Re: Unpacking tree node parameters

2006-09-13 Thread Audrey Tang 


在 Sep 4, 2006 2:11 AM 時，Gaal Yahas 寫到：

Unless I'm mistaken, this doesn't cast back to subroutine signature  
land

very well:

  sub f1 (Dog ($fido, $spot))   { ... }
  sub f2 (Dog $ ($fido, $spot)) { ... }
  sub f3 (Dog :($fido, $spot))  { ... }


Correct.


Unless Audrey's latest S03 patch pertains here as well, and Dog
distributes; in that case it would be as if the programmer had written

  sub f1 ([Dog $fido, Dog $spot])   { ... }


Yes, except there's no [] there; it's two positional parameters.

as described in the preceding section, Unpacking array  
parameters. If

this is *not* the case, then f1 is not ambiguous!


TimToady++ just confirmed on IRC that it is indeed the case.

Please clarify, so I can clean up the confusing paragraph and  
introduce

the optionality of the colon less jarringly.


Woot! :-)

Audrey

Unpacking tree node parameters

2006-09-03 Thread Gaal Yahas 
  sub traverse  ( BinTree $top ( $left, $right ) ) { ... }

LS06/Unpacking tree node parameters has two infelicities I wanted to
fix with a patch. The first is that this style of unpacking is really
syntactically a Signature, but most the examples in this section
abbreviate out the optional colon. (The above is more formally spelled:

   sub traverse  ( BinTree $top :( $left, $right ) ) { ... }  # or even...
   sub traverse :( BinTree $top :( $left, $right ) ) { ... }

I'm ignoring the final version here.)

The paragraph explaining this point, however, uses for its justification
an example from the variable declaration world:

  You may omit the top variable if you prefix the parentheses with a colon
  to indicate a signature.  Otherwise you must at least put the sigil of
  the variable, or we can't correctly differentiate:

 my Dog ($fido, $spot)   := twodogs();  # list of two dogs
 my Dog $ ($fido, $spot) := twodogs();  # one twodog object
 my Dog :($fido, $spot)  := twodogs();  # one twodog object

Unless I'm mistaken, this doesn't cast back to subroutine signature land
very well:

  sub f1 (Dog ($fido, $spot))   { ... }
  sub f2 (Dog $ ($fido, $spot)) { ... }
  sub f3 (Dog :($fido, $spot))  { ... }

  f1(twodogs());  # really a list of two dogs?
  f2(twodogs());  # one twodog object
  f3(twodogs());  # one twodog object

Unless Audrey's latest S03 patch pertains here as well, and Dog
distributes; in that case it would be as if the programmer had written

  sub f1 ([Dog $fido, Dog $spot])   { ... }

as described in the preceding section, Unpacking array parameters. If
this is *not* the case, then f1 is not ambiguous!

Please clarify, so I can clean up the confusing paragraph and introduce
the optionality of the colon less jarringly.

-- 
Gaal Yahas [EMAIL PROTECTED]
http://gaal.livejournal.com/