On Sat, 29 Jun 2002, Shlomi Fish wrote:
No. But I want to generate an irregular series, which I determine the intervals between two consecutive numbers myself. E.g:
let (num1, next1) = (counter 5) (num2, next2) = (next1 100) (num3, next3) = (next2 50) in [num1,num2,num3]
Will have the numbers [5, 105, 155].
The other answers you've received really question whether this is a good way to use Haskell at all --- usually, if you want to generate a sequence of values, it's a good idea just to represent them explicitly as a (lazy) list. For example, you can compute more-or-less the same result as you want just using standard list processing functions: Main> scanl (+) 0 [5,100,50] [0,5,105,155] However, you can also do almost exactly what you suggest. Not quite exactly, because the function you describe would have a recursive type type Counter = Int -> (Int, Counter) and Haskell requires that type recursion involve a data or newtype. So, here's a solution: newtype Counter = Counter (Int -> (Int, Counter)) count :: Counter -> Int -> (Int, Counter) count (Counter c) = c counter n = (n, Counter next) where next k = counter (n+k) xs = let (num1, next1) = (counter 5) (num2, next2) = (next1 `count` 100) (num3, next3) = (next2 `count` 50) in [num1,num2,num3] Main> xs [5,105,155] The only difference from your idea is that a counter is not a function; we have to use the function count to invoke it. And that's forced on us by the need to avoid type recursion. John Hughes