Hello list, hello David,

You wrote:

> I get this error also.  It comes when I use the \arpeggio without
> specifically assigning a chord with it.  In other words:  <c e g>\arpeggio =
> no error while something like
> <<  e\arpeggio \\ c\arpeggio >> will generate the error.  The output looks
> correct with the arpeggios being connected (when you \set
> PianoStaff.connectArpeggios = ##t).  I'd call it a bug but the output works
> so I haven't worried about it.

This is according to my own experiences.
I wouldn't be able to explain it in a better way ...

Best Regards           Roland


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