RE: RE: Re: YAlQ: Yet Another local() Question.

i was just emphasizing the fact that my provides deep binding and local provides shallow binding. it wasn't clear (to me) from your post that you knew this.

jeff

Update: when you implement the transformation, the closure works again. apparently the transformation you propose doesn't actually work. or is that your point? that local provides no binding at all, but transforming it into an implementation with my does provide binding. whatever the case, local's binding is called "shallow binding" in the documentation, so thats what i'll call it here.

Comment on RE: RE: Re: YAlQ: Yet Another local() Question.

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local doesn't bind, in my opinion. by BlaisePascal (Monk) on Aug 03, 2000 at 18:45 UTC
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.	[reply] [d/l] [select]

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local doesn't bind, in my opinion.
by BlaisePascal (Monk) on Aug 03, 2000 at 18:45 UTC

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;
}
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