[clean-list] New Clean Programmer

[John M. Gogolin]

Hi folks,

As is going to become painfully evident shortly, I'm rather new to the
language of Clean.

I've been trying to run through the exercises at the ends of the
chapters in The Clean Book with mixed results.  The one I'm currently
working on is the second from chapter 3, which states:

    Define the function MakeFrequenceTable
    MakeFrequenceTable [a] -> [(a, Int)] | == a

    That makes a frequence table. The table consists of a list of
tuples. Each tuple consists
    of an element from the original list and its frequency (percentage).
E.g.
    Frequentietabel [1,2,3,2] = [(1,25),(2,50),(3,25)]

Where I run into to trouble is the "| == a" clause.  I can define the
function as:

    MakeFrequenceTable :: [a] -> [(a, Int)] | Eq a
    MakeFrequenceTable [] = []
    MakeFrequenceTable [x:xs] = take n (removeDup([(x, Frequence x
[x:xs])] ++ MakeFrequenceTable (xs ++ [x])))
        where n = length (removeDup [x:xs])

with the support functions:

    CountOccurences :: a [a] -> Int | == a
    CountOccurences f [] = 0
    CountOccurences f [x:xs] 
        | f == x = 1 + (CountOccurences f xs)
	           = CountOccurences f xs

    Frequence :: a [a] -> Int | Eq a
    Frequence f x = 100 * (CountOccurences f x) / length x

If I try to use "| == a", however, the compiler generates the error
"internal overloading of "removeDup" could not be solved.

After digging around some, I came up with a much shorter example that
generates the same error:

    ShortenedList :: [a] -> [a] | == a
    ShortenedList list = removeDup list

How is "| == a" in conflict with "| Eq a" in these cases?  What
fundamental piece am I not getting?  Any help here would be greatly
appreciated.  Also, given that difference, how would one suggest writing
the MakeFrequenceTable function as per requested in the text without
rewriting a bunch of the library code?

Thanks in advance,
John