Sven Panne writes:
David Menendez wrote:
[...] If that turned out to be a performance bottleneck, you could factor out pair and write f directly: [...]
.... or directly complain to your compiler supplier if the compiler in question does not do this simple transformation for you. :-)
Sure. It's a longer-term strategy, though. :-)
Let's e.g. have a look at the code generated by GHC for the "inefficient" version:
Neat! Are you getting this from -ddump-simpl?
--------------------------------------------------------------------- helper :: (Fitness, a) -> (Fitness, a) -> (Fitness, a) helper = \ds eta -> case ds of (f, a) -> case eta of (g, b) -> (g `plusInteger` f, b)
rw :: Population a -> Population a rw = \ds -> case ds of Population xs -> Population (case xs of (x:xs1) -> scanl helper x xs1 [] -> []) ---------------------------------------------------------------------- <snip> What has happened? `pair', `id', and `scanl1' were completely inlined and instead of the overloaded (+), an Integer-specific addition was chosen (`plusInteger', it's not the real name, just something for presentation purposes). Although these are all only O(1) optimizations (nothing "really clever"), one can hardly do better by hand... So keep the program in a form which is most comprehensible, even if this seems to imply some "superfluous" things at first. This enables you to have more time for more interesting things which could really have some effect on performance, like clever algorithms etc.
I completely agree.
Actually, I might have suggested using (***) or first from Control.Arrow
rather than defining pair, since they effectively do the same thing. But
looking at the Core output I see that:
rw2 :: Population a -> Population a
rw2 (Population xs) = Population (scanl1 f xs)
where
f (n,a) = (+n) *** id
yields (if I'm interpreting it right):
helper2 = \ds eta ->
case ds of
(n,a) -> ( case eta of
(x,y) -> x `plusInt` y
, case eta of
(x,y) -> y
)
rw2 = \ds ->
case ds of
Population xs ->
Population (case xs of
(x:xs1) -> scanl helper2 x xs1
[] -> []
)
I don't know what to make of that. Semantically, helper2 is identical to
helper, but I'm not brave enough to look at the C output to see if they
produce different results.
--
David Menendez