[Haskell-cafe] Re: How to "Show" an Operation?
uzytkownik2 at gmail.com
Thu Jun 10 16:10:08 EDT 2010
On Thu, 2010-06-10 at 19:44 +0200, Martin Drautzburg wrote:
> On Thursday, 10. June 2010 00:08:34 Luke Palmer wrote:
> > Or just:
> > apply = val_of
> > So, to summarize: if you have something that isn't a function and you
> > want to use it like a function, convert it to a function (using
> > another function :-P). That's all. No syntax magic, just say what
> > you're doing.
> Thanks Luke
> The reason I was asking is the following: suppose I have some code which uses
> some functions, and what it primarily does with those functions is CALL them
> in different orders.
> Now at a later point in time I decide I need to give names to those functions
> because at the end I need to print information about the functions which
> together solved a certain problem. Think of my problem as "In which order do
> I have to call f,g,h such that (f.g.h) 42 = 42?".
> I don't want to change all places where those functions are called
> into "apply" style. Therefore I was looking for some idiom like the python
> __call__() method, which, when present, can turn just about anything into a
> I could change the *definition* of my original functions into "apply" style
> and the rest of the code would not notice any difference. But that does not
> really help, because in the end I want to Show something like [g,h,f], but my
> functions would no longer carry names.
> Alternatively I could associate functions with names in some association
> function, but that function simply has to "know to much" for my taste.
> The thing is, I only need the names at the very end. Throughout the majority
> of the computation they should stay out of the way.
data Named a = Named String a
instance Functor Named where
f `fmap` (Named s v) = Named s (f v)
instance Applicative Named where
pure x = Named "" x
(Named s f) <*> (Named t v) = Named (s ++ "(" ++ t ++ ")") (f v)
instance Eq a => Eq (Named a) where
(Named _ x) == (Named _ y) = x == y
instance Show (Named a) where
show (Named s _) = s
namedPure :: Show a => a -> Named a
namedPure x = Named (show x) x
test :: Num a
=> (a -> a) -> (a -> a) -> (a -> a) -> [String]
test f g h = do
[f', g', h'] <- permutations [Named "f" f, Named "g" g, Named "h" h]
guard $ namedPure 42 == f' <*> g' <*> h' <*> namedPure 42
return $ show f' ++ " . " ++ show g' ++ " . " ++ show h'
(code is not tested but it should work)
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