The 6th of the 99 Haskell questions on wiki.haskell.org presents a monadic way to test whether a list (something of type [a]
) is a palindrome:
isPalindromeM :: (Eq a) => [a] -> Bool
isPalindromeM = reverse >>= (==)
(here reverse :: [c] -> [c]
takes a list and outputs a list in backwards order).
The bind operation implies a monad is at play here, but what is this monad?
Since isPalindrome involves lists, my first guess is that the monad is related to the List one, but I don't see how to draw the connection.
It took me some time to find the monad, and since it is not related to List
, I figured I'd post an answer.
The expression reverse >>= (==)
implies the type of reverse
is m a
, for some monad m
, hence m a
is the type [c] -> [c]
. We also need isPalindromeM
to have type [c] -> Bool
, and the bind expression implies m b
is identical to [c] -> Bool
. Finally this context requires (==) :: [c] -> [c] -> Bool
to have the type a -> m b
.
Therefore we deduce a
is [c]
and b
is Bool
, and the monad m
takes a type a
and sends it to the function type [c] -> a
. This suggests the monad at play is Monad ((->) [c])
. defined here
If there's some moral to the story, perhaps it's "you can make a monad out of anything".
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