[Haskell-beginners] Sequence function

[Jimbo]

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 <jimbo4350 at gmail.com> 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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