Fwd: [Haskell-cafe] General function to count list elements?

[Alberto G. Corona]

here goes a test of  pointer equality (valid for functions too):

samepointer :: a -> a -> IO Bool
samepointer f1 f2 = do
      p1<- varHash f1
      p2<- varHash f2
      return $  p1 == p2
  where
   varHash x=  do
      sn <- makeStableName x
      return  $ hashStableName sn


2009/4/18  <nowgate at yahoo.com>:
> Hi John,
>
> Yes, I see now what the hullabaloo was about.
>
> I've only been hacking Haskell for a few weeks, so I'm still seeing the
> "forest" rather than the "trees". But I'm coming along.
>
> Thanks.
>
> Michael
>
> --- On Sat, 4/18/09, John Dorsey <haskell at colquitt.org> wrote:
>
> From: John Dorsey <haskell at colquitt.org>
> Subject: Re: [Haskell-cafe] General function to count list elements?
> To: "michael rice" <nowgate at yahoo.com>
> Cc: haskell-cafe at haskell.org
> Date: Saturday, April 18, 2009, 2:36 PM
>
> Michael,
>
>> I had a count function that worked fine for an enum type, but thought
>> why not go polymorphic with it. So I changed it and got the error I
>> originally inquired about.
>
> For variety, I'll go a slightly different direction.
>
> If you generalize count to use any predicate, instead of always
> equality...
>
>     gcount :: (a -> a -> Bool) -> a -> [a] -> Int
>     gcount pred x0 xs = length (filter (pred x0) xs)
>
>     count = gcount (==)
>
> This will work with any type that you can write a predicate for with the
> type (a -> a -> Bool).  I can even use this with functions, if I'm
> careful.
>
>     ghci> gcount (\f g -> True)       (*2) [id,(const 1),(*3)]
>     3
>     ghci> gcount (\f g -> f 1 == g 1) (^2) [id,(const 1),(*3)]
>     2
>
> By the way, do you see why everyone's bothing you about comparing
> functions?  The type you gave count, which didn't have an Eq constraint,
> was an assertion that you could compare two values of *any* type.  If
> there's a type that's not comparable, then count's type was wrong.
> Functions are the canonical example of an incomparable type.
>
> When you're bored some time, read a bit about the Curry-Howard
> correspondence.  It's interesting, even if (like me) you don't grok all
> of its implications.
>
> Regards,
> John
>
>
>
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