On 6/9/16 6:06 AM, Dan Douglas wrote:
> $ bash -c 'typeset -n a=b b; b=a[1]; a=foo; typeset -p a b' # bash 4.3
> declare -a a='([1]="foo")'
> declare -n b="a[1]"
> $ ./bash -c 'typeset -n a=b b; b=a[1]; typeset -p a b; a=foo' # 4.4
> declare -n a="b"
> declare -n b="a[1]"
> ./bash: line 0: `a[1]': not a valid identifier
>
> (That's the confusing error mentioned previously btw)
There are a few ways we can go on this. Using your script
$ cat x1
typeset -n a=b b
b=a[1]
typeset -p a b
a=foo
typeset -p a b
we can
1. Honor the nameref attribute and reject the assignment, as the current
devel git branch does:
$ ../bash-20160603/bash ./x1
declare -n a="b"
declare -n b="a[1]"
./x1: line 4: `a[1]': not a valid identifier
declare -n a="b"
declare -n b="a[1]"
2. Honor the assignment and remove the nameref attribute, treating the
variable as a scalar:
$ ./bash ./x1
declare -n a="b"
declare -n b="a[1]"
./x1: line 4: warning: a: removing nameref attribute
declare -a a=([0]="b" [1]="foo")
declare -n b="a[1]"
3. Honor the assignment and delete the nameref variable, creating a new
one, like bash-4.3:
$ ../bash-4.3-patched/bash ./x1
declare -n a="b"
declare -n b="a[1]"
declare -a a='([1]="foo")'
declare -n b="a[1]"
4. Honor the assignment and remove the nameref attribute and value, but
with a warning:
$ ./bash ./x1
declare -n a="b"
declare -n b="a[1]"
./x1: line 4: warning: a: removing nameref attribute
declare -a a='([1]="foo")'
declare -n b="a[1]"
5. The bizarre ksh93 behavior is always an option:
$ ksh93 ./x1
typeset -n a=.deleted
typeset -n -a b=('b[1]')
./x1: line 4: .deleted: is read only
--
``The lyf so short, the craft so long to lerne.'' - Chaucer
``Ars longa, vita brevis'' - Hippocrates
Chet Ramey, ITS, CWRU [email protected] http://cnswww.cns.cwru.edu/~chet/
