A Haskell "Foldable" quiz

21 August, 2025 ⋅ haskell

How does GHCi respond to fold [] 42?

Answer

ghci> fold [] 42
()

How does GHCi respond to fold [] 42 == []?

Answer

ghci> fold [] 42 == []
True

What is the type of fold [] 42?

Answer

ghci> :t fold [] 42
fold [] 42 :: Monoid t => t

What the hell is going on!?

Answer

fold has type (Foldable t, Monoid a) => t a -> a, so fold [] has type Monoid a => a.

fold [] is mempty, which means fold [] 42 is equal to mempty 42. This is well-typed because functions have a Monoid instance.

mempty for functions is const mempty. So mempty 42 is const mempty 42 is mempty :: Monoid b => b.

In GHCi, b is defaulted to (), which is why the first question prints ().

When we evaluate fold [] 42 == [], the list monoid is chosen instead, reducing to [] == [].

Why am I even thinking about this?

Answer

I just found a bug due to replacing fromMaybe Set.empty (which has type Maybe (Set a) -> Set a) with fold Set.empty (which has type Monoid b => a -> b and is equivalent to const mempty). This meant that all my Justs containing non-empty sets were being ignored.

I meant to replace fromMaybe Set.empty with fold, which really are the same.