X-Git-Url: https://git.donarmstrong.com/?a=blobdiff_plain;f=lily%2Fduration.cc;h=0b80b1ce1f33dba7ed7ec97910edfdd33913e5b7;hb=6a18123d7fcd2f4ee57835c5ea70835ceabb5c39;hp=413ac889ec497e7447b5e443c8e783b3704513dd;hpb=31568c504806f35aac420a394c9eab07abd9faa7;p=lilypond.git diff --git a/lily/duration.cc b/lily/duration.cc index 413ac889ec..0b80b1ce1f 100644 --- a/lily/duration.cc +++ b/lily/duration.cc @@ -3,7 +3,7 @@ source file of the LilyPond music typesetter - (c) 1997--2006 Jan Nieuwenhuizen + (c) 1997--2009 Jan Nieuwenhuizen Han-Wen Nienhuys */ @@ -34,6 +34,56 @@ Duration::Duration (int log, int d) factor_ = Rational (1, 1); } +Duration::Duration (Rational r, bool scale) +{ + factor_ = Rational (1, 1); + + if (r.num () == 0.0) + { + durlog_ = 0; + dots_ = 0; + } + else + { + /* we want to find the integer k for which 2q/p > 2^k >= q/p. + It's simple to check that k' = \floor \log q - \floor \log p + satisfies the left inequality and is within a factor of 2 of + satistying the right one. Therefore either k = k' or k = k'+1 */ + + int p = r.num (); + int q = r.den (); + int k = intlog2 (q) - intlog2 (p); + if (shift_left(p, k) < q) + k++; + + assert (shift_left(p, k) >= q && shift_left(p, (k-1)) < q); + + /* If we were to write out log (p/q) in base 2, then the position of the + first non-zero bit (ie. k in our notation) would be the durlog + and the number of consecutive 1s after that bit would be the number of + dots */ + p = shift_left(p, k) - q; + dots_ = 0; + while ((p *= 2) >= q) + { + p -= q; + dots_++; + } + + /* we only go up to 64th notes */ + if (k > 6) + { + durlog_ = 6; + dots_ = 0; + } + else + durlog_ = k; + + if (scale || k > 6) + factor_ = r / get_length (); + } +} + Duration Duration::compressed (Rational m) const { @@ -91,7 +141,7 @@ Duration::print_smob (SCM s, SCM port, scm_print_state *) Duration *r = (Duration *) SCM_CELL_WORD_1 (s); scm_puts ("#to_string ().c_str ()), port); + scm_display (ly_string2scm (r->to_string ()), port); scm_puts (" >", port); return 1;