问题
I have this lovely fixana function here that performs about 5 times faster than her sister ana. (i have a criterion report to back me on this)
ana alg = Fix . fmap (ana alg) . alg
fixana alg = fix $ \f -> Fix . fmap f . alg
Can I express their cousin cata in the same fashion?
cata f = f . fmap (cata f) . unFix
It seems to me that I cannot, but I have been humbled by my S.O. fellows quite a few times in the past.
回答1:
Actually, this has nothing to do with catamorphisms.
Whenever a function is defined as
f = (... f ...) -- some expression involving f
one can rewrite it using fix as
f = fix $ \g -> (... g ...)
In the posted code we have a slight variant
f x = (... (f x) ...)
We can regard the above as f being defined recursively. However, it's simpler if we regard f x (rather than f) being defined recursively.
f x = fix $ \g -> (... g ...)
This should be more efficient than the plain translation
f = fix $ \g x -> (... (g x) ...)
since we don't need to call g over and over again with the same argument x.
来源：https://stackoverflow.com/questions/48705438/can-one-express-catamorphism-through-data-function-fix