Question

这可以在for循环中完成吗？

        TickEventArgs targs1 = new TickEventArgs(lbl1_up_time, _elapsedTime_up1);
        timer_up1.Tick += (sender, e) => Tick(targs1);

        TickEventArgs targs2 = new TickEventArgs(lbl2_up_time, _elapsedTime_up2);
        timer_up2.Tick += (sender, e) => Tick(targs2);

        TickEventArgs targs3 = new TickEventArgs(lbl3_up_time, _elapsedTime_up3);
        timer_up3.Tick += (sender, e) => Tick(targs3);

        TickEventArgs targs4 = new TickEventArgs(lbl4_up_time, _elapsedTime_up4);
        timer_up4.Tick += (sender, e) => Tick(targs4);

        TickEventArgs targs5 = new TickEventArgs(lbl5_up_time, _elapsedTime_up5);
        timer_up5.Tick += (sender, e) => Tick(targs5);

这不起作用，因为我超出界限（5）

        targs[0] = new TickEventArgs(lbl1_up_time, _elapsedTime_up1);
        targs[1] = new TickEventArgs(lbl2_up_time, _elapsedTime_up2);
        targs[2] = new TickEventArgs(lbl3_up_time, _elapsedTime_up3);
        targs[3] = new TickEventArgs(lbl4_up_time, _elapsedTime_up4);
        targs[4] = new TickEventArgs(lbl5_up_time, _elapsedTime_up5);

        timers[0] = timer_up1;
        timers[1] = timer_up2;
        timers[2] = timer_up3;
        timers[3] = timer_up4;
        timers[4] = timer_up5;

        int i = 0;

        for (i = 0; i <= 4; i++)
        {
            timers[i].Tick += (sender, e) => Tick(targs[i]);
        }

Answer 1

这是来自lambda表达式; i is shared between all of them。当函数执行时，它们基本上被称为timers[i].Tick += (sender, e) => Tick(targs[5])。

要避免这种情况，请创建一个本地范围的变量（int locali = i）并在您的行中使用它。这将确保每个lambda表达式实际上获得您期望的值。

for (i = 0; i <= 4; i++)
{
    int locali = i;
    timers[locali].Tick += (sender, e) => Tick(targs[locali]);
}

退出前，

i从循环的最后一次迭代变为5。当然，您没有targs[5]元素，因此它会抛出IndexOutOfRangeException。

从技术上讲，您不需要locali用于timers[i].Tick部分，因为它会立即进行评估，但我个人觉得将两者混合会让人感到困惑。

关于概念的一些额外阅读：

The foreach identifier and closures

Closing over the loop variable considered harmful

Answer 2

在这种情况下只有一个i，并且所有lambda都捕获相同的值。使用作用于循环的本地，以便每个lambda具有不同的副本

for (i = 0; i <= 4; i++)
{
  int j = i;
  timers[j].Tick += (sender, e) => Tick(targs[j]);
}

for循环中事件处理程序的匿名方法

2 个答案: