Re: consistency probs var & function re-use

[L A Walsh]

Chet Ramey wrote:
>  You are misunderstanding what that is supposed to do, or ignoring it.
>  `declare -p' quotes its output in a way that allows it to be reused as
>  shell input. Executing the output of `declare -p' will recreate the
>  variable with an identical value.

Re-use as shell input?  That's a bit vague.  As the right-hand
expression for an assignment? or something one reads in?
There are plenty of examples that could qualify as being "shell input"
that don't work:


exec of output:

   > x=$'foo\nbar'
   > exec $(declare -p x)
   bash: exec: declare: not found


Assign & exec w/ quotes:

   > a=$(declare -p x)
   > exec "$a"
   bash: exec: declare -- x="foo
   bar": not found


Assign&exec w/o quotes:

   > exec $a
   bash: exec: declare: not found


Assign & try direct interpretation of value:

   > $a
   bash: declare: `bar"': not a valid identifier

Same w/quoteS:

   > "$a"
   bash: $'declare -- x="foo\nbar"': command not found


Reading output:

   > declare -p x|while read def; do
   > echo "def=$def"
   > done
   def=declare -- x="foo
   def=bar"

Reading it via proc-subst:

   > read a< <(declare -p x)
   Ishtar:law> echo "$a"
   declare -- x="foo

(I'm running out of ideas)...

It's not very easy to discern what you mean.  I'm sure
it is easy for someone who already knows the answer --
trivial even, but what type of bash-input are you referring to?