[Haskell-beginners] foldl on Bool:s

[Kostiantyn Rybnikov]

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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