Jan-Peter Voigt <jp.vo...@gmx.de> writes:
> Hello Pedro,
>
> Am 04.12.18 um 04:28 schrieb Pedro Pessoa:
>> Hello!
>> I want a function that takes a string as arg an from that produces a valid
>> variable reference, as follows:
>>
>> %%% pseudo
>> Nabc={a1 d e f}
>> Nxyz={b1 e a d}
>>
>> fun=
>> #(define-music-function (x)(string?)
>> #{
>> <<
>> #(concat x "abc")
>> \\
>> #(concat x "xyz")
>> >>
>> #})
>>
>> \fun "N" %produces parallel music with Nabc and Nxyz
>> %%%
>>
>> ---
>>
>> I've ran this test:
>>
>> %%%
>> \Nabc={some music}
>> (display (string->symbol (string-append "N" "abc")))
>> %%%
>>
>> It outputs "Nabc", not the music content of Nabc, as I expected.
>> Why is that? How do I make it point to the actual music?
> The string is converted to a symbol and a symbol is a primitive datatype
> in guile-scheme. To receive the value of the variable you have to ask
> the parser. To place the result in the music you should use an instant
> scheme expression (introduced by '$' not '#').
>
> HTH:
>
> fun=
> #(define-music-function (x)(string?)
> #{
> <<
> $(ly:parser-lookup (string->symbol (string-append x "abc")))
> \\
> $(ly:parser-lookup (string->symbol (string-append x "xyz")))
> >>
> #})
In this particular case, # would have worked though $ tends to work in
more cases. However, $ also creates a _copy_ of the music while #
doesn't. If the music ends up in a \relative or \transpose or other
construct modifying its content in place, having a copy is important so
that the original variable does not get changed.
\xabc creates a copy like $xabc does, while #xabc doesn't.
--
David Kastrup
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