Re: [perl #22369] $1 dynamic scoping breaks with recursion

"frederik@ugcs.caltech.edu (via RT)" <perlbug-followup@perl.org> wrote:
> 
> sub r
> { 
>   shift =~ /(.*)/;
>   if(shift) {
>     r("bar", 0);
>     print "$1\n";
>   }
> }
> r("foo", 1);
> 
> # prints: bar
> # (expected: foo)

As I *know* how the digit variables are implemented, I actually expect
"bar" there, but you're right, most people probably expect "foo".

Basically regexp vars ($1 and alii) are associated to regexps. There's
one set of such variables per regexp. In addition, there's some magic
to change the "current" regexp -- the one in scope -- at scope exit.
There's only one regexp in your code snippet I quoted, thus only one $1
variable slot, so it's normal that you get only one value out of $1.

This could be better documented.

• [perl #22369] $1 dynamic scoping breaks with recursion by perlbug-followup
• Re: [perl #22369] $1 dynamic scoping breaks with recursion by Rafael Garcia-Suarez