On 8/6/18 10:13 PM, Grisha Levit wrote:
> A few more problematic test cases in this vein (tested against latest
> devel snapshot 20180803).
Thanks for the report. These are all the same issue, with the same fix.
Chet
--
``The lyf so short, the craft so long to lerne.'' - Chaucer
A few more problematic test cases in this vein (tested against latest
devel snapshot 20180803).
f() { local -n ref=var; local -A ref=([1]=); ref=([2]=); declare -p ref var; }
unset ref var; f
# declare -An ref=()
# declare -A var=([1]="" )
declare -p ref
# declare -a ref=([2]="")
unset ref;
On 7/29/18 4:04 PM, Grisha Levit wrote:
> I think there's a related regression in the latest devel commit. Creating a
> local variable with the same name as a higher-scoped nameref pointing to an
> unset variable creates a local variable with the name of the target of the
> nameref.
I think
I think there's a related regression in the latest devel commit. Creating a
local variable with the same name as a higher-scoped nameref pointing to an
unset variable creates a local variable with the name of the target of the
nameref.
Starting with:
declare -n ref=var
f() { local ref=Y;
On 7/25/18 4:27 PM, Grisha Levit wrote:
> In the latest devel this issue is fixed for the case that the local nameref
> points to a non-existent variable but there is still a bug if the variable
> pointed to by a local nameref already exists.
Thanks for the report. The original fix was too
In the latest devel this issue is fixed for the case that the local nameref
points to a non-existent variable but there is still a bug if the variable
pointed to by a local nameref already exists. In such a case, `declare'
commands after the initial `declare -n' end up modifying the value and/or
On Wed, Jul 18, 2018 at 8:23 AM Greg Wooledge wrote:
> FYI that one is already fixed in 5.0-alpha:
>
> wooledg:~$ declare -n ref=var; declare -a ref=(X); declare -p ref var
> declare -n ref="var"
> declare -a var=([0]="X")
> wooledg:~$ f() { declare -n ref=var; declare ref=(X); declare -p ref
On Tue, Jul 17, 2018 at 05:47:20PM -0400, Grisha Levit wrote:
> At global scope this works as expected:
>
> $ declare -n ref=var; declare -a ref=(X); declare -p ref var
> declare -n ref="var"
> declare -a var=([0]="X")
>
> But in a function, we end up with the nameref variable having both the
>
At global scope this works as expected:
$ declare -n ref=var; declare -a ref=(X); declare -p ref var
declare -n ref="var"
declare -a var=([0]="X")
But in a function, we end up with the nameref variable having both the
`a' and `n' attributes and nothing in the target:
$ f() { declare -n ref=var;
