Question

我正在尝试编写一个并行排序的Qsort函数，并将此代码直接转换为Elixir，但出于某种原因，当我运行pqsort:pqsort([5, 4, 3, 10, 1]).时，它会因此错误而失败：

** exception error: bad argument
     in function  spawn_link/3
        called as spawn_link(pqsort,#Fun<pqsort.0.102886275>,[<0.33.0>,[3,2,1]])
     in call from pqsort:pqsort_worker/2 (pqsort.erl, line 6)

这是我的代码：

-module(pqsort).
-export([pqsort/1, pqsort_worker/2]).

pqsort_worker(Listener, []) -> Listener ! {self(), []};
pqsort_worker(Listener, [H | T]) ->
    PID_1 = spawn_link(pqsort, fun pqsort_worker/2, [self(), [ X || X <- T, H >= X ]]),
    PID_2 = spawn_link(pqsort, fun pqsort_worker/2, [self(), [ Y || Y <- T, H < Y ]]),
    receive
        {PID_1, List_1} ->
            receive
                {PID_2, List_2} -> Listener ! {self(), List_1 ++ [H] ++ List_2}
            end;
        {PID_2, List_2} ->
            receive
                {PID_1, List_1} -> Listener ! {self(), List_1 ++ [H] ++ List_2}
            end
    end.

pqsort(List) -> element(2, pqsort_worker(self(), List)).

Answer 1

如果你去documentation，你会注意到第二个参数不是函数而是原子

只需更改

PID_1 = spawn_link(pqsort, fun pqsort_worker/2, [self(), [ X || X <- T, H >= X ]]),
PID_2 = spawn_link(pqsort, fun pqsort_worker/2, [self(), [ Y || Y <- T, H < Y ]]),

到

PID_1 = spawn_link(pqsort, pqsort_worker, [self(), [ X || X <- T, H >= X ]]),              |(emacs@Mac-mini-de-Rodrigo)1> c("/Users/rorra/erlang/pqsort", [{outdir, "/Users/rorra/erlang/"}\
PID_2 = spawn_link(pqsort, pqsort_worker, [self(), [ Y || Y <- T, H < Y ]]),

Erlang spawn_link / 3的错误参数错误

1 个答案: