Myujth

Having fun working through a Haskell tutorial...

One problem posed is as you see in the subject line... this is one I'd really like to figure out, but I'm at a loss... I've used any like so:

ghci >any (==55) [15,25,35,45,55,65,75,85,95]
True
ghci >any (==55) [15,25,35,45,54,65,75,85,95]
False
ghci >all even [2,4,6,8]
True
ghci >all even [1,3,5,7,9]
False

and it seems like checking if a list element is divisible by 42 or not, should be fairly easy...

I mean, you would check if any of the numbers in the list are n `mod` 42 == 0, right?

But how do you state that in the expression? Or do you need to write a 'helper' function?

Composing (0==) and (`mod 42`):

f :: [Integer] -> Bool
f = any ((0==).(`mod` 42))

futher reducing parenthesis noise:

f :: [Integer] -> Bool
f = any $ (0==).(`mod` 42)

Helper function:

f :: [Integer] -> Bool
f = any div42
 where
 div42 n = n `mod` 42 == 0

Stylistically, for this function, either way seems fine. However, say you wanted to check if any values are divisible by 42 OR 52, then utilizing composition may become more obfuscated/complex/futile. Whereas using a helper function keeps things readable: div42Or52 n = n `mod` 42 == 0 || n `mod` 52 == 0. Note, I've used a where clause above, but a let in expression or a lambda are possible alternative ways to structure helper functions.

Ultimately, its up to the developer to balance concise code with understandable code.

You can define function inplace:

any (n -> n `mod` 42 == 0) [1, 2, 42]