This is what the documentation says: https://docs.perl6.org/syntax/WHAT You can override it, but we'll pay no attention anyway, basically. So you can't achieve it otherwise, I guess.
El mar., 12 jun. 2018 a las 21:14, JJ Merelo (<jjmer...@gmail.com>) escribió: > > > El mar., 12 jun. 2018 a las 21:11, Brandon Allbery (<allber...@gmail.com>) > escribió: > >> I should clarify this, but I'm not recalling full details at the moment >> which is why I didn't originally. >> >> Perl uses a metaobject protocol (MOP, which you'll see in various places >> in the docs). The "macro" to access the metaobject is the .HOW >> pseudo-method. If you do this for a normal class or object of that class, >> you get Perl6::Metamodel::ClassHOW back. This is what the .^method syntax >> is accessing; it's short for (thing).HOW.method((thing), ...). The >> metaclass doesn't magically know its children, so the object has to be used >> once to get at its metaclass and a second time to tell the metaclass what >> it is to introspect. >> >> I'm not seeing documentation for what .WHAT actually does; it (correctly) >> notes that it's implemented specially within the compiler (hence "macro") >> but not how you achieve it otherwise. Then again, .HOW has the same issue; >> there's a bit of a bootstrapping issue with getting at the metamodel, you >> need to have it first. Which is why it's wired into the compiler and gets >> those uppercase pseudo-method names. >> > > All the metamodel is not exactly part of the language; it's part of the > compiler. So it's in the gray NOT-SPECCED zone regarding documentation of > "Perl 6" the language, as oposed to "Perl 6, the implementation by Rakudo". > But it's a gray zone and sometimes you fall short of documenting things > like WHAT. I'll see what we can in that area. > > JJ > > -- JJ
