Haskell: List append operation

Question

I'm looking at 99 problems solved in Haskell . I'm trying to understand few basics here.

    combinations :: Int -> [a] -> [[a]]
    combinations _ [] = [[]]
    combinations 0 _  = [[]]
    combinations k (x:xs) = x_start ++ others
    where x_start = [ x : rest | rest <- combinations (k-1) xs ]
          others  = if k <= length xs then combinations k xs else []

consider, combinations 1 [3, 4]

pattern matches with the combinations k (x:xs) . We will start with computing x_start which will be x_start = [3 : combinations 0 [4]]

combinations 0 [4] will pattern match with combinations 0 [4] and "return" [[]] (a list of 1 elem empty list)

we now see x_start = [3 : combinations 0 [4]] ===> x_start = [3 : [[]]] . Is this correct? if so what would x_start be at this point, going by list prepend operator : , will x_start = [[3, []]] which is not same as [[3]]?

when I try to print this in ghci, it complains

No instance for (Num [t0]) arising from a use of ‘it’
In a stmt of an interactive GHCi command: print it

Answer 1

We will start with computing x_start which will be x_start = [3 : combinations 0 [4]]

This is not true, because x_start and others are of type [[a1]] . So it's more likely that x_start = [3] : rest , where rest is coming (since list comprehension here is just a syntax sugar with List Monad <- usage) from the combinations 0 [4] . Because of definition combinations 0 [4] is a [[]] , thus there is only one value came from it - [] . Finally, [3] : [] is [[3]] .

As it mentioned in the comments, [3 : [[]]] expression could not be typed by ghci well, because first element is Num a and second element is list of lists, not lists of Num s. This is what compiler is trying to say.