[Haskell-beginners] foldl on Bool:s

[goforgit .]

Interesting, I did not know that, thank you!

On Sat, Sep 26, 2015 at 12:16 AM, Rein Henrichs <rein.henrichs at gmail.com>
wrote:

> Note that you would like the elem function to stop at the matching element
> and return True rather than checking the rest of the list needlessy, which
> can be done with foldr but not with foldl. The actual implementation of
> elem does this, so you can say:
>
> > elem 1 [1..]
> True
>
> which would fail to terminate with the foldl version.
>
> On Fri, Sep 25, 2015 at 11:10 AM goforgit . <teztingit at gmail.com> wrote:
>
>> Thanks I got it now :)
>>
>> On Thu, Sep 24, 2015 at 9:45 PM, Kostiantyn Rybnikov <k-bx at k-bx.com>
>> wrote:
>>
>>> Hi.
>>>
>>> Your function gets passed numbers one by one in the place of x, and its
>>> previous result in the place of acc, and it returns a Bool. Initial value
>>> in place of acc parameter ("previous result") is put as False (since you
>>> begin with answer "no" to question "is it elem?").
>>>
>>> Hope this helps.
>>> 24 вер. 2015 19:04 "goforgit ." <teztingit at gmail.com> пише:
>>>
>>>> Reading http://learnyouahaskell.com/higher-order-functions
>>>>
>>>> I understand that with the function
>>>>
>>>> sum' :: (Num a) => [a] -> a
>>>> sum' = foldl (+) 0
>>>>
>>>> the call
>>>>
>>>> ghci>>> sum' [1,2,3]
>>>>
>>>> will be evaluated as
>>>>
>>>> 0 + 1 + 2 + 3 = 6
>>>>
>>>> But what about the function
>>>>
>>>> elem' :: (Eq a) => a -> [a] -> Bool
>>>> elem' y ys = foldl (\acc x -> if x == y then True else acc) False ys
>>>>
>>>> and calling it with
>>>>
>>>> ghci>>> elem' 3 [1,2,3]
>>>>
>>>> How is that evaluated to True by foldl in elem'?
>>>>
>>>> Thanks in advance for any explanation to this!
>>>>
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