Am So., 7. Okt. 2018 um 22:28 Uhr schrieb David Kastrup <d...@gnu.org>:
>
> Thomas Morley <thomasmorle...@gmail.com> writes:
>
> > Am So., 7. Okt. 2018 um 18:00 Uhr schrieb David Kastrup <d...@gnu.org>:
> >>
> >> Thomas Morley <thomasmorle...@gmail.com> writes:
> >>
> >> > If you have too much time, you may have a look at ‘scheme-engraver.ly’
> >> > from
> >> > http://lilypond.org/doc/v2.13/input/regression/collated-files.html The
> >> > newer file uses the great 'make-engraver'-macro (by David Kastrup).
> >> > Otoh, this macro hides an engraver _is_ a list (which is obvious, if
> >> > you look in the older file).
> >>
> >> An engraver is not a list, just like it isn't a string like
> >> "Note_engraver". That's just how you _specify_ an engraver. If it's a
> >> list, the Scheme_engraver type will be used for actually instantiating
> >> an engraver (which is bound to a particular context), initialized using
> >> the list contents.
> >
> > Hi David,
> >
> > I don't understand the subtleties.
> > Could you elaborate a bit using the examples below.
> >
> > testI =
> > #(list
> > (cons 'initialize
> > (lambda (trans)
> > (write trans))))
> >
> > testII =
> > #(lambda (ctx) testI)
> >
> > \layout {
> > \context {
> > \Voice
> > \consists #testI
> > \consists #testII
> > }
> > }
> >
> > { R1 }
> >
> > testI is a list, testII a procedure using testI.
> > Both return #<Translator Scheme_engraver >, though.
>
> They don't "return" the Translator but rather cause it to be created
> when specified after \consists , like a Notehead_engraver instance is
> created when specifying the string "Notehead_engraver".
>
> Basically there is the _specification_, \consists looks up (or in your
> example creates) a suitable translator creator given a specification.
> This translator creator is called with a suitable context to actually
> create a translator instance when a context is instantiated, and that
> creator instance is what you see listed rather unspecifically as
> #<Translator Scheme_engraver > or #<Engraver Notehead_engraver> or
> whatever else (to get to see those, create a suitable Scheme engraver
> and look at the "source-engraver" argument of one of your grob
> acknowledgers).
>
> One of the more problematic bits in history was making a
> \consists/\removes pair work for ad-hoc defined engravers (rather than
> those registered under a name) since at the time of the \removes call,
> the translator creator did not carry information about its originating
> expression (particularly so if its list expression was the result of
> calling a lambda function) and creating a fresh translator creator from
> the same expression led to a different translator creator unsuitable for
> looking up the preexisting translator creator.
>
> --
> David Kastrup
Hi David,
thanks a lot for your explanations.
I try to repeat/summarize in my own words:
Defining some code like #(define my-code ...) will not create a
translator, before \consists has done its work in some context.
But with (for example)
\layout {
\context {
\Voice
\consists #my-code
}
}
an unnamed and unregistered translator is created. Assumed 'my-code'
is suitable, ofcourse.
Below some musing.
Don't read unless you're very interested ;)
So the two codings below will not create any engraver.
testI =
#(list
(cons 'initialize
(lambda (engraver)
(newline)
(pretty-print (list 'from-initialize: (cons 'engraver engraver)))))
(cons 'acknowledgers
(list
(cons 'note-head-interface
(lambda (engraver grob source-engraver)
(pretty-print
(list
'from-acknowledgers:
(cons 'engraver engraver)
(cons 'source-engraver source-engraver))))))))
testII =
#(lambda (ctx) testI)
But as soon as I do:
\layout {
\context {
\Voice
\consists #testI
\consists #testII
}
}
Two translators are created.
Which can be checked with some code you once tailored for me
http://lilypond.1069038.n5.nabble.com/catching-layout-content-tc213339.html
Ofcourse adjusted to fit here, thanks again for it.
#(define all-Voice-translators
(assoc
'Voice
(map
(lambda (x)
(if (ly:context-def? (cdr x))
(cons (car x) (ly:context-def-lookup (cdr x) 'consists))
x))
(sort
(ly:module->alist (ly:output-def-scope $defaultlayout))
(lambda (p q) (symbol<? (car p) (car q)))))))
#(newline)
#(format #t "\nAll Voice-translator:\n~y"
(cdr all-Voice-translators))
Though:
#(pretty-print (ly:get-all-translators))
doesn't show them.
Is this expected?
%%%%%
%%%%%
Things are different for registered translators:
#(ly:register-translator
testI 'testI
'((grobs-created . ())
(events-accepted . ())
(properties-read . ())
(properties-written . ())
(description . "")))
#(ly:register-translator
testII 'testII
'((grobs-created . ())
(events-accepted . ())
(properties-read . ())
(properties-written . ())
(description . "")))
Now (ly:get-all-translators) sees them as:
testII:
#<procedure #f (ctx)>
testI:
((initialize . #<procedure #f (engraver)>)
(acknowledgers
(note-head-interface
.
#<procedure #f (engraver grob source-engraver)>)))
Their names can be found with
#(format #t "\nAll translator-names:\n~y"
(sort
(map ly:translator-name (ly:get-all-translators))
symbol<?))
So _registering_ them makes testI/II translators, \consists may look them up.
Many thanks,
Harm
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