Question

I am doing Problem 15. Which states:

(**) Replicate the elements of a list a given number of times.

    Example:

    * (repli '(a b c) 3)
    (A A A B B B C C C)

    Example in Haskell:

    > repli "abc" 3
    "aaabbbccc"

My plan was to do something like this:

repli :: [a] -> Integer -> [a]

repli [] y = []

repli (x:xs) y | appendNo x y == [] = repli(xs) y
               | otherwise = appendNo x y : (x:xs)
               where
               appendNo :: a -> Integer -> [a]
               appendNo a 0 = []
               appendNo a y = a:appendNo a (y-1)

Where I would make a function called appendNo that returns a list of 1 element y times then append it to the original list. Then take the body of the list and repeat this process until there are no more body elements left. But, I get the error:

H15.hs:6:30:
    Couldn't match type `a' with `[a]'
      `a' is a rigid type variable bound by
          the type signature for repli :: [a] -> Integer -> [a] at H15.hs:3:1
    In the return type of a call of `appendNo'
    In the first argument of `(:)', namely `appendNo x y'
    In the expression: appendNo x y : (x : xs)
Failed, modules loaded: none.

6:30 is at the on the p in appendNo in this line:

           | otherwise = appendNo x y : (x:xs)

Ok thanks dave4420 I was able to figure it out by doing:

repli :: [a] -> Integer -> [a]

repli [] y = []

repli (x:xs) y = appendNo x y ++ repli(xs) y
               where
               appendNo :: a -> Integer -> [a]
               appendNo a 0 = []
               appendNo a y = a:appendNo a (y-1)

Answer 1

       | otherwise = appendNo x y : (x:xs)

There is a type error in this line. So ask yourself:

1. What is the type of appendNo x y?
2. What is the type of (x:xs)?
3. What is the type of (:)?

Then you should be able to see why they don’t match up.

If you still can’t see why they don’t match up, ask yourself

1. What is the type of x?
2. What is the type of xs?
3. What is the type of (:)?

Bear in mind that this time the types do match up.