What is enum2 doing in all of this - it appears to be ignored.
2009/6/18 Jake McArthur
Jake McArthur wrote:
Generally, you can transform anything of the form:
baz x1 = a =<< b =<< ... =<< z x1
into:
baz = a <=< b <=< ... <=< z
I was just looking through the source for the recently announced Hyena library and decided to give a more concrete example from a real-world project. Consider this function from the project's Data.Enumerator module[1]:
compose enum1 enum2 f initSeed = enum1 f1 (Right initSeed) >>= k where f1 (Right seed) bs = ... k (Right seed) = ...
First, I would flip the `(>>=)` into a `(=<<)` (and I will ignore the `where` portion of the function from now on):
compose enum1 enum2 f initSeed = k =<< enum1 f1 (Right initSeed)
Next, transform the `(=<<)` into a `(<=<)`:
compose enum1 enum2 f initSeed = k <=< enum1 f1 $ Right initSeed
We can "move" the `($)` to the right by using `(.)`:
compose enum1 enum2 f initSeed = k <=< enum1 f1 . Right $ initSeed
Finally, we can drop the `initSeed` from both sides:
compose enum1 enum2 f = k <=< enum1 f1 . Right
I didn't test that my transformation preserved the semantics of the function or even that the type is still the same, but even if it's wrong it should give you the idea.
- Jake
[1] http://github.com/tibbe/hyena/blob/9655e9e6473af1e069d22d3ee75537ad3b88a732/... _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe