[erlang-questions] foldl vs foldr and ++

Hynek Vychodil <>
Thu May 10 10:53:19 CEST 2012


When I see foldr implementation I wonder why it is not simple

foldr(F, Acc, L) -> foldl(F, Acc, lists:reverse(L, [])).

There should be less memory consumption. lists:reverse is BIF which I
believe consume less memory than stack frame per each list element
what current foldr implementation do. It should be faster and both
accepts only proper lists. Only difference which I can found current
implementation consumes memory on stack but above implementation would
consume heap and affect GC behavior. Current foldr implementation
would make sense in Haskell but I don't understand its reason in
Erlang.

On Thu, May 10, 2012 at 10:26 AM, Robert Virding
<> wrote:
> The documentation is referring to the *implementation* of the foldl/foldr functions, foldl uses a tail-recursive function while foldr cannot do that. The difference in speed comes from what you do inside the fold function. As others have pointed out the efficiency of ++ depends solely on its left argument.
>
> Robert
>
> ----- Original Message -----
>> Aha I just looked at my own explanation and realized it made no
>> sense. Of course the slow one is slow for exactly the reason I
>> expected. Sorry for the waste, but thank you for the clear
>> complexity analysis.
>>
>> Andrew Ledvina
>>
>> On May 9, 2012, at 2:57 PM, "Richard O'Keefe" <>
>> wrote:
>>
>> >
>> > On 10/05/2012, at 7:00 AM, Andrew Ledvina wrote:
>> >
>> >> Okay, so I was profiling something and it led me to write this:
>> >>
>> >> from_left(X) ->
>> >>   lists:foldl(fun(Y,R) -> R ++ [{Y+1,Y*2}] end, [], X).
>> >>
>> >> from_right(X) ->
>> >>   lists:foldr(fun(Y,R) -> [{Y+1,Y*2}] ++ R end, [], X).
>> >
>> > Immediate reaction:  from_left/1 is going to be atrociously slow.
>> >
>> >> My understanding is that foldl is preferred over foldr because it
>> >> is tail recursive.
>> >
>> > ***other things being equal***
>> >
>> > In this case, they are not.  It should be clear that from_right/1
>> > is
>> > O(|X|) because it does O(1) work per element of X, while
>> > from_left/1 is O(|X|**2) because it does O(|R|) work per element of
>> > X, and R grows from empty to 2|X|-2 elements long.
>> >
>> > Write [{Y+1,Y*2} || Y <- X] and let the compiler worry about the
>> > best way to do it.
>> >
>> >> I was also to understand that putting the accumulator on the
>> >> left-hand-side of the ++ operation is super bad. So, the
>> >> following results were a bit of a surprise at first:
>> >
>> > If you understood that long++short is more expensive than
>> > short++long,
>> > you should NOT have been surprised by your result.
>> >
>> >
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Hynek Vychodil
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