I have a simple haskell program that returns the index of a specific number (say variable "i" from "mylist", and should return the index like.. "(Ind i)". "Num" is of type Int. nums is declared as a list.
This may be easy for some, but could you help me, what is wrong with my simple code? It says "parse error on input `)'".
list1 :: Int -> Int
list1 (Num i) = (do let j = [nums] !! i; return (Ind j));
....
I'm a newbie in haskell. :( help!
You probably want something like this:
find :: Eq a => a -> [a] -> Int
find i [] = -1 -- not found
find i (x:xs) = if i==x then 0 else 1 + find i xs
The above is not the most elegant solution, but uses only basic notions.
Another variant could be
find :: Eq a => a -> [a] -> Int
find i list = go 0 list
where go _ [] = -1 -- not found
go n (x:xs) = if i==x then n else go (n+1) xs
You should return Nothing instead of -1, though, even if this requires changing the return type from Int to Maybe Int. This is also done by the elemIndex library function, which solves this task.
If you want to use the single-line do notation, it needs to look like do {x; y; z} - parentheses do not suffice for delimiting it. But also this doesn't look like it needs do at all - in fact I can't figure out at all what your function is supposed to do, so I can't fix it for you. What is the intended purpose of this function?
There is no need for do notation, since this should be a pure function. What is nums? Should it perhaps be an argument to list1? Is Ind a constructor, or are you trying to return a string that reads "Ind "++ show j? Last, using !! is not appropriate here - it will look up the ith item in the list, while I think you want to find an item equal to i in the list.
I would suggest solving this with manual recursion for practice, then looking at Data.List for a more idiomatic solution. Here's a simple recursive version:
list1 :: Eq a => [a] -> a -> Maybe Int
list1 ls item = helper ls item 0
where helper :: Eq a => [a] -> a -> Int -> Maybe Int
helper [] _ _ = Nothing
helper (x:xs) m n
| x == m = Just n
| otherwise = helper xs m (n+1)
This is how I would approach the problem: limit the do notation to the final call, in this case in the main function. The rest is pure.
nums = [1,2,3]
list1 aList aNumber = list2 aList aNumber 0
list2:: [Int] -> Int -> Int -> Int
list2 [] aNumber aCount = 0
list2 (h:xs) aNumber aCount = if h == aNumber then aCount + 1
else list2 xs aNumber (aCount + 1)
main = do
let index = list1 [1,2,3] 3
putStrLn $ show index
Will this work for you?
doconstruct and monads in general. Get used to the pure functional style first. - ThomasNumandIndyourself? Please show those datatypes/constructors - Bergi