[Haskell-beginners] Example needed

Dear all, I wonder why haskell do not take foldl' as default foldl if foldl is not used in production often. I was told that because foldl is lazy evaluated. I know what the advantage of lazy evaluation: do not evaluate the expression which is not needed to, and can work on infinite list. Can you give me a example about how foldl can work on an infinite list? Thanks. Best regards, Zhi-Qiang Lei zhiqiang.lei@gmail.com