Implementation of maximum in haskell

Question

I was just looking up the implementation of the maximum function in haskell:

maximum :: forall a . Ord a => t a -> a
maximum = fromMaybe (errorWithoutStackTrace "maximum: empty structure") .
   getMax . foldMap (Max #. (Just :: a -> Maybe a))

Could somebody explain this piece of code to me?

In particular it is not clear to me how this part

Max #. (Just :: a -> Maybe a)

works. What does #. mean? To me it looks like something similar to function composition but typing

:t #.

in the REPL gives an error. Also I think the argument for foldMap should be of the form

Monoid m => a -> m

But I can't see where the monoid structure comes from here.

(#.) is just an optimized version of the composition operator (.) which only works in some cases, such as in Max . Just. Pretend it's plain composition, and note that Max a is a monoid with binary "max" for its operation (roughly, there's a Maybe wrapper so that we can define the unity to be Nothing). — chi
– chi, Commented Jul 27, 2016 at 18:46
@jberryman, no you don't. But you need :t (#.), not :t #.. And it's normally imported from Data.Profunctor.Unsafe, but Data.Foldable has its own specialized copy. — dfeuer
– dfeuer, Commented Jul 27, 2016 at 19:51

dfeuer · Accepted Answer · 2016-07-27 20:30:31Z

4

#. is an operator that can make coercions easier to use.

f #. g

means

g, coerced to the type of f . g

Similarly,

f .# g

means

f, coerced to the type of f . g

In most cases, f #. g only makes sense if f is some combination of newtype constructors, newtype accessors, and fmap.

The reason for using #. in maximum us probably over-zealousness on my part. It can almost certainly be replaced with . without any problem.

edited Jul 27, 2016 at 20:30

answered Jul 27, 2016 at 20:09

dfeuer

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