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