How is foldl lazy?

Question

There are lots of good questions and answers about foldl, foldr, and foldl\' in Haskell.

So now I know that:
1) foldl<

Answer 1

You know, that by definition:

foldl op start (x1:x2:...:xN:[]) = ((start `op` x1) `op` x2) ...

the line in foldl that does this is:

foldl op a (x:xs) = foldl op (a `op` x) xs

You are right that this is tail recursive, but observe that the expression

(a `op` x)

is lazy and at the end of the list, a huge expression will have been build that is then reduced. The difference to foldl' is just that the expression above is forced to evaluate in every recursion, so at the end you have a value in weak head normal form.