Re: Derive the correct print function of a grob

[Thomas Morley]

2015-01-20 22:30 GMT+01:00 tisimst <address@hidden>:
> 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