Isaac _______________________________________________ Haskell-prime mailing list Haskell-prime@haskell.org http://www.haskell.org/mailman/listinfo/haskell-prime
In class Integral, divMod has a default in terms of quotRem.
(quot,rem,div,mod all have defaults as the both-function they're part
of.) I'm sure divMod is more natural than quotRem to implement for some
types... so why doesn't quotRem have a default in terms of divMod? it
has no default! Then the "minimal to implement" will change from
(toInteger and quotRem) to (toInteger and (quotRem or divMod)).
- default for quotRem in terms of divMod? Isaac Dupree
- default for quotRem in terms of divMod? Isaac Dupree