我们可以 运行 池中的每个任务不是以同步方式吗?

Can we run every task from a pool not in a synchronous manner?

我是 Ada 新手。

我已经声明了我的新任务类型,并将其中三个存储在一个池中。然后,我想运行循环中的每个任务。

预期的行为是所有这些都同时执行。

实际情况是,他们一个接一个被处决。因此,tasks(2) 刚一执行完,tasks(1) 就终止了。事实上,task(2) 将永远不会被执行,因为它由于 select 约束而终止。

我的代码:

with Counter;
procedure Main is
    task type CounterTask is
        entry Execute(t:in Counter.Timeout; d:in Duration);
    end CounterTask;

    task body CounterTask is
        begin MyLoop: loop 
            select
                accept Execute(t:in Counter.Timeout;d:in Duration) do
                    Counter.Run(t, d);
                end Execute;
            or
                delay 2.0;
                exit;
            end select;
        end loop MyLoop;
    end CounterTask;
    tasks:Array(1..3) of CounterTask;
begin
    for i in Integer range 1..3 loop
        tasks(i).Execute(Counter.Timeout(10*i), Duration(0.5 * i));
    end loop;
end Main;

欢迎任何提示或想法!

当您的主程序调用 accept 语句时

accept Execute(t:in Counter.Timeout;d:in Duration) do
   Counter.Run(t, d);
end Execute;

直到 end Execute 才被阻止。你没有显示Counter.Run,但我猜那里有一个delay t(或d?)。

需要将Execute的参数复制到accept语句中的局部任务变量中,然后才调用Counter.Run;这样,主程序和 Countertask 都可以自由进行。

task body CounterTask is
   Timeout : Counter.Timeout;
   Dur : Duration;
begin 
MyLoop:
   loop 
      select
         accept Execute(t:in Counter.Timeout;d:in Duration) do
            Timeout := T;
            Dur := D;
         end Execute;
         Counter.Run (Timeout, Dur);
      or
         delay 2.0;
         exit;
      end select;
   end loop MyLoop;
end CounterTask;

除了将 Counter.Runaccept 块中取出(正如 Simon Wright 所说),您可能还想考虑使用同步屏障(另请参阅 ARM D.10.1 ):

with Counter;
with Ada.Synchronous_Barriers;

procedure Main is

  use Ada.Synchronous_Barriers;

  Num_Tasks : Positive := 3;

  Sync : Synchronous_Barrier (Num_Tasks);

  task type Counter_Task is
      entry Execute (T : in Counter.Timeout; D : in Duration);
  end Counter_Task;

  task body Counter_Task is
      Notified     : Boolean;
      The_Timeout  : Counter.Timeout;
      The_Duration : Duration;
  begin
      MyLoop : loop
        select

            accept Execute (T : in Counter.Timeout; D : in Duration) do
              The_Timeout  := T;
              The_Duration := D;
            end Execute;

            --  Synchronize tasks: wait until all 3 tasks have arrived at this point.
            Wait_For_Release (Sync, Notified);

            Counter.Run (The_Timeout, The_Duration);    
        or
            delay 2.0;
            exit;
        end select;
      end loop MyLoop;
  end Counter_Task;

  Tasks : array (1 .. Num_Tasks) of Counter_Task;

begin
  for K in Tasks'Range loop
      Tasks (K).Execute
        (Counter.Timeout (K * 10),
         Duration (Duration (0.5) * K));
  end loop;
end Main;