Am Freitag 29 Januar 2010 22:38:11 schrieb Han Joosten:
knyttr wrote: <snip> fun x [] [] = ... fun x y [] = ... fun x y z = ... would make more sence.
however, there still is overlap. (ghc could generate a warning for this)
now the where part: there are several options:
fun x y z
| z == [] = ... | y == [] = ... | otherwise = ...
where ...
Better use fun x y z | null z = ... | null y = ... | otherwise = .... where ....
or using a case expression...
fun x y z = case (x,y,z) of (_ , _ , []) -> ... (_ , [] , [_:_]) -> ... (_ , [_:_], [_:_]) -> ... where ... pattern matches do not overlap this way. you could use this pattern matches also in the previous solution.
Then your second question:
2. how can I change a field in a list in the most easy way? the only solution I found is below. I am not very happy with it - it does too many operations just to change one field.
changeIn array index value = (take (index-1) array)++[value]++(drop index array)
Remember, in haskell you do not 'change' values in variables. Thers is no such thing as a variable. You probably want a function that takes a list, an index and a value. It produces a list with all elements from the first list, only the n-th element is replaced with the given value. I would specify this as follows:
f x:xs i v
| i == 1 = v:xs | i > 1 = x : (f xs (i-1) v)
but there are probably dozens of other ways to do this. While you are new to Haskell, do not worry too much about performance.
True, but also *** be aware that lists are NOT arrays *** trying to use them as such is sure to land you in a deep hole of horrible performance sooner or later. If what you need is an array, use arrays (immutable or mutable, depending on the problem at hand). If you need a set, use sets (Data.Set e.g.), if a list is appropriate for what you want to do, use the versatile [].
(ghc will surly surprise you =^D) The main thing is to think functions. If you do that, you will appreciate Haskell for sure.