Assign to outer parameter from function called by zargs

Assign to outer parameter from function called by zargs

[Eric Nielsen]

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Hi. I'm trying to assign to a parameter set by an outer function from a
function called by zargs.

Accessing instead of assigning works:

access_outer() {
  print ${outer}${1}
}
() {
  autoload -Uz zargs
  local outer=a
  zargs -n 1 -P 0 -- b c -- access_outer
}

results in:
ab
ac
as expected.

But assigning does not work as I was expecting:

assign_outer() {
  outer=${outer}${1}
}
() {
  autoload -Uz zargs
  local outer=a
  zargs -n 1 -P 0 -- b c -- assign_outer
  print ${outer}
}

results in:
a
and I was expecting to get:
abc
(or acb, I don't care about the order since I'm using -P 0 and I expect
paralellism could interfere with the order)

What am I missing?

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Re: Assign to outer parameter from function called by zargs

[Roman Perepelitsa]

On Tue, Mar 28, 2023 at 6:25 PM Eric Nielsen <ericbn@hey.com> wrote:
>
> assign_outer() {
>   outer=${outer}${1}
> }
> () {
>   autoload -Uz zargs
>   local outer=a
>   zargs -n 1 -P 0 -- b c -- assign_outer
>   print ${outer}
> }
>
> results in:
> a
> and I was expecting to get:
> abc

assign_outer runs in a subshell, so it cannot modify the parameters of
the parent shell. It's similar to this:

    outer=a
    ( outer+=b ) &
    ( outer+=c ) &
    wait
    print $outer  # still "a"

Roman.

Re: Assign to outer parameter from function called by zargs

[Bart Schaefer]

On Tue, Mar 28, 2023 at 10:50 AM Roman Perepelitsa
<roman.perepelitsa@gmail.com> wrote:
>
> assign_outer runs in a subshell, so it cannot modify the parameters of
> the parent shell.

More specifically, there's this comment in zargs:

# Everything has to be in a subshell so that we don't "wait" for any
# unrelated jobs of the parent shell.

The "wait" is necessary to set the return status of zargs in (as
nearly as possible) the same way that xargs would do, so you can't run
current-shell actions with zargs.

What's the context of your question and your desired end result?

Re: Assign to outer parameter from function called by zargs

[Eric Nielsen]

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>> assign_outer runs in a subshell, so it cannot modify the parameters
of
>> the parent shell.

> More specifically, there's this comment in zargs:
>
> # Everything has to be in a subshell so that we don't "wait" for any
> # unrelated jobs of the parent shell.
>
> The "wait" is necessary to set the return status of zargs in (as
> nearly as possible) the same way that xargs would do, so you can't run
> current-shell actions with zargs.

Thanks Roman and Bart for the explanation. Makes perfect sense.

> What's the context of your question and your desired end result?

My use case might be too specific: I wanted to somehow capture the end
result of the processing of all the calls made by zargs. I didn't want
to rely on the zargs return code when zargs is used with -P as it was
missing values before Zsh 5.9 and it's intermittently failing in Zsh 5.9
and macOS. (https://www.zsh.org/mla/workers/2022/msg00611.html)
So I though having the called functions assign a status code in an outer
parameter could work. It was not a good idea...  :- )


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