Re: Scheme help request: How can else of an if-statement be made to do nothing?

[Aaron Hill]

On 2020-12-11 12:30 pm, Matthew Fong wrote:
> Hello Aaron,
>
> > .< Oh boy, that is *simple*. I went off the deep end on this, trying 
> > to
> make another variable that would get assigned the color. That clearly 
> is
> not the way Scheme works. The inline conditional is a thing of beauty.
>
> Looks like I need to spend more time studying Scheme syntax.

Defining a variable would make sense if you needed the value in a few 
places, since that would cut down on redundant expressions.  But even if 
you only needed the value once, it sometimes makes sense to use 
variables to help keep other expressions simpler.  The \markup below is 
arguably easier to follow without the embedded Scheme expression:

%%%%
print-if-defined =
#(define-void-function
  (foo sym text)
  ((boolean? #f) symbol? markup?)
  (let ((color (if foo '(0.8 0.2 0.2) '(0.2 0.8 0.2))))
   (if (defined? sym)
       (add-text #{ \markup \with-color #color #text #}))))

symA = "Something"

\print-if-defined symB "Text"     % hidden
\print-if-defined symA "Text"     % shown, green
\print-if-defined ##t symA "Text" % shown, red
%%%%

NOTE: LilyPond's parser is able to interpret "symA" as a symbol on its 
own without needing to escape to Scheme syntax by writing #'symA.  Just 
something to keep in mind as I think it results in cleaner usage.


-- Aaron Hill