Re: tie problem

[Rainer Hahnekamp]

Hi Kieren,

thank you very much vor answer. Your snippet solved the problem and btw. it
makes completely sense. Should really be better integrated into the
documentation.

Greetings,
Rainer

Kieren MacMillan wrote:

> Hi, Rainer:
> 
>> I've following code:
>> <c d g>1\arpeggio ~ | << <c d g>1\fermata \\ {r2 r4 b'32 (c d e f g
>> a b)} >>
>>
>> According to the documentation the two "<c d g>" should be tied.
>> Unfortunately this is not the case. Could you help me please?
> 
> Well, you've run into a subtlety of Lilypond that could, perhaps, be
> better documented...  ;-)
> 
> Consider the following code snippet:
> 
> %%% BEGIN SNIPPET %%%
> 
> \version "2.9.17"
> 
> \paper
> {
> indent = 0\in
> line-width = 3\in
> }
> 
> theMusic = \relative c'
> {
> <c d g>1 ~ | << <c d g>1\fermata \\ { s2 c4 c } >>  \break
> <c d g>1 ~ | << { <c d g>1\fermata } \new Voice { s2 c4 c } >>
> }
> 
> \score
> {
> \theMusic
> }
> 
> %%% END SNIPPET %%%
> 
> Notice that the tie does not work "as expected" in the first example,
> but does in the second.
> 
> This is because, in the first example, the << \\ >> construct
> explicitly instantiates TWO voices, BOTH of which are in addition to
> the one which contains the <c d g> that starts the tie -- as a
> result, the tie doesn't know where to end, because its Voice doesn't
> continue on into the <<>> block.
> 
> In the second example, the \\ is replaced by an explicit (manual)
> instantiation using \new Voice -- this ensures that anything before
> the \new Voice command is considered part of the Voice that existed
> before the <<>> block began, and so the tie knows where to terminate.
> 
> Does that make sense?
> Or, at the very least, does it explain why you're seeing what you're
> seeing?  =)
> 
> Best regards,
> Kieren.