This looks good. I found a way to break it, but only with perverse input.
http://codereview.appspot.com/5495076/diff/1004/lily/beaming-pattern.cc File lily/beaming-pattern.cc (right): http://codereview.appspot.com/5495076/diff/1004/lily/beaming-pattern.cc#newcode174 lily/beaming-pattern.cc:174: int tuplet_count = (int) factor.den (); The name 'tuplet_count' is used differently later in this file. This usage, 3 for triplets, seems to fit the name better. http://codereview.appspot.com/5495076/diff/1004/lily/beaming-pattern.cc#newcode210 lily/beaming-pattern.cc:210: int tuplet_count = factor.num (); The other use of the name: how many notes would fill the time if they were not in a tuplet http://codereview.appspot.com/5495076/diff/1004/lily/beaming-pattern.cc#newcode273 lily/beaming-pattern.cc:273: Moment ratio = ((tuplet_count == 1) ? dt : tuplet_dt) / options.base_moment_ / tuplet_moment; I would think you want tuplet_count = 3 here for { \times 1/3 { b16 b b } b8 b8 \times 1/3 { b16 b b } } but you get the other definition with tuplet_count = 1 and the beamlets get pointed wrongly. http://codereview.appspot.com/5495076/ _______________________________________________ lilypond-devel mailing list [email protected] https://lists.gnu.org/mailman/listinfo/lilypond-devel
