On 2022/12/11 20:47, Lawrence Velázquez wrote:
This happens because "declare"/"typeset" creates local variables
within functions. Using -g works around this...
$ Export() { declare -gx "$@"; }
$ Export -r foo=1
$ declare -p foo
declare -rx foo="1"
...but now "Export" always creates global variables, rather than
scoping as "declare" and your alias-based version does. On the
other hand, "export" also creates global variables, so in a sense
the workaround version is more consistent.
$ f() { export "$@"; }
$ f var=1
$ declare -p var
declare -x var="1"
I see, but you still can't use "-r" w/export, though I think
the -r flag would get dropped in any exported shell, though
in that case, one might expect an error if one uses "typeset -xr"
along the lines of "-r flag won't be exported".
Side curious: If one uses -g, does it cause the var to be defined
in all intervening functions as well as the top(global)
and current scope?
NOTE: The original question was about allowing "-r" with export:
$ f() { export "$@"; }
$ f -r var=1
export: -r: invalid option
export: usage: export [-fn] [name[=value] ...] or export -p
Seems like it would get rid of an unnecessary error message
and maybe, an inconsistency with "typeset -xr".
Thanks for the info on using -g in the function. I haven't
used -g too much since one of my machines still used bash-v3 and
I don't think -g appeared until 4.x (don't quote me on that though).
