to group the entire right side in the above example.
ie.

$ echo 10 | (read a; echo .$a)
.10
$

That is interesting.  I did not know zsh did that by default.
However, I am not sure you are correct about zsh forking a sub-shell
for the left side of the pipe.  If so, then local shell variables from
the parent shell should not be accessible unless they are exported,
but they are.

$ x=foo 
$ echo $x | read a; echo .$a
.foo
$

Vincent

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Vincent Stemen
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