Seg, 2008-06-09 às 17:51 -0700, Larry Wall escreveu: > On Sat, Jun 07, 2008 at 09:49:03PM +0100, Daniel Ruoso wrote: > : 2) Assume the capture-translation and define that > : $foo.HOW.can($foo,'bar') keeps the $how as the invocant and must receive > : the referring object as first argument. > I prefer this approach, I think.
I took the liberty to write a patch to S12 with this new context. The extended format $foo.HOW.methods($foo) becomes a little awkward, specially for Class based OO. Maybe there should be a more extended modification to promote the .^methods syntax, since $foo.HOW is then more usefull for prototype-based OO, where $foo.HOW might be, for instance, Prototype::Delegation::C3 or Prototype::Concatenation. This actually brings me to another issue, which is how much do we expect for this prototype-based meta implementations to be exchangeable between Perl 6 implementations? The point is that they *must* be representation independent, so that you can use it with whatever low-level object metainstance (in the Moose jargon) you like, and for that to be possible, we need a representation API. SMOP is already pointing in that direction, take a look at: * http://www.perlfoundation.org/perl6/index.cgi?smop_oo_api * http://svn.pugscode.org/pugs/v6/smop/src-s1p/P6Meta.pm daniel __DATA__ Index: S12.pod =================================================================== --- S12.pod (revision 14546) +++ S12.pod (working copy) @@ -123,11 +123,15 @@ C<HOW> method. A "class" object is just considered an "empty" instance in Perl 6, more properly called a "prototype" object, or just "protoobject". -The actual class object is the metaclass object pointed to by the -C<HOW> syntax. So when you say "C<Dog>", you're referring to both a -package and a protoobject, that latter of which points to the -actual object representing the class via C<HOW>. The protoobject -differs from an instance object not by having a different + +In a class-based OO implementation, the actual class object is the +metaclass object pointed to by the C<HOW> syntax. So when you say +"C<Dog>", you're referring to both a package and a protoobject, that +latter of which points to the actual object representing the class via +C<HOW>. In a prototype-based OO implementation, on the other hand, the +metaclass object is probably going to be shared by several types and +the methods are most likely going to be stored in the prototypes. The +protoobject differs from an instance object not by having a different type but rather in the extent to which it is defined. Some objects may tell you that they are defined, while others may tell you that they are undefined. That's up to the object, and depends on how the @@ -1892,23 +1896,29 @@ $x === $y $obj.bless(%args) -Every class has a C<HOW> function/method that lets you get at the -class's metaobject, which lets you get at all the metadata properties -for the class (or other metaobject protocol) implementing the objects -of the class: +Every object, defined or not, has a C<HOW> function/method that lets +you get at the metaobject, which lets you get at all the metadata +properties for the type (or other metaobject protocol) implementing +the objects of the type: - MyClass.methods() # call MyClass's .methods method (error?) - MyClass.HOW.methods() # get the method list of MyClass + MyClass.methods() # call MyClass's .methods method (error?) + MyClass.HOW.methods(MyClass) # get the method list of MyClass The C<^> metasyntax is equivalent to C<.HOW>: - MyClass.HOW.methods() # get the method list of MyClass - ^MyClass.methods() # get the method list of MyClass - MyClass.^methods() # get the method list of MyClass + MyClass.HOW.methods(MyClass) # get the method list of MyClass + ^MyClass.methods(MyClass) -Each object of the class also has a C<.HOW> or C<.^> method: +When using the C<^> metasyntax in the method invocation, it also +implies a translation in the capture, passing the invocant as first +argument: - $obj.HOW.methods(); + MyClass.HOW.methods(MyClass) # get the method list of MyClass + MyClass.^methods() # equivalent shorter syntax + +Each object of the type also has a C<.HOW> or C<.^> method: + + $obj.HOW.methods($obj); $obj.^methods(); Class traits may include: @@ -1964,9 +1974,9 @@ Strictly speaking, metamethods like C<.isa()>, C<.does()>, and C<.can()> should be called through the meta object: - $obj.HOW.can("bark") - $obj.HOW.does(Dog) - $obj.HOW.isa(Mammal) + $obj.HOW.can($obj, "bark") + $obj.HOW.does($obj, Dog) + $obj.HOW.isa($obj, Mammal) or @@ -1992,7 +2002,7 @@ actually calls: - $obj.HOW.does(Dog) + $obj.HOW.does($obj, Dog) which is true if C<$obj> either "does" or "isa" C<Dog> (or "isa" something that "does" C<Dog>). If C<Dog> is a subset, any additional
