Re: [Lilypond-auto] Issue 4290 in lilypond: Patch: Make Grob::name accessible to Scheme

[lilypond]

Comment #4 on issue 4290 by address@hidden: Patch: Make Grob::name  
accessible to Scheme
https://code.google.com/p/lilypond/issues/detail?id=4290

> I had a look in my files where I used a definition for 'grob-name' (more  
> than 30 >times I defined it) and it's use-cases.

Goodness!

> Mostly I compare with an other grob-name.
> Comparing via eq? is faster than equal? or string=?

> In 'all-grob-descriptions' the grob-name is a symbol.

> This would point to a symbol as output.

Yes, this is the way I'm leaning too.

> Otoh and just trying to be thoroughly, you could let it as is and write  
> another >definiton in lily-library.ly to convert it.

Hmm, I think what I will end up doing is rewriting the function to return a  
symbol.  Grob::name takes a symbol, converts it into a string usable by  
C++, then
ly:grob-name converts this string into a Scheme type.  I'd hate to write  
another function which converts that Scheme string back into the symbol we  
started out with!  It makes a lot of sense to simply return the symbol, and  
if someone really wants a string--and, come to think of it, I don't believe  
I ever have--use symbol->string.

> A similiar case is 'ly:stencil-in-color' taking single list-entries for a  
> rgb->color and 'stencil-with-color' from stencil.scm taking a predefined  
> color.
> (Ok, not exactly the same case, but two function aiming the same, taking  
> >different kind of arguments)

This is really confusing.  ly:stencil-in-color takes r g b values as  
separate arguments. meaning that you can't use the color variables, like  
red and green.  However, it actually converts these arguments to a list  
when it makes the stencil.

stencil-with-color takes the direct route and accepts the color as a list.

Maybe there should just be ly:stencil-in-color which takes a list?
Nevertheless, all in all I'd vote for outputting a symbol.

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