If you type foldr instead of foldl it won't complain.

In most cases it will because they have different types:

    foldl :: (a -> b -> a) -> a -> [b] -> a
    foldr :: (a -> b -> b) -> b -> [a] -> b

So the only time it will be the same is if a and b are the same type.

> So the only time it will be the same is if a and b are the same type.

Which they will be in most of the example code he used. They're not the best examples.