CODEWITHSUNDEEP
haskell
Ninjanoel
Ninjanoel

Reputation: 2914

99 Haskell Questions #9

https://wiki.haskell.org/99_questions/1_to_10

regarding the solution to Question 9

pack :: Eq a => [a] -> [[a]]                        -- problem 9
pack [] = []
pack (x:xs) = (x:first) : pack rest
     where
       getReps [] = ([], [])
       getReps (y:ys)
               | y == x = let (f,r) = getReps ys in (y:f, r)
               | otherwise = ([], (y:ys))
       (first,rest) = getReps xs

--input
pack ['a', 'a', 'a', 'a', 'b', 'c', 'c', 'a','a', 'd', 'e', 'e', 'e', 'e']
--output
["aaaa","b","cc","aa","d","eeee"]

whats going on here: (x:first), i can see that rest is being passed pack, but i don't understand the bit before that.

Upvotes: 0

Views: 65

Answers (1)

chepner
chepner

Reputation: 532408

getReps is a function specific to each iteration of pack, as it closes over the current values of x and xs.

The first call to pack "aaaabccaadeeee" results in a call to getReps "aaabccaadeeee". getReps will return ("aaa", "bccaadeeee"). Effectively, pack self splits off the first 'a' (the value of x), and getReps separates the rest of the 'a's from the value of xs. Thus, the final "value" of pack "aaaabccaadeeee" is ('a':"aaa") : pack "bccaadeeee".

Upvotes: 2

Related Questions