[Haskell-cafe] foldl vs foldl'

[Jonathan Cast]

On Wed, 2008-11-05 at 10:01 -0800, Daryoush Mehrtash wrote:
> Lets assume we don't have undefined in the list, are there functions
> (or properties in the function) that would cause foldl to have
> different results than foldl'?

If the function is partial on some elements of the list.

(3 /), for example, if the list contains 0.

If f is total over the elements of the list (whether the elements of the
list are partial or total) and f z /= _|_, then foldl' f z = foldl f z.

jcc