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@gmail.com> wrote:

Thanks I got it now :)

On Thu, Sep 24, 2015 at 9:45 PM, Kostiantyn Rybnikov <k-bx@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@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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