4

Consider the following code snippet:

-- list_1 = [1, 2, 3]
-- list_2 = [4, 5, 6] 
final_list =  list_1 ++ list_2
result = map (+1) final_list

Is the time spent by it proportional just to the length of final_list, and the price of list concatenation is not paid?

My idea is that concatenation is done lazily; to quote the source of GHC Base.hs,

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

Since map constantly chips away the head of the list, I suppose that the concatenation as done by the recursive call in the last line is always done in lockstep with the map execution, so list_1 is effectively scanned only once.

Is this correct?

3

That's correct. It's easy to test your theory by making list_2 infinite:

list_1 = [1, 2, 3]
list_2 = repeat 4
final_list = list_1 ++ list_2
result = map (+1) final_list

print $ take 10 result
-- Outputs [2,3,4,5,5,5,5,5,5,5]

Congratulations! You just ran an infinite loop in a few microseconds! (Not really).

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