OK, I see.

Well, if I rewrite your ladd based on the transform I proposed, I get: 

sub ladd($)
{
   my $hidden_add = $add;
   $add = $_[ 0 ];
   return sub
   {
      my $hidden_num = $num;
      $num = $_[ 0 ];
      my $hidden_sum = $sum 
      $sum = $num + $add;
      print "$num + $add = $sum\n";
      ($num,$sum) = ($hidden_num,$hidden_sum);
   };
   $add = $hidden_add;
}

[download]
All the variables have global lexical scope except the hidden ones, which are never directly used. I don't see local providing any "binding" at all.

In reply to local doesn't bind, in my opinion. by BlaisePascal
in thread YAlQ: Yet Another local() Question. by BlaisePascal

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