Hi Claude, Thanks a lot for the example. Btw, is this where you are trying in-place replacement? modifyAtIndex :: (a -> a) -> Nat -> List a -> List a modifyAtIndex f i as = let ias = zip nats as g (Tuple2 j a) = case i `eq` j of False -> a True -> f a in map g ias modifyAtIndex2 :: (a -> a) -> Nat -> Nat -> List (List a) -> List (List a) modifyAtIndex2 f i j = modifyAtIndex (modifyAtIndex f i) j Doesn't modfiyAtIndex return a new list? On Fri, Jul 16, 2010 at 2:28 PM, Claude Heiland-Allen < claudiusmaximus@goto10.org> wrote:
Hi,
On 16/07/10 07:35, C K Kashyap wrote:
Haskell without using any standard library stuff?
For example, if I wanted an image representation such as this
[[(Int,Int.Int)]] - basically a list of lists of 3 tuples (rgb) and wanted to do in place replacement to set the pixel values, how could I go about it.
Break the problem down into parts:
1. replace a single pixel 2. modify an element in a list at a given index using a given modification function 3. modify an element in a list of lists at a pair of given indices using a given replacement function
I had a stab at it. Without any standard library stuff I couldn't figure out how to print any output, though - so who knows if the code I wrote does what I intended.
The point is, it's libraries all the way down - so use them, study them where necessary for understanding, and write them and share them when you find something missing.
Claude -- http://claudiusmaximus.goto10.org
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-- Regards, Kashyap