Re: [Haskell-cafe] why does a foldr not have a space leak effect?

On 12-07-25 09:06 AM, Mathieu Boespflug wrote:

"Albert Y. C. Lai" writes:

...

foldr (+) and foldl (+) for Int have the same asymptotic costs, both time and space. See my http://www.vex.net/~trebla/haskell/lazy.xhtml

Therefore, I do not understand why they are labeled opposite space-leakness.

That may be true, but if the function argument supplied to foldr is lazy in its second argument, then the story can be quite different.

In my http://www.vex.net/~trebla/haskell/lazy.xhtml which I mentioned yesterday, I have an example for that, too. foldr (||).