Question

Data.Lists.Split is actually not working winHugs interpreter. which can be get by splitEvery 3 ['a'..'z'] function

is it possible to achieve this without Data.Lists.Split ?
list of separate integers [5,5,5,6,6,6] -> to be [555,666] to concat every three numbers !

should be [Int]->[Int]

Answer 1

Possible Solution:

Here is one possible solution, using only function from the Prelude:

group3 :: [Int] -> [Int]
group3 [] = []
group3 xs = toSingleNum (take 3 xs): (group3 (drop 3 xs))
     where toSingleNum ys = read $ concat $ map show ys

There are, of course, innumerable ways to do this. This is one.

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Explanation:

group3 works by using the take and drop functions and natural recursion to split the list into groups of 3, and applying toSingleNum to each of the groups.

Ignoring the application of take and drop, the function works roughly as below:

group3 [1,2,3,4,5,6,7,8,9]
toSingleNum [1,2,3] : group3 [4,5,6,7,8,9]
toSingleNum [1,2,3] : toSingleNum [4,5,6] : group3 [7,8,9]
toSingleNum [1,2,3] : toSingleNum [4,5,6] : toSingleNum [7,8,9] : group3 []
toSingleNum [1,2,3] : toSingleNum [4,5,6] : toSingleNum [7,8,9] : []

After toSingleNum is applied, we’d have:

123 : 456 : 789 : []
[123, 456, 789]

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toSingleNum converts a list of numbers into a single number.

toSingleNum [1, 3, 4]
>>> 134
toSingleNum [12, 13, 14]
>>> 121314

The conversion in toSingleNum is done by converting the numbers to a String, then combining them, then converting back to a number.

toSingleNum [1, 3, 4]
read $ concat $ map show [1, 3, 4]
read $ concat $ ["1", "3", "4"]
read $ "134"
134

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Further Reading:

If you’re interested in Haskell, I’d recommend Graham Hutton’s excellent “Programming in Haskell”, or one of the tutorials listed on the Haskell Wiki.