Hello Carajillu, Monday, September 18, 2006, 1:51:34 PM, you wrote:
Hi, I'm a student and I have to do a program with Haskell. This is the first time I use this languaje, and I'm having problems with the indentation. I want to check if this function is correct, but when I try to make the GHCi interpret it, I get "line 18:parse error (possibly incorrect indentation)"
i want to add to Andrea's answer:
The function is:
-- Replaces a wildcard in a list with the list given as the third argument substitute :: Eq a => a -> [a] -> [a] -> [a] substitute e l1 l2= [check_elem c | c <- l1] where check_elem::Eq a => a -> [a] check_elem x = if x == e then return l2
Haskell functions are like mathematical ones - first, you don't need to use 'return' to denote result - just write it. second, you should provide results for any possible input values, otherwise function will just fail on return values for which you don't provided result. 'if' construct ensures this by forcing to write both 'then' and 'else' parts. so, your function, probably, should return l2 if comparison succeeds, and x otherwise:
check_elem x = if x == e then l2 else x
next problem is that x and l2 had different types and type-checking will not allow you to write this. It's right - you can't construct list, whose some elements are, for example, -- i've got a break to rescue a bat that was flied into our house :) Int and some - [Int]: [1,[0,0,0],3,4] is impossible if your need to replace all elements of l1 that are equal to e, with _several_ list elements, i.e. substitute 2 [1,2,3,4] [0,0,0] = [1,0,0,0,3,4] you should use the following technique: replace all elements of l1 with either l2 or [x] (where x is original list element). result of such operation: substitute1 2 [1,2,3,4] [0,0,0] = [[1],[[0,0,0],[3],[4]] i.e. now all elements are lists, although some have just one element. then you should use concat operation to transform this list into what you need: concat [[1],[[0,0,0],[3],[4]] = [1,0,0,0,3,4] i think that you are interested to write the complete solution yourself. just note that there is concatMap function that will allow you to further simplify the code
-- Tries to match two lists. If they match, the result consists of the sublist -- bound to the wildcard in the pattern list. (line 18) match :: Eq a => a -> [a] -> [a] -> Maybe [a] match _ _ _ = Nothing {- TO BE WRITTEN -}
this definition is perfectly ok :) -- Best regards, Bulat mailto:Bulat.Ziganshin@gmail.com