open-questions.txt

page 167 ex2
foo :: a -> a -> a
foo x y = (x y) | (y x) //are those the two cases?

page 167 ex3.
bar :: a -> b -> b
bar a b = b ?? identity function with extra argument to fill?

is (Int a, Char b) => a -> [b]
the same as Int -> [Char]
Asnwer: the former is called constrained polimorphic value, the second one is concrete type type constructor

so have a look below, even with the type
signatures commented,  the type of multiBy10
turns out to be Integer -> Integer,
why is this when the (*) operator is Num type
{-multiBy10 :: Int -> Int-}
multiBy10 = compose multiby5 multiby2
  where multiby5 = (*) 5 {-:: Int -> Int-}
        multiby2 = (*) 2 {-:: Int -> Int-}
compose :: (a2 -> a3) -> (a1 -> a2) -> (a1 -> a3)
compose f1 f2 = \x -> f1 (f2 x)

let's go again over exercise 6 before chapter 7.12

i gave up on "Fixing divideBy", what's the cleanest way to update the new Data tapa to support negate?

I am not sure what is the correct way to implement splitBy on page 332, check my solution at chapter-9 take-drop.hs

WTF Chapter 10.7 code snippet
  So the second example:
  ```
  foldl (flip const) 0 (reverse (repeat 1 ++ [1,2,3]))
  ```
  I think the example fails to demonstrate how foldl will go wrong as explained. Because if you try to reverse an endless list which in this case is (repeat 1) then no matter if you use foldl or foldr it will hang for sure.