Probably is a homework question, so HINT: you can do it using 2 and only 2 prelude functions and no additional function definitions of your own. -- Hal Daume III "Computer science is no more about computers | hdaume@isi.edu than astronomy is about telescopes." -Dijkstra | www.isi.edu/~hdaume On Wed, 5 Jun 2002, Andy Fugard wrote:

...

===== Original Message From "xoo" ===== hi.. i was just wondering if some body could give a simple equation for the following situation.other than recursion plz..

occurrences :: Eq a => a -> [a] -> [a] --occurrences xs ys returns the number of times that xs occurs in ys

You may find it easier if you make

occurrences :: Eq a => a -> [a] -> Integer

since it would seem it is to return a number, and not another list!

Also I would guess the function will have a form something like

occurrences x xs = foldr (countOp x) 0 xs where countOp :: Eq a => a -> a -> Integer -> Integer ...

I'm not sure if I should go any further, in case this is a homework question...

Andy

-- [ Andy Fugard ] [ +44 (0)7901 603075 ]

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At 20:11 05/06/2002 +0200, Hannah Schroeter wrote:

Why not combine filter and length appropriately?

Pass -- I'm also a newbie, and haven't got as far as filter. *Looks up filter* Well, there's his homework done :-)