delphi - 如何在Delphi中将循环内的上下文传递到TTask.IFuture？

Question

我有一组任务来执行它们在一个数组中。我想遍历数组并为每个数组启动一个任务。

下面的例子很简单（计算平方）来说明这个问题。

program FutureSquares;

{$APPTYPE CONSOLE}

{$R *.res}

uses
  System.SysUtils,
  System.Threading,
  System.Classes;
const
  nums: array[0..9] of Integer = (1,2,3,4,5,6,7,8,9,10);
var
  i, contextIndex: Integer;
  f: array[0..9] of ITask;
  answer, futureAnswer: Integer;
  matchDisplay: string;
  futureFunc: TFunc<Integer>;
begin

  try
    for i := Low(f) to High(f) do
    begin
      contextIndex := i;

      futureFunc := function: Integer
        begin
          Sleep(Random(1000));
          Result := nums[contextIndex]*nums[contextIndex]; // trying  because a reference to it is captured. How to capture the actual value?
        end;

      f[i] := TTask.Future<Integer>(futureFunc);
      f[i].Start;
    end;

    //verify results in sequential manner
    for i := Low(f) to High(f) do
    begin
      answer := nums[i]*nums[i];
      futureAnswer := IFuture<Integer>(f[i]).Value;
      if futureAnswer = answer then
        matchDisplay := 'match'
      else
        matchDisplay := 'MISMATCH';
      writeln(Format('%d * %d = %d. Future<Integer> = %d - %s', [nums[i],nums[i],answer, futureAnswer, matchDisplay]));
    end;

    readln;
  except
    on E: Exception do
      Writeln(E.ClassName, ': ', E.Message);
  end;
end.

该程序的输出如下：

1 * 1 = 1. Future<Integer> = 16 - MISMATCH
2 * 2 = 4. Future<Integer> = 100 - MISMATCH
3 * 3 = 9. Future<Integer> = 100 - MISMATCH
4 * 4 = 16. Future<Integer> = 100 - MISMATCH
5 * 5 = 25. Future<Integer> = 100 - MISMATCH
6 * 6 = 36. Future<Integer> = 100 - MISMATCH
7 * 7 = 49. Future<Integer> = 100 - MISMATCH
8 * 8 = 64. Future<Integer> = 100 - MISMATCH
9 * 9 = 81. Future<Integer> = 100 - MISMATCH
10 * 10 = 100. Future<Integer> = 100 - match

我如何实现我的目标？

我注意到TTask.Future不允许我将上下文传递给它，这将很有用。

Answer 1

您需要捕获值而不是匿名方法中的变量。最简单的方法是将匿名方法放在独立函数中，并传递您要捕获的索引。这样，将捕获值而不是变量。

function CaptureFuture(const aTheIndex: Integer): TFunc<Integer>;
begin
  Result := function: Integer
    begin
      Sleep(Random(1000));
      Result := nums[aTheIndex]*nums[aTheIndex]; // trying  because a reference to it is captured. How to capture the actual value?
    end;
end;

然后，当您分配futureFunc时，您的代码将如下所示：

for i := Low(f) to High(f) do
begin
  futureFunc := CaptureFuture(i);

  f[i] := TTask.Future<Integer>(futureFunc);
  f[i].Start;
end;

这将为您提供所需的结果：

1 * 1 = 1. Future<Integer> = 1 - match
2 * 2 = 4. Future<Integer> = 4 - match
3 * 3 = 9. Future<Integer> = 9 - match
4 * 4 = 16. Future<Integer> = 16 - match
5 * 5 = 25. Future<Integer> = 25 - match
6 * 6 = 36. Future<Integer> = 36 - match
7 * 7 = 49. Future<Integer> = 49 - match
8 * 8 = 64. Future<Integer> = 64 - match
9 * 9 = 81. Future<Integer> = 81 - match
10 * 10 = 100. Future<Integer> = 100 - match

有关变量与值捕获的更多信息，请参见以下问题：
Anonymous methods - variable capture versus value capture