c# - 对正确使用 Async/Await 的困惑

Question

这是我在 Windows 窗体应用程序中实现的 async/await

async Task<int> DoAysnc1()
{
    await Task.Delay(3000);
    return 3000;
}

async Task<int> DoAsync2()
{
    await Task.Delay(5000);
    return 5000;
}

private async void button1_Click(object sender, EventArgs e)
{
    this.textBox1.Text = "";
    var doAsync1 = DoAysnc1();
    var doAsync2 = DoAysnc2();
    var async1 = await doAsync1;
    var async2 = await doAsync2;
    this.textBox1.Text = $"{async1} & {async2}";
}

5 秒后，TextBox 中的结果为“3000 & 5000”。

但是当我像这样修改 button1_Click 时:

private async void button1_Click(object sender, EventArgs e)
{
    this.textBox1.Text = "";
    var async1 = await DoAysnc1();
    var async2 = await DoAysnc2();
    this.textBox1.Text = $"{async1} & {async2}";
}

结果相同，但需要 8 秒。

为什么 button1_Click 的第二个版本是同步的？

Answer 1

区别的解释如下:

this.textBox1.Text = "";
var doAsync1 = DoAysnc1();         // <--- Run DoAsync1
var doAsync2 = DoAysnc2();         // <--- Run DoAsync2
var async1 = await doAsync1;       // <--- wait for DoAsync1 to finish
var async2 = await doAsync2;       //<--- wait for DoAsync2 to finish
this.textBox1.Text = $"{async1} & {async2}";

同时:

this.textBox1.Text = "";
var async1 = await DoAysnc1();          // <-- Run DoAsync1 and wait for it to be done
var async2 = await DoAysnc2();          // <-- Run DoAsync2 and wait for it to be done
this.textBox1.Text = $"{async1} & {async2}";

所以在第一个版本中，两个任务是同时运行的。在第二个版本中，您永远不会运行第二个任务，直到第一个任务完成。

我想读这个Article对您的知识有很大帮助。