[Haskell-beginners] No accumulation of partially applied functions allowed?
jays at panix.com
Tue Jun 26 23:19:28 CEST 2012
On Tue, 26 Jun 2012, Brent Yorgey <byorgey at seas.upenn.edu> wrote:
> On Tue, Jun 26, 2012 at 10:08:49PM +0200, Obscaenvs wrote:
>> Sorry if this is a less than stellar question.
>> The problem:
>> Given a function f :: a -> a -> a -> b, make it work on a list
>> instead: f `applyTo`[x,y,z] where [x,y,z] :: [a].
>> My stab at a general solution was
>> applyTo f  = error "no arg"
>> applyTo f (x:xs) = go (f x) xs
>> go acc  = acc
>> go acc (y:) = acc y
>> go acc (y:ys) = go (acc $ y) ys
>> I thought this would work, functions being "first class citizens" but
>> ghci complains:
>> "Occurs check: cannot construct the infinite type: t1 = t0 -> t1
>> In the return type of a call of `acc'
>> Probable cause: `acc' is applied to too many arguments
>> In the expression: acc y
>> In an equation for `go': go acc (y : ) = acc y"
>> The 'probable cause' isn't the real cause here, but something to do
>> with the fact that it's impossible to accumulate functions in this
>> Or am I just too tired too make it work? I can see that the type of
>> `go` could be a problem, but is it insurmountable?
> The type of `go` is exactly the problem. In particular, the type of
> acc's first parameter. In the third clause of go's definition, we can
> see that `acc` and (acc $ y) are both used as the first argument to
> go, hence they must have the same type. However, this is impossible
> -- if acc has type (t0 -> t1), then y must have type t0, and (acc $ y)
> has type t1, so it would have to be the case that t1 = t0 -> t1 --
> hence the error message.
> It is not possible in Haskell to define `applyTo`.* I know this
> function gets used a lot in lisp/scheme, but Haskell style is
> different. If you explain the context in which you wanted this
> function, perhaps we can help you figure out a better way to structure
> things so it is not needed.
> * At least not without crazy type class hackery.
What is the difficulty?
Is the difficulty at the level of "syntax"?
Or is it that the type "Haskell expression", perhaps "Haskell
form", to use an old and often confusing Lisp term, does not
exist in the Haskell System of Expression? Here "exist" should be
read as "exist at the right level", right level for attaining
These alternatives, I think, need not be disjoint.
I am ignorant of Haskell, but sometimes I write Perl in Lisp, and
the blurb for my last public rant mentioned a specific lambda
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