[Erlang 0128] Term sharing in Erlang/OTP 下篇

继续昨天的话题,昨天提到io:format对数据共享的间接影响,如果是下面两种情况恐怕更容易成为"坑", 呃,恰好我都遇到过;

如果是测试代码是下面这样,得到的结果会是怎样?猜!

s2()->
  L=[1,2,3,4,5,6],
  L2=[L,L,L,L],
erlang:display( {{erts_debug:size(L),erts_debug:flat_size(L)},{erts_debug:size(L2),erts_debug:flat_size(L2)}}
).

结果是

5> d:s2().

{{12,12},{56,56}}

　　

这个结果出来之后,我足足用了5分钟用来怀疑人生,为什么和期望的结果不一样呢?是因为我现在用的最新版本(17.2)吗?是实现已经修改掉但是没有更新文档吗?出于好奇,我还是按照之前探索问题的套路,生成了一下to_core文件,真相大白:

's2'/0 =
    %% Line 11
    fun () ->
        let <_cor5> =
            %% Line 14
            call 'erts_debug':'size'
                ([1|[2|[3|[4|[5|[6]]]]]])
        in  let <_cor4> =
                %% Line 14
                call 'erts_debug':'flat_size'
                    ([1|[2|[3|[4|[5|[6]]]]]])
            in  let <_cor3> =
                    %% Line 14
                    call 'erts_debug':'size'
                        ([[1|[2|[3|[4|[5|[6]]]]]]|[[1|[2|[3|[4|[5|[6]]]]]]|[[1|[2|[3|[4|[5|[6]]]]]]|[[1|[2|[3|[4|[5|[6]]]]]]]]]])
                in  let <_cor2> =
                        %% Line 14
                        call 'erts_debug':'flat_size'
                            ([[1|[2|[3|[4|[5|[6]]]]]]|[[1|[2|[3|[4|[5|[6]]]]]]|[[1|[2|[3|[4|[5|[6]]]]]]|[[1|[2|[3|[4|[5|[6]]]]]]]]]])
                    in  %% Line 14
                        call 'erlang':'display'
                            ({{_cor5,_cor4},{_cor3,_cor2}})

修改一下代码:

s3(L)->
    L2=[L,L,L,L],
    {{erts_debug:size(L),erts_debug:flat_size(L)},{erts_debug:size(L2),erts_debug:flat_size(L2)}}
.

　　

对应的s3的代码是

's3'/1 =
    %% Line 18
    fun (_cor0) ->
        let <L2> =
            %% Line 19
            [_cor0|[_cor0|[_cor0|[_cor0|[]]]]]
        in  let <_cor5> =
                %% Line 20
                call 'erts_debug':'size'
                    (_cor0)
            in  let <_cor4> =
                    %% Line 20
                    call 'erts_debug':'flat_size'
                        (_cor0)
                in  let <_cor3> =
                        %% Line 20
                        call 'erts_debug':'size'
                            (L2)
                    in  let <_cor2> =
                            %% Line 20
                            call 'erts_debug':'flat_size'
                                (L2)
                        in  %% Line 20
                            {{_cor5,_cor4},{_cor3,_cor2}}

　　

换句话,在编译阶段s2方法里面的常量数据就已经展开了,所以L2无论是size还是flat_size都是一样的.之所以要先把这个测试做了,就是避免后面的测试误入陷阱.

这个怎么破呢?除了上面传入参数的方法之外,还有一个路子:换成函数调用即可,如下:

s4()->
  L=lists:seq(1,6),
  L2=[L,L,L,L],
erlang:display( {{erts_debug:size(L),erts_debug:flat_size(L)},{erts_debug:size(L2),erts_debug:flat_size(L2)}}
).

　　

对应的代码为:

's4'/0 =
    %% Line 24
    fun () ->
        let <L> =
            %% Line 25
            call 'lists':'seq'
                (1, 6)
        in  let <L2> =
                %% Line 26
                [L|[L|[L|[L|[]]]]]
            in  let <_cor5> =
                    %% Line 27
                    call 'erts_debug':'size'
                        (L)
                in  let <_cor4> =
                        %% Line 27
                        call 'erts_debug':'flat_size'
                            (L)
                    in  let <_cor3> =
                            %% Line 27
                            call 'erts_debug':'size'
                                (L2)
                        in  let <_cor2> =
                                %% Line 27
                                call 'erts_debug':'flat_size'
                                    (L2)
                            in  %% Line 27
                                call 'erlang':'display'
                                    ({{_cor5,_cor4},{_cor3,_cor2}})

　　

不要小看这个问题,这样一个常量优化在极端情况下会有"大惊喜",论文里面给了这样一个例子:

show_compiler_crashes() ->

L0 = [0],

L1 = [L0, L0, L0, L0, L0, L0, L0, L0, L0, L0],

L2 = [L1, L1, L1, L1, L1, L1, L1, L1, L1, L1],

L3 = [L2, L2, L2, L2, L2, L2, L2, L2, L2, L2],

L4 = [L3, L3, L3, L3, L3, L3, L3, L3, L3, L3],

L5 = [L4, L4, L4, L4, L4, L4, L4, L4, L4, L4],

L6 = [L5, L5, L5, L5, L5, L5, L5, L5, L5, L5],

L7 = [L6, L6, L6, L6, L6, L6, L6, L6, L6, L6],

L8 = [L7, L7, L7, L7, L7, L7, L7, L7, L7, L7],

L9 = [L8, L8, L8, L8, L8, L8, L8, L8, L8, L8],

L = [L9, L9, L9, L9, L9, L9, L9, L9, L9, L9],

L.

　　

影响有多大呢?看结果:

 

After a bit more of 45 minutes of struggling, the compiler tries to allocate 3.7 GB of memory and gives up:

$ erlc demo.erl
Crash dump was written to: erl_crash.dump
eheap_alloc: Cannot allocate 3716993744 bytes of
memory (of type "heap_frag").
Abort

好吧,勇于自黑,由于上面遇到这样让人恼火的问题,我决定在Shell中完成后续的测试,然后,我一脚踏进"新坑":

 

陷阱2 Shell ! Shell !

Eshell V6.0  (abort with ^G)
1> L=[1,2,3,4,5,6,7,8,9,10].
[1,2,3,4,5,6,7,8,9,10]
2>  L2=[L,L,L,L,L,L].
[[1,2,3,4,5,6,7,8,9,10],
[1,2,3,4,5,6,7,8,9,10],
[1,2,3,4,5,6,7,8,9,10],
[1,2,3,4,5,6,7,8,9,10],
[1,2,3,4,5,6,7,8,9,10],
[1,2,3,4,5,6,7,8,9,10]]
3> erts_debug:size(L2).
32
4> erts_debug:flat_size(L2).
132
5> io:format("~p",[L2]).
[[1,2,3,4,5,6,7,8,9,10],
[1,2,3,4,5,6,7,8,9,10],
[1,2,3,4,5,6,7,8,9,10],
[1,2,3,4,5,6,7,8,9,10],
[1,2,3,4,5,6,7,8,9,10],
[1,2,3,4,5,6,7,8,9,10]]ok
6> erts_debug:size(L2).
32
7> erts_debug:flat_size(L2).
132

　　

一开始启动shell的时候,Shell的Pid是<0.33.0>.然后我们在中间故意执行一个不存在的方法 fake:fake().这时查看一下,Shell已经重启,Pid变成<0.40.0>.注意再执行erts_debug:size(L2).结果已经变成了132了,换句话说,这里L2数据已经展开了.

Eshell V6.0  (abort with ^G)
1> self().
<0.33.0>
2>  L=[1,2,3,4,5,6,7,8,9,10].
[1,2,3,4,5,6,7,8,9,10]
3>  L2=[L,L,L,L,L,L].
[[1,2,3,4,5,6,7,8,9,10],
[1,2,3,4,5,6,7,8,9,10],
[1,2,3,4,5,6,7,8,9,10],
[1,2,3,4,5,6,7,8,9,10],
[1,2,3,4,5,6,7,8,9,10],
[1,2,3,4,5,6,7,8,9,10]]
4>  erts_debug:size(L2).
32
5> erts_debug:flat_size(L2).
132
6> fake:fake().
** exception error: undefined function fake:fake/0
7> self().
<0.40.0>
8>  erts_debug:size(L2).
132
9> erts_debug:flat_size(L2).
132
10>

　　

那为什么会触发数据展开(expand ,flattening)呢? 看下面的代码,在Shell启动的时候,会把之前已经绑定的变量作为spawn_link参数以启动新的shell.

erl6.2\lib\stdlib-2.2\src

start_eval(Bs, RT, Ds) ->
    Self = self(),
    Eval = spawn_link(fun() -> evaluator(Self, Bs, RT, Ds) end),
    put(evaluator, Eval),
    Eval.

　　

换句话说,Erlang中使用spawn创建进程,传入的参数(包括函数闭包),需要拷贝到新进程的heap,换句话说进程创建的时候需要考虑参数的大小.

OK,这个问题差不多了,休息.