Re: Scheme function with boolean argument.

[David Kastrup]

Pierre Perol-Schneider <address@hidden> writes:

> Hi,
>
> I'd like to understand why the following function has no effect :
>
> %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
>
> \version "2.19.3"
>
>
> myFunctionFalse = \override Score.BarNumber.break-visibility = ##(#f #f #f)
>
> myFunctionTrue = \override Score.BarNumber.break-visibility = ##(#f #t #t)
>
>
> myFunction =
>
> #(define-music-function (parser location arg) (boolean?)
>
>    #{
>
>      \override Score.BarNumber.break-visibility = ##(#f $arg $arg)
>
>    #})

The symbol $arg is likely a mistake, but the symbol arg would not be
better.  Please note that #(...) is the syntax for a _constant_ vector.
All its contents are quoted.  With #f, this does not make a difference.
But arg remains arg.

So you probably want to write ... = #(vector #f arg arg) instead.  That
way arg is not quoted.

-- 
David Kastrup