呼叫链接的性能提升？

Question

Do you gain any performance, even if it's minor, by chaining function calls as shown below or is it just coding style preference?

execute() -> 
   step4(step3(step2(step1())).

Instead of

execute() ->
   S1 = step1(),
   S2 = step2(S1),
   S3 = step3(S2),
   step4(S3).

I was thinking whether in the 2nd version the garbage collector has some work to do for S1 , S2 , S3 . Should that apply for the 1st version as well?

Answer 1

They are identical after compilation. You can confirm this by running the erl file through erlc -S and reading the generated .S file:

$ cat a.erl
-module(a).
-compile(export_all).

step1() -> ok.
step2(_) -> ok.
step3(_) -> ok.
step4(_) -> ok.

execute1() ->
   step4(step3(step2(step1()))).

execute2() ->
   S1 = step1(),
   S2 = step2(S1),
   S3 = step3(S2),
   step4(S3).
$ erlc -S a.erl
$ cat a.S
{module, a}.  %% version = 0

...

{function, execute1, 0, 10}.
  {label,9}.
    {line,[{location,"a.erl",9}]}.
    {func_info,{atom,a},{atom,execute1},0}.
  {label,10}.
    {allocate,0,0}.
    {line,[{location,"a.erl",10}]}.
    {call,0,{f,2}}.
    {line,[{location,"a.erl",10}]}.
    {call,1,{f,4}}.
    {line,[{location,"a.erl",10}]}.
    {call,1,{f,6}}.
    {call_last,1,{f,8},0}.


{function, execute2, 0, 12}.
  {label,11}.
    {line,[{location,"a.erl",12}]}.
    {func_info,{atom,a},{atom,execute2},0}.
  {label,12}.
    {allocate,0,0}.
    {line,[{location,"a.erl",13}]}.
    {call,0,{f,2}}.
    {line,[{location,"a.erl",14}]}.
    {call,1,{f,4}}.
    {line,[{location,"a.erl",15}]}.
    {call,1,{f,6}}.
    {call_last,1,{f,8},0}.

...

As you can see, both execute1 and execute2 result in identical code (the only thing different are line numbers and label numbers.