Haskell foldl' poor performance with (++)

Question

I have this code:

import Data.List

newList_bad  lst = foldl' (\acc x -> acc ++ [x*2]) [] lst
newList_good lst = foldl' (\acc x -> x*2 : acc) [] lst

These functions return lists with each element multiplied by 2:

*Main> newList_bad [1..10]
[2,4,6,8,10,12,14,16,18,20]
*Main> newList_good [1..10]
[20,18,16,14,12,10,8,6,4,2]

In ghci:

*Main> sum $ newList_bad [1..15000]
225015000
(5.24 secs, 4767099960 bytes)
*Main> sum $ newList_good [1..15000]
225015000
(0.03 secs, 3190716 bytes)

Why newList_bad function works 200 times slower than newList_good? I understand that it's not a good solution for that task. But why this innocent code works so slow?

What is this "4767099960 bytes"?? For that simple an operation Haskell used 4 GiB??

After compilation:

C:\1>ghc -O --make test.hs
C:\1>test.exe
225015000
Time for sum (newList_bad [1..15000]) is 4.445889s
225015000
Time for sum (newList_good [1..15000]) is 0.0025005s

Answer 1

Classic list behavior.

Recall:

(:)  -- O(1) complexity
(++) -- O(n) complexity

So you are creating an O(n^2) algo, instead of an O(n) one.

For this common case of appending to lists incrementally, try using a dlist, or just reverse at the end.