5 List Handling
Lists can only be built starting from the end and attaching list elements at the beginning. If you use the "++" operator as follows, a new list is created that is a copy of the elements in List1, followed by List2:
List1 ++ List2
Looking at how lists:append/1 or ++ would be implemented in plain Erlang, clearly the first list is copied:
append([H|T], Tail) -> [H|append(T, Tail)]; append(, Tail) -> Tail.
When recursing and building a list, it is important to ensure that you attach the new elements to the beginning of the list. In this way, you will build one list, not hundreds or thousands of copies of the growing result list.
Let us first see how it is not to be done:
bad_fib(N) -> bad_fib(N, 0, 1, ). bad_fib(0, _Current, _Next, Fibs) -> Fibs; bad_fib(N, Current, Next, Fibs) -> bad_fib(N - 1, Next, Current + Next, Fibs ++ [Current]).
Here more than one list is built. In each iteration step a new list is created that is one element longer than the new previous list.
To avoid copying the result in each iteration, build the list in reverse order and reverse the list when you are done:
tail_recursive_fib(N) -> tail_recursive_fib(N, 0, 1, ). tail_recursive_fib(0, _Current, _Next, Fibs) -> lists:reverse(Fibs); tail_recursive_fib(N, Current, Next, Fibs) -> tail_recursive_fib(N - 1, Next, Current + Next, [Current|Fibs]).
Lists comprehensions still have a reputation for being slow. They used to be implemented using funs, which used to be slow.
In recent Erlang/OTP releases (including R12B), a list comprehension:
[Expr(E) || E <- List]
is basically translated to a local function:
'lc^0'([E|Tail], Expr) -> [Expr(E)|'lc^0'(Tail, Expr)]; 'lc^0'(, _Expr) -> .
In R12B, if the result of the list comprehension will obviously not be used, a list will not be constructed. For example, in this code:
[io:put_chars(E) || E <- List], ok.
or in this code:
... case Var of ... -> [io:put_chars(E) || E <- List]; ... -> end, some_function(...), ...
the value is not assigned to a variable, not passed to another function, and not returned. This means that there is no need to construct a list and the compiler will simplify the code for the list comprehension to:
'lc^0'([E|Tail], Expr) -> Expr(E), 'lc^0'(Tail, Expr); 'lc^0'(, _Expr) -> .
lists:flatten/1 builds an entirely new list. It is therefore expensive, and even more expensive than the ++ operator (which copies its left argument, but not its right argument).
In the following situations, you can easily avoid calling lists:flatten/1:
- When sending data to a port. Ports understand deep lists so there is no reason to flatten the list before sending it to the port.
- When calling BIFs that accept deep lists, such as list_to_binary/1 or iolist_to_binary/1.
- When you know that your list is only one level deep, you can use lists:append/1.
... port_command(Port, DeepList) ...
... port_command(Port, lists:flatten(DeepList)) ...
A common way to send a zero-terminated string to a port is the following:
... TerminatedStr = String ++ , % String="foo" => [$f, $o, $o, 0] port_command(Port, TerminatedStr) ...
... TerminatedStr = [String, 0], % String="foo" => [[$f, $o, $o], 0] port_command(Port, TerminatedStr) ...
> lists:append([, , ]). [1,2,3] >
> lists:flatten([, , ]). [1,2,3] >
In Section 7.2, the following myth was exposed: Tail-Recursive Functions are Much Faster Than Recursive Functions.
To summarize, in R12B there is usually not much difference between a body-recursive list function and tail-recursive function that reverses the list at the end. Therefore, concentrate on writing beautiful code and forget about the performance of your list functions. In the time-critical parts of your code (and only there), measure before rewriting your code.
This section is about list functions that construct lists. A tail-recursive function that does not construct a list runs in constant space, while the corresponding body-recursive function uses stack space proportional to the length of the list.
For example, a function that sums a list of integers, is not to be written as follows:
recursive_sum([H|T]) -> H+recursive_sum(T); recursive_sum() -> 0.
sum(L) -> sum(L, 0). sum([H|T], Sum) -> sum(T, Sum + H); sum(, Sum) -> Sum.