Re: [Haskell-beginners] Sequence function

Thank you very much. Final question, in the line: return (1 : []) -- Just [1] Does the value ([1] in this case) get wrapped in Just because of the type signature of sequence? I.e sequence :: Monad m => [m a] -> m [a] On 26/09/2017 1:49 PM, David McBride wrote:

Remember that foldr has flipped operator order from foldl.

...

:t foldl foldl :: Foldable t => (b -> a -> b) -> b -> t a -> b :t foldr foldr :: Foldable t => (a -> b -> b) -> b -> t a -> b

That means that you should expand them in the opposite order from how it seems to read.

p >>= \x -> -- Just 1 >>= \ 1 q >>= \y -> -- return [] >>= \ [] return (1 : []) -- Just [1]

On Tue, Sep 26, 2017 at 12:59 PM, Jimbo wrote:

...

Hello everyone,

Just trying to understand the sequence function as follows:

sequence [Just 1]

-- evaluates to Just [1]

sequence = foldr mcons (return []) where mcons p q = p >>= \x -> q >>= \y -> return (x:y)

-- I'm trying to walk through the code as follows, I understand what is below isn't -- haskell code

p >>= \x -> [] q >>= \y -> Just 1 return (x:y) -- [] : Just 1

Am I thinking of sequence correctly here?

Best regards,

Jim

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