编译后它们是相同的。您可以通过erlc -S
运行ERL文件和读取生成.S
文件证实了这一点:
$ 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}.
...
正如你所看到的,execute1
和execute2
结果相同的代码(唯一不同,是行号和标签号。
我无法找到与文档中汇编代码生成相关的'-S'选项的任何引用。它声明'.S'文件包含汇编代码,但它不会告诉您如何生成它们。 http://erlang.org/doc/man/erlc.html – ipinak