How can foldMap do the same as foldr when the latter does not have anything to do with monoids?

Question

foldr and foldMap can be used to define each other as I understand. But how is that possible, as the latter uses monoids, while the former does not? Do we have any guarantees that the stuff the foldr works on can have a monoid?

You might want to read [Tom Ellis' "What is `foldr` made of?"](http://web.jaguarpaw.co.uk/~tom/blog/posts/2012-11-04-what-is-foldr-made-of.html) and [Brent Yorgey's response "`foldr` is made of monoids"](https://byorgey.wordpress.com/2012/11/05/foldr-is-made-of-monoids/). — Luis Casillas, May 31 '16 at 21:06

score 8 · Accepted Answer · edited May 29 '16 at 20:06

foldr :: (a -> b -> b) -> b -> [a] -> b

Note that a -> b -> b is a -> (b -> b). Functions b -> b form a monoid under composition.

Note how this resembles

foldMap :: (..omitted..) => (a -> m) -> f a -> m

The only difference is that foldMap doesn't use the "zero" argument of type b of fold and returns an m, which in terms of foldr would be b->b. Now just apply one to the other and you've recovered foldr from foldMap.

How can foldMap do the same as foldr when the latter does not have anything to do with monoids?

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