Question

使用+=运算符时，操作是同步还是异步？

Action actions = () => Console.WriteLine("First Action");

actions += () => Console.WriteLine("Second Action");
actions += () => Console.WriteLine("Third Action");
actions += () => Console.WriteLine("Fourth Action");

actions();

此代码块打印：

First Action
Second Action
Third Action
Fourth Action

Answer 1

此操作将同步运行，只需执行此操作：

Action actions = () => Console.WriteLine("First Action");

actions += () => Console.WriteLine("Second Action");
actions += () => {
                    Thread.Sleep(2000);
                    Console.WriteLine("Third Action")
                 };
actions += () => Console.WriteLine("Fourth Action");
actions();

你可以检查线程等待写下最后一个

Answer 2

您的问题的答案简短而简单：行动同步。就像事件句柄一样，操作按照分配/订阅的顺序依次按顺序执行。

您可能已经注意到，该块已按顺序打印了所有字符串。如果你定义这样的东西：

Action actions = () => Console.WriteLine("2");
actions += () => Console.WriteLine("3");
actions += () => Console.WriteLine("1");
actions += () => Console.WriteLine("4");

无论发生什么情况，您都会看到2 3 1 4被打印出来。

多个操作分配添加是否同步运行？

2 个答案: