Haskell own concat method with recursion

Question

Im practicing recursion on Haskell and need to do concat method with own impelentation on a nested list.

I've tried

    myConcat :: [[a]] -> [a]
    myConcat [[]] = []
    myConcat ((x:xs)) = x : myConcat ((xs))

But it doesnt seem to work.

myConcat [[1,2],[3,4,5],[6]] == [1,2,3,4,5,6]

This is the goal.

The main problem is that i dont really know how to work with nested lists.

SInce x is a list, you can not here use x : ..., unless you again want to create a list of lists. — willeM_ Van Onsem
– willeM_ Van Onsem, Commented Oct 23, 2019 at 12:04
Do you mind posting the type error for future searchability? — luqui
– luqui, Commented Oct 25, 2019 at 1:00

willeM_ Van Onsem · Accepted Answer · 2019-10-23 12:07:25Z

5

Since x is a list here (it has type [a]), you can not use x : … here, unless you again want to create a list of lists, but then your myConcat would act like an id function for lists.

You here need to append x with the rest of the list, so you can use (++) :: [a] -> [a] -> [a], indeed:

myConcat :: [[a]] -> [a]
myConcat [] = []
myConcat (x:xs) = x ++ myConcat xs

Note that you forgot the base case []. By writing [[]] you match a list with one element: the empty list. That is something different.

answered Oct 23, 2019 at 12:07

willeM_ Van Onsem

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Comments

nader.h · Accepted Answer · 2019-10-23 12:07:52Z

1

hello you have just to replace your x : myConcat xs by x ++ myConcat xs

answered Oct 23, 2019 at 12:07

nader.h

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2 Comments

dfeuer Over a year ago

That's not enough. The pattern matches remain incomplete.

nader.h Over a year ago

yes sure the case of empty List! My solution isnt complete, i just wanted to highlight the difference between : and ++

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