Viscosity in the tongue

Question

I understand that if I had the operator c = a AND b, if a was false, then the compiler would not evaluate b, it would know the result, because a is already false.

However, if I had a function that was recursive, and ANDing was recursive calls.

So myfunc input1 input2 = and[myfunc(input1),myfunc(input2)]

if a function returns from any point in the above recursive function tree returned false, will the calls to the recursive function end and the value false will simply be evaluated at the origin of the call?

In other words, will he use the lazy score above?

Answer 1

Yes. In fact, one implementation andis recursive (and has no added rigor):

and :: [Bool] -> Bool
and []     =  True
and (x:xs) =  x && and xs

, , False and ,

Prelude> and (repeat False)
False

, True s, False, .

Answer 2

, , Haskell .

&& : :

(&&) :: Bool -> Bool -> Bool
True  && y = y
False && _ = False

Haskell , &&, , .

, and, :

and :: [Bool] -> Bool
and []     = True
and (b:bs) = b && and bs

, Haskell , b bs : && b, False, bs .

, - , Haskell : , , , .

Answer 3

Haskell, , , , , , true, .

, , my_func - , , my_func.

, , . , my_func , . , , , . , .

, .

Answer 4

, false, , false ?

, . , exists OCaml true, p true :

let rec exists p = function
  | [] -> false
  | x::xs -> p x || exists p xs

, p true .

, ?

, , .