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

[Matthew Fong]

Hello Aaron,

This makes sense, and the let statement is something I was missing (!)

mattfong

On Fri, Dec 11, 2020 at 12:54 PM Aaron Hill <lilypond@hillvisions.com> wrote:

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