2015-01-20 22:30 GMT+01:00 tisimst <tisimst.lilyp...@gmail.com>:
> Is it possible to derive the correct default print function for a grob?
>
> For example, if I want to get the stencil of a hairpin, I know the function
> is called "ly:hairpin::print", but for accidentals, it's
> "ly:accidental-interface:print".
>
> I'd like to be able to get the right one automatically (in scheme) without
> needing to specify them all and use conditionals to figure out the right
> one. Any suggestions? I'm not sure if this is even possible, and I can deal
> with it if it's not, but I'd rather not if I can help it.
>
> Thanks,
> Abraham
Hi Abraham,
it's not that hard ;)
Though, one Problem might be a previous applied stencil-override...
The code below will return the name of the default and the actual
print-procedure.
\version "2.19.15"
#(define new-stil
(lambda (grob)
(grob-interpret-markup grob "xy")))
{
\override NoteHead.stencil = #new-stil
\override NoteHead.after-line-breaking =
#(lambda (grob)
(display "\nactual-stencil:\t\t")
(display
(procedure-name
(assoc-get 'stencil
(ly:grob-basic-properties grob))))
(display "\ndefault-stencil:\t")
(display
(procedure-name
(assoc-get 'stencil
(reverse (ly:grob-basic-properties grob)))))
)
c''1
}
HTH,
Harm
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