Let's see, I am quite new to it, so this is a check to see if I understood the things correctly ;) Joel Reymont wrote:
Folks,
I understand that arbitrary defines the possible values.
How do I generally come up with co-arbitrary, though? you need to change a generator depending on the value of your type.
variant changes the seed of the generator adding an integer to it, so if you can map your object to mostly different integers (like the hash function) you can modify the generators using something like coarbitrary a = variant (hash a) then variant 1 . coarbitrary a is simply modify a little (variant 1) the modification that coarbitrary a would do. as to why you need this, it is to generate functions that have as starting domain your type. Fawzi
Would someone kindly explain the choice of co-arbitrary in the following cases, specially the very last bit with "variant 1 . coarbitrary a"?
instance Arbitrary Char where arbitrary = elements ([' ', '\n', '\0'] ++ ['a'..'h']) coarbitrary c = variant (fromEnum c `rem` 4)
instance Arbitrary Ordering where arbitrary = elements [LT, EQ, GT] coarbitrary LT = variant 0 coarbitrary EQ = variant 1 coarbitrary GT = variant 2
instance Arbitrary a => Arbitrary (Maybe a) where arbitrary = frequency [ (1, return Nothing) , (3, liftM Just arbitrary) ] coarbitrary Nothing = variant 0 coarbitrary (Just a) = variant 1 . coarbitrary a
Thanks, Joel
_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe