使“A || B && C” 在 Haskell 中无点

Question

I'm tackling a simple leap year exercise in Haskell, and I'd like to make my solution point-free. Starting from this:

isLeapYear :: Integer -> Bool
isLeapYear year = divisibleBy 400 year || divisibleBy 4 year && not (divisibleBy 100 year)
  where
    divisibleBy m y = (== 0) $ flip mod m y

I tried using liftA3 , with a function doing (x || (y && z)) following this , but the tests do not finish and I don't know why.

So, then, I have 3 questions:

• Do you have any pointers as to how could I achieve this?
• In my first solution, what's preventing divisibleBy to be point-free? (Typechecker complains if I remove the arguments)
• As I mentioned before, I tried something like liftA3 (\\xyz -> x || (y && z)) (divisibleBy 400) (divisibleBy 4) (indivisibleBy 100) , but the tests hang. Why does that happen? I'm not getting how liftA3 works.

Thanks a lot for your help.

Answer 1

In my first solution, what's preventing divisibleBy to be point-free? (Typechecker complains if I remove the arguments)

You might think that these are equivalent (I am writing flipmod as one function for simplicity's sake):

divisibleBy  m y = (== 0) $ flipmod m y
divisibleBy'     = (== 0) . flipmod

But in actuality, divisibleBy' is now an (invalid) function that takes an argument x and then compares flipmod x to zero:

    ((==0) . flipmod) 5
→   (==0) (flipmod 5)
→   flipmod 5 == 0

Comparing a function ( flipmod 5 ) and a number is certainly no good.

You'd need to write something more sophisticated, namely:

divisibleBy = ((== 0) .) . flipmod

So that now, properly:

    divisibleBy 5 6
→   (((== 0) .) (flipmod 5)) 6
→   ((== 0) . flipmod 5) 6
→   (== 0) (flipmod 5 6)
→   flipmod 5 6 == 0

This construction (f.).g can also be written as ((.).(.)) fg , and that operator is sometimes called dot . I don't think writing things like this is a very good idea, but it might answer your question.

---

The tests hang. Why does that happen?

I don't know. You'll probably need to provide an mcve here, because this works fine for me as a full program:

import Control.Applicative

isLeapYear :: Integer -> Bool
isLeapYear = liftA3 (\x y z -> x || (y && z))
                    (divisibleBy 400)
                    (divisibleBy 4)
                    (not . divisibleBy 100)
  where
    divisibleBy m y = (== 0) $ flip mod m y

main = print (filter isLeapYear [1850..1950])