hi. The following just manifested itself while experimenting with code. I find it quite neat since it makes for quite natural looking constants: class duration_log { int8_t log; public: constexpr duration_log(int log) : log(log) {} constexpr operator int() const { return log; } template<char...> friend duration_log constexpr operator"" _th(); }; duration_log constexpr maxima = 3; duration_log constexpr longa = 2; duration_log constexpr breve = 1; duration_log constexpr whole = 0; duration_log constexpr half = -1; duration_log constexpr quarter = -2; template<> duration_log constexpr operator"" _th<'8' >() { return -3; } template<> duration_log constexpr operator"" _th<'1', '6' >() { return -4; } template<> duration_log constexpr operator"" _th<'3', '2' >() { return -5; } template<> duration_log constexpr operator"" _th<'6', '4' >() { return -6; } template<> duration_log constexpr operator"" _th<'1', '2', '8'>() { return -7; } template<> duration_log constexpr operator"" _th<'2', '5', '6'>() { return -8; } --- (there is actually no need for the wrapper class duration_log, but it makes it easier to play with overloaded functions. You could of course just replace duration_log with int, if you do not care for type safety.) Now I can write things like "16_th" or "64_th". It turns out that there is a quite neat formula for the typical music notation duration calculation. Given that the base rhythmic type is expressed as a log (as in LilyPond for instance), the rational number of the duration with a certain amount of dots and time modification (tuple) can be written like this: #include <boost/rational.hpp> template<typename Rational = boost::rational<int>> inline Rational augmented_rational( int log, unsigned dots , int numerator, int denominator ) { return log < 0 ? Rational { ((1 << (dots + 1)) - 1) * numerator , denominator << -log << dots } : Rational { (((1 << (dots + 1)) - 1) << log) * numerator , denominator << dots }; } --- This is quite efficient since there is no need to multiply or manipulate rational numbers in any way. You just get the final value straight out of the expression. Comments? -- CYa, ⡍⠁⠗⠊⠕