zsh: Variável é desativada sem motivo

Isso foi um bug mesmo. Você fez a coisa certa ao denunciá-lo.

Foi então corrigido por este commit: https://sourceforge.net/p/zsh/code/ci/d946f22a4cd2eed0f3a67881cfa57c805703929c/ que será incluído na próxima versão.

E aqui está a explicação do zshmantenedor do :

On Wed, 2019-08-14 at 10:37 +0100, Stephane Chazelas wrote:
> 2019-08-08 20:38:05 +0430, Aryn Starr:
> Now, that being said, as discussed on U&L it looks like a bug
> indeed and a shorter reproducer is:
> 
> $ zsh -xc 'v=1; f() { local v; v=2 true; }; f; typeset -p v'
> +zsh:1> v=1
> +zsh:1> f
> +f:0> local v
> +f:0> v=2 +f:0> true
> +zsh:1> typeset -p v
> zsh:typeset:1: no such variable: v
> 
> Most likely, that's the "v=2 true" (where "true" is a builtin) that ends up
> unsetting the "v" from the global scope.

Yes, the saved version of "v" that we restore after the builtin is
missing the pointer back to the version of v in the enclosing scope.  So
it's not only not shown as set, it will leak memory.

This simply preserves that pointer in the copy, but this assumes we've
correctly blocked off the old parameter from being altered inside the
function scope --- if we haven't that preserved old pointer is going to
get us into trouble.  However, if we haven't that's already a bug, so
this shouldn't make things worse.

pws

[patch ignorado]