[Erlang-0013][OTHERS] reductions计数

 

缘起坚强哥分享霸爷对于公平调度的解答。

 

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如果函数A会调用函数B， 函数B每次调用都会递归3次。那么每调用一次函数A，是否就是4个reduction？

到底是怎样自己动手测一下吧：

 

先是不嵌套的

Eshell V5.9  (abort with ^G)

1> A = spawn(fun() -> one:start() end).

<0.33.0>

2> process_info(A, reductions).

{reductions,46}

3> process_info(A, reductions).

{reductions,46}

4> process_info(A, reductions).

{reductions,46}

5> process_info(A, reductions).

{reductions,46}

6> A!go.

go

7> process_info(A, reductions).

{reductions,49}

 

 代码：

-module(one).

-compile(export_all).

 

start() ->

    receive

        go ->

            aaa(3)

    end,

    receive

        stop ->

            ok

    end,

    ok.

 

aaa(1) ->

    ok;

aaa(N) ->

    aaa(N-1).

========================================================

改为嵌套调用：

1> A = spawn(fun() -> one:start() end).

<0.33.0>

2> process_info(A, reductions).

{reductions,46}

3> A!go.

go

4> process_info(A, reductions).

{reductions,50}

 

代码：

-module(one).

-compile(export_all).

 

start() ->

    receive

        go ->

                bbb()

    end,

    receive

        stop ->

            ok

    end,

    ok.

bbb() ->

    aaa(3).

aaa(1) ->

    ok;

aaa(N) ->

    aaa(N-1).

 

===================================================

receive会不会有影响呢？

不会的。。。

 

代码：

-module(one).

-compile(export_all).

 

start() ->

    receive

        go ->

                bbb()

    end,

    receive

        rec ->

            ok

    end,

    receive

        stop ->

            stop

    end,

    ok.

bbb() ->

    aaa(3).

aaa(1) ->

    ok;

aaa(N) ->

    aaa(N-1).

 

综上，reduction是函数调用，如果函数A会调用函数B， 函数B每次调用都会递归3次。那么每调用一次函数A，就是4个reduction。