Re: Is this the expected behaviour for nameref in Bash 4.4 now?

[Jack Kuan]

Thanks for replying.

I was expecting that in the second iteration of the loop, the
local var=var_123  command would make var a normal variable again with value "var_123"

and then the local -n var=$var would make var again a nameref to var_123. This seems

to be the behavior of Bash 4.3.

But anyway, it's easy to work around by introducing another variable rather than redefining the nameref var:

set -x

var_123=123
f() {
    while (( $# )); do
        shift
        local var=var_123
        local -n var_ref=$var
    done
}

f one two

On Thu, Oct 20, 2016 at 4:02 PM, Dan Douglas <ormaaj@gmail.com> wrote:

Yes that was an intentional change to require valid identifiers. I can't
say it will always be that way or that there won't at some point be a
workaround. You can stiill use `${!param}' for now to refer to positional
parameters as you always could, but as usual that isn't useful if you
want to assign by reference.