Let me try again.

In your root node you show how leaving a scope appears to restore the old values of the match variables and explain that this actually happens by virtue of the outer scope's opnodes pointing to a different set of match variables from the inner scope's.

But this doesn't explain how an inner scope manages to have the same value for match values as the outer scope when they haven't been overwritten. For example: 

for(  qw( foobar foobaz )  ) {
    /(oo)/;
    {
        /(ar)/;
        print "($1) ";
    }
    print "[$1]\n";
}

[download]
outputs: 
(ar) [oo]
(oo) [oo]

[download]
and your root node explains why "[oo]" is output both times but doesn't explain how "(oo)" could be output.

That is, how did the inner scope manage to get $1 a value of "oo" when that was set in the outer scope's match variable? I recall hearing that this is done by copying from one set of match variables to the other.

                - tye

In reply to Re^3: Zen and the Art of Match Variables (copy?) by tye
in thread Zen and the Art of Match Variables by Elian

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