On 10/20/16 2:45 PM, kjk...@gmail.com wrote:
> set -x
>
> var_123=123
> f() {
> while (( $# )); do
> shift
> local var=var_123
> local -n var=$var; : status is $?
> local -p
> : var is $var
> done
> }
>
> f one two
>
>
> Running above script gives the follow output:
>
> + var_123=123
> + f one two
> + (( 2 ))
> + shift
> + local var=var_123
> + local -n var=var_123
`var' is a local nameref with value var_123
> + : status is 0
> + local -p
> var=var_123
> + : var is 123
Since var is a nameref, it expands to $var_123, which is 123
> + (( 1 ))
> + shift
> + local var=var_123
var doesn't get unset; it's still a nameref with value var_123. Setting
the local attribute on an existing local variable doesn't unset any of
the variable's attributes. You have to actually unset the variable to
do that.
> + local -n var=123
> ./x.sh: line 10: local: `123': invalid variable name for name reference
So now you try to assign `123' as the value of a nameref. That would
cause subsequent attempts to assign to var to attempt to create and
assign to a variable named `123', which is invalid. Bash-4.3 didn't
have enough error checking and would create variables with invalid names
in situations like this.
> + : status is 1
> + local -p
> var=var_123
This is probably where the confusion comes. `local -p' doesn't print other
variable attributes. If you had used `declare -p | grep var=' between the
two calls to `local' you would have seen that `var' had the nameref
attribute.
> + : var is 123
> + (( 0 ))
Chet
--
``The lyf so short, the craft so long to lerne.'' - Chaucer
``Ars longa, vita brevis'' - Hippocrates
Chet Ramey, UTech, CWRU c...@case.edu http://cnswww.cns.cwru.edu/~chet/
