[Haskell-beginners] question about list processing

[Alex Hammel]

With the join it's the same thing as:

computeHead :: (a -> [b]) -> [a] -> [b]
computeHead f xs = take 1 xs >>= f


On Thu, 12 Nov 2015 at 06:58 Dennis Raddle <dennis.raddle at gmail.com> wrote:

> Oh I needed join.
>
> computeHead :: (a -> [b]) -> [a] -> [b]
> computeHead f = join . take 1 . fmap f
>
> That's because
>
> f :: a -> [b]
>
> not
>
> f: a -> b
>
>
> On Thu, Nov 12, 2015 at 5:22 AM, Martin Vlk <martin at vlkk.cz> wrote:
>
>> How about:
>>
>> computeHead f = take 1 . fmap f
>>
>> Martin
>>
>> Dennis Raddle:
>> > What would be an elegant way of writing this
>> >
>> > computeHead :: (a -> [b]) -> [a] -> [b]
>> >
>> > Where when [a] is null, it returns a null list, but when [a] contains
>> one
>> > or more elements, it applies the given function to the head of a and
>> > returns that? Is there some existing typeclass operator that facilitates
>> > this?
>> >
>> > You can write
>> >
>> > computeHead _ [] = []
>> > computeHead f (x:_) = f x
>> >
>> > But that first line seems suspicious to me... it makes me think about
>> how
>> > in the list Monad, an empty list falls through. But I can't quite make
>> it
>> > work.
>> >
>> >
>> >
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