2015-03-15 21:48 GMT+01:00 Kevin Barry <[email protected]>:
> Hi LilyPond experts,
>
> I'm trying to make a function that will draw a curved line given only a
> destination point (with some math I'll add later), but I've hit an early
> stumbling block: I can't seem to substitute variables for values in the
> path/curveto command list. The following code produces a
> `wrong-argument-type' error, but it points to scm/stencil.scm which isn't
> very helpful for figuring out what's wrong. Help appreciated!
>
> \version "2.18.2"
>
> #(define-markup-command (draw-curved-line layout props points)
> (number-pair?)
> (let ((xpt (car points))
> (ypt (cdr points)))
> (interpret-markup layout props
> (markup #:path 1 '((curveto 0 ypt 0 ypt xpt ypt))))))
>
> \relative {
> b_\markup { \draw-curved-line #'(5 . 5) }
> }
>
>
Hi Kevin,
in cases where i have no clue what's wrong and don't understand the
error-message, I boil down the code and display all kind of data, values,
ets
In your case I'd do:
#(define-markup-command (draw-curved-line layout props points)
(number-pair?)
(let ((xpt (car points))
(ypt (cdr points)))
(display '((curveto 0 ypt 0 ypt xpt ypt)))
(newline)
(display (caddar '((curveto 0 ypt 0 ypt xpt ypt))))
(newline)
(display (symbol? (caddar '((curveto 0 ypt 0 ypt xpt ypt)))))
(interpret-markup layout props
(markup
;#:path 1 '((curveto 0 ypt 0 ypt xpt ypt))
"xy"
))
))
The problem should be clear now: the variables in '((curveto 0 ypt 0 ypt
xpt ypt)) are not evaluated but taken as symbols.
You wrote a quoted list, but need some elements of the list be unquoted.
Shortest: use a combi of ` and ,
#(define-markup-command (draw-curved-line layout props points)
(number-pair?)
(let ((xpt (car points))
(ypt (cdr points)))
(interpret-markup layout props
(markup
#:path 1 `((curveto 0 ,ypt 0 ,ypt ,xpt ,ypt))
))))
HTH,
Harm
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