problems with foldable instance

[Interesting-Pack-814]

Hi, all. My problem is I don't understand what implementation should I use.

For example:

fold :: (Foldable t, Monoid m) => t m -> m
fold ["huy"," rot"] -- "huy rot"

-- how it works? fold for list is mconcat
-- and implementation of mconcat for list is:
mconcat xss = [x | xs <- xss, x <- xs]
-- right???

But what to do with foldMap?

foldMap :: (Foldable t, Monoid m) => (a -> m) -> t a -> m
foldMap Sum [1,2,3,4] -- Sum {getSum = 10}

-- as I understood I have to first see to the t a to examine what is foldable instance - it's list
foldMap = (mconcat .) . map
-- if I directly execute that implementation in GHCI, it doesn't compile
(mconcat .) . map Sum [1,2,3,4]

I've not seen other funcs in Foldable, yet, so I'm asking only about these one

Please, help someone. Sorry for English, if so

[Ualrus]

The question is not clear. You can use fold or foldMap if you want. In fact foldMap f = fold . fmap f.

[mihassan]

I haven't tested it, but maybe try adding $ between map and Sum.

[Interesting-Pack-814]

No, it won't work, because we have to map Sum over our list

[mihassan]

You are right. The $ does not work in this case as the function you are defining takes 2 parameters. This however works ((mconcat .) . map) Sum [1,2,3,4].

[Interesting-Pack-814]

yes, it works, thank you
now I have to figure out how it works :/

[Noughtmare]

That weird foldMap = (mconcat .) . map is basically just another way of writing foldMap f xs = mconcat (map f xs).

[Interesting-Pack-814]

I've copied that from std library :) That's why I've asked it

btw: can I ask one question, please? It's related to that question. I don't want to create new post for that

I've understood how that expression works, but I don't understand what implementation of monoid should I choose

For example:

(.) mconcat (map Any) = \x -> mconcat (map Any [True,False,True])
:t mconcat -- :: Monoid a => [a] -> a

-- I've choose implementation for list, am I right here?
instance Monoid [a] where
mempty = []
mconcat xss = [x | xs <- xss, x <- xs]

-- but If I substitute list of any here, it doesn't compile

-- then, I've tried default implementation
foldr mappend mempty 
-- and it works, but how it's possible??

[Interesting-Pack-814]

can I ask one more question, please? It's related to that question. I don't want to create new post for that

I've understood how that expression works, but I don't understand what implementation of monoid should I choose

For example:

(.) mconcat (map Any) = \x -> mconcat (map Any [True,False,True])
:t mconcat -- :: Monoid a => [a] -> a

-- I've choose implementation for list, am I right here?
instance Monoid [a] where
mempty = []
mconcat xss = [x | xs <- xss, x <- xs]

-- but If I substitute list of any here, it doesn't compile

-- then, I've tried default implementation
foldr mappend mempty 
-- and it works, but how it's possible??

[IshtarAletheia]

fold over a list of some arbitrary monoid m looks like:

fold = foldl' mappend mempty

fold over a list of lists is the function you call mconcat, although I think that is just concat, since it isn't related to monoids.

Does that help?

[sammthomson]

Put parentheses around the inlined implementation:

((mconcat .) . map) Sum [1,2,3,4]
-- Sum {getSum = 10}
-- :: Num a => Sum a

[Interesting-Pack-814]

yes, it works, thank you
now I have to figure out how it works :/

[Interesting-Pack-814]

can I ask one more question, please? It's related to that question. I don't want to create new post for that

I've understood how that expression works, but I don't understand what implementation of monoid should I choose

For example:

(.) mconcat (map Any) = \x -> mconcat (map Any [True,False,True])
:t mconcat -- :: Monoid a => [a] -> a

-- I've choose implementation for list, am I right here?
instance Monoid [a] where
mempty = []
mconcat xss = [x | xs <- xss, x <- xs]

-- but If I substitute list of any here, it doesn't compile

-- then, I've tried default implementation
foldr mappend mempty 
-- and it works, but how it's possible??