Re: [Haskell-beginners] Sequence function

Monadic bind has this signature (Monad m => m a -> (a -> m b) -> m b. Note that m is the same in both arguments and also the return value. So when you see p >>= \_ -> q ..., that means both p and q must be (m Something), where the m is the same. So when you go (Just 1 >>= \x -> return [] >>= \y -> return (x:y)) you know that return [] and return (x:y) are both using the Maybe Monad instance because in Just 1, the m is Maybe. So return [] is then equivalent to Just [], and return (x:y) is equivalent to Just (x:y). I hope that made sense. On Tue, Sep 26, 2017 at 2:10 PM, Jimbo wrote:

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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