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

The short answer is that in foldl f, it's not necessarily the case that f is strict, so it might be too eager to reduce the thunks up front. However, in practice it usually is, so you nearly always want to be using foldl'.

I wrote a more in-depth explanation of how the evaluation order of foldl and foldl' works on another question. It's rather long but I think it should clarify things a bit for you.