At 06:14 PM 8/1/2002 +0200, Jerome Vouillon wrote:
> On Wed, Jul 31, 2002 at 11:40:39AM -0600, Jonathan Sillito wrote:

>  > So here is my take on a slightly simpler example:

>  >

>  >   sub foo {

>  >     my $x = 13;

>  >     return sub { print "$x\n"; };

>  >   }

>  >

>  >   $foo()

> 

> Melvin, I think it would really help if you could explain us how you

> would compile this code.  Also, you should describe precisely what

> "invoke" and "new_pad" (and maybe the other scratchpad-related

> opcodes) do as far as scratchpads are concerned.


Here is an attempt. I'm assuming the stack of pads is a singly-linked list
of pads with children pointing to parents, and multiple children can refer
to the same parent. If they are created on the fly, the creation of
a Sub or Closure will simply hold a pointer to its parent, and pads will
be garbage collected.

Disclaimer: I'm only speaking my ideas which may not coincide with
what Dan has in mind.

# Assuming we have no symbol table or pad "descriptor"
# then to do it dynamically with the new_pad/store_lex ops...

foo:
  new_pad                   # lexical scope 1
  new P0, .PerlInt
  set P0, 13
  store_lex "$x", P0        # or store_lex 0, P0
  new P1, .PerlSub          # captures the current lexical pad tree/stack
  set P1, anon_closure      # set bytecode address
  store P1                  # return value
  ret                       # ret may close current pad which now is only
                            #   referred to by anon_closure

anon_closure:
  new_pad                   # activate new pad which inherits previous pad
                            # for any lexicals.
                            # lexical scope 2
  fetch_lex P0, "$x"        # or fetch_lex P0, 0
  print P0
  print "\n"
  ret


  new P0, .PerlSub, foo     # lexical scope 0
  invoke P0
  restore P1                # get closure sub returned from foo into P1
  invoke P1                 # closure contains a lexical stack of:   [0,1] 
and will
                                   # push lex scope 2 which in this example 
doesn't
                                   # create any new lexicals.

end