奥尔良的单线程性质不受 ContinueWith 的尊重

Question

我有以下代码（https://github.com/avinash0161/OrleansExperiments/tree/c0155b4b0c8c1bfe60aea8624f2cc83a52853dc7）：

// Client code
Console.WriteLine("Client making a call");
var hashGenerator = client.GetGrain<IGrainA>(0);
hashGenerator.Call_A_ToTemp();
await Task.Delay(1000);
hashGenerator.Call_B_ToTemp();
// GrainA code
public async Task Call_A_ToTemp()
{
   Console.WriteLine("Making call A to a fellow grain");
   IGrainB grain = this.GrainFactory.GetGrain<IGrainB>(1);
   grain.CallA().ContinueWith((t)=>
   {
     if(t.IsFaulted)
     {
       // Silo message timeout is 32s so t.IsFaulted is true
       Console.WriteLine("Task Faulted");
       Call_A_ToTemp();
     }
    });
}
public async Task Call_B_ToTemp()
{
   Console.WriteLine("Making call B to a fellow grain");
   IGrainB grain = this.GrainFactory.GetGrain<IGrainB>(1);
   await grain.CallB();
}
// GrainB code
public async Task CallA()
{
   Console.WriteLine("Call A came to GrainB");
   await Task.Delay(34000);  // more than timeout for the caller
}
public Task CallB()
{
   Console.WriteLine("Call B came to GrainB");
   return Task.CompletedTask;
}

此代码的输出为：

Client making a call
Making call A to a fellow grain
Call A came to GrainB
Making call B to a fellow grain
Task Faulted                       <---------------- This comes after Call_B_ToTemp executes
Making call A to a fellow grain

如我们所见，Call_B_ToTemp在Call_A_ToTemp完全执行之前执行（ContinueWith 部分Call_A_ToTemp稍后执行）。这是意料之中的，是否违反了颗粒的单线程性质？

---

当我将 Call_A_ToTemp（） 中的代码替换为：

public async Task Call_A_ToTemp()
{
    Console.WriteLine("Making call A to a fellow grain");
    IGrainB grain = this.GrainFactory.GetGrain<IGrainB>(1);
    bool isSuccess = false;
    while (! isSuccess)
    {
       try
       {
          await grain.CallA();
          isSuccess = true;
       } catch(TimeoutException){
            Console.WriteLine("task faulted");
       }
    }
}

代码现在保留了单线程性质，并且在执行Call_A_ToTemp（） 的所有部分之前不会调用Call_B_ToTemp ContinueWith。控制台输出如下所示：

Client making a call
Making call A to a fellow grain
Call A came to GrainB
Task Faulted                       
Making call A to a fellow grain

谁能解释一下？当有ContinueWith时是否违反了单线程性质？

Answer 1

单线程性质没有被违反。项目中的编译警告使问题的根源一目了然。特别是： This async method lacks 'await' operators and will run synchronously. Consider using the 'await' operator to await non-blocking API calls, or 'await Task.Run(...)' to do CPU-bound work on a background thread.

该方法async Task Call_A_ToTemp()永远不会等待对谷物 B 的调用。相反，它会在发出调用后立即返回。由于 Call_A_ToTemp() 返回的Task会立即完成，因此允许对粒度执行另一个调用。一旦grain.CallA()完成，延续（ContinueWith(...)）将尽快在谷物的TaskScheduler上执行（例如，当谷物等待另一个调用或闲置时）。

相反，如果调用已等待，或者如果从方法中删除async并且代码更改为返回grain.CallA().ContinueWith(...)调用，则将观察到预期的行为。即，将代码更改为此代码将为您提供预期的结果：

// removed 'async' here, since we're not awaiting anything.
// using 'async' is preferred, but this is to demonstrate a point about
// using ContinueWith and un-awaited calls
public Task Call_A_ToTemp()
{
   Console.WriteLine("Making call A to a fellow grain");
   IGrainB grain = this.GrainFactory.GetGrain<IGrainB>(1);
   // Note the 'return' here
   return grain.CallA().ContinueWith((t)=>
   {
     if(t.IsFaulted)
     {
       // Silo message timeout is 32s so t.IsFaulted is true
       Console.WriteLine("Task Faulted");
       Call_A_ToTemp();
     }
    });
}