[Haskell-beginners] Pointfree style
Bas van Gijzel
nenekotan at gmail.com
Tue Nov 11 14:49:21 EST 2008
Wow, thanks for the thorough answer :). I already understood quite a bit
about composition and eta reduction, but your response made it clear for me
that this example couldn't be solved that easily.
Thanks for the information, it really helps.
On Tue, Nov 11, 2008 at 7:51 PM, Brent Yorgey <byorgey at seas.upenn.edu>wrote:
> On Tue, Nov 11, 2008 at 12:40:19PM +0100, Bas van Gijzel wrote:
> > Hi,
> > I'm trying to understand pointfree style better, but it's not coming
> > as well as I'd like it to.
> > The thing I can't get to work is to reduce an argument that is used more
> > than once in a function.
> > My function looks like this now (which works like it should):
> > f x = g ((h . i) x) x
> > But I'd like to reduce the last argument x. I've looked at the wiki
> but I
> > couldn't find a systematic way to obtain pointfree functions when they
> > more complicated.
> > Any pointers to pages or papers with more examples of obtaining pointfree
> > functions are appreciated.
> If you're doing this just to learn, great. If you're doing this
> because you think a pointfree style is somehow 'better', you should
> know that there are limits. =) In the case of your function f above
> (and, in general, with any function that uses its argument more than
> once) I would leave it as it is. The really important things to know
> are composition, i.e.
> f x = g (h (x)) becomes f x = (g . h) x
> and eta-reduction, i.e.
> f x = foo x becomes f = foo.
> There are other things that can be nice, such as flip, and various
> Arrow combinators (such as (&&&), (***)) for the (->) instance of
> Arrow  for use with functions involving tuples. Going much beyond that
> is often just obfuscation, IMO.
> But to answer your question, a systematic way to transform functions
> which use their argument more than once into pointfree versions is to
> use the ((->) e) (aka reader) monad :
> f x = g (h x) x becomes f x = (h >>= g) x
> Essentially, in the ((->) e) monad, (>>=) is a combinator to do
> exactly what you are asking about -- compose two functions with a
> duplicated input. Of course, if you dig into the implementation of
> (>>=) for the ((->) e) monad, you will eventually find a function
> which is not point-free -- this is unavoidable at some level; you
> can't actually duplicate an arbitrary thing without giving it a name.
> Applying this to your example:
> f x = g ((h . i) x) x
> f x = ((h . i) >>= g) x
> f = (h . i) >>= g
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