John Goerzen writes: > timeDiffToSecs :: TimeDiff -> Integer > timeDiffToSecs td = > (fromIntegral $ tdSec td) + > 60 * ((fromIntegral $ tdMin td) + > 60 * ((fromIntegral $ tdHour td) + > 24 * ((fromIntegral $ tdDay td) + > 30 * ((fromIntegral $ tdMonth td) + > 365 * (fromIntegral $ tdYear td)))))
I was wondering: Does this calculation account for leap years? Does it have to? As I see it, it's not possible to convert a TimeDiff accurately without knowing the point in time from which to "diff" from. How many days is "4 years from now"? It's almost certainly not 4*365 days. Uh, did I mention leap seconds? ;-) Peter _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe