c# 在 Lambda 表达式中声明变量

Question

以下代码输出 33 而不是 012。我不明白为什么在每次迭代中都没有捕获新变量 loopScopedi 而不是捕获相同的变量。

Action[] actions = new Action[3];

for (int i = 0; i < 3; i++)
{

   actions [i] = () => {int loopScopedi = i; Console.Write (loopScopedi);};
}

foreach (Action a in actions) a();     // 333

Hopwever，这段代码产生了012，两者有什么区别？

Action[] actions = new Action[3];

for (int i = 0; i < 3; i++)
{
    int loopScopedi = i;
    actions [i] = () => Console.Write (loopScopedi);
}

foreach (Action a in actions) a();     // 012

Answer 1

这称为“访问修改后的闭包”。基本上，只有一个 i 变量，所有三个 lambda 都引用它。最后，i 变量增加到 3，因此所有三个操作都打印 3。 (请注意，lambda 中的 int loopScopedi = i 仅在您稍后调用 lambda 时运行。)

在第二个版本中，您为每次迭代创建一个新的 int loopScopedi，并将其设置为 i 的当前值，即 0 和 1 和 2，对于每次迭代。

您可以尝试想象内联 lambda 以更清楚地看到发生了什么:

foreach (Action a in actions)
{
   int loopScopedi = i; // i == 3, since this is after the for loop
   Console.Write(loopScopedi); // always outputs 3
}

对比:

foreach (Action a in actions)
{
   // normally you could not refer to loopScopedi here, but the lambda lets you
   // you have "captured" a reference to the loopScopedi variables in the lambda
   // there are three loopScopedis that each saved a different value of i
   // at the time that it was allocated
   Console.Write(loopScopedi); // outputs 0, 1, 2
}