Question

I understand if I had the statement c = a AND b, if a was false then the compiler wouldnt bother evaluating b, it would know the result because a is already false.

However, what if I had a function which was recursive and ANDing together the recursive calls.

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

if a function return from any point in the above recursive function-calling tree returned false, would the recursive function calls terminate and the value false would just be evaluated at the original calling point?

In other words, would it use lazy evaluation above?

Answer 1

Yes. In fact, one implementation of and is recursive (and doesn’t have any strictness added):

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

To show that this works, you can pass an infinite list of Falses to and, and see that it returns

Prelude> and (repeat False)
False

Note, however, this does not work with an infinite list of Trues, because it will forever look for a False, but never find one.