有一个问题here,其中讨论了在ContinueWith
的上下文中执行Orleans任务。我对该示例进行了一些修改,以检查是否可以在不使用[Reentrant]标志的情况下使两个奥尔良相同任务的任务交错。整个代码为here。该代码的重点如下:
// Client code
var hashGenerator = client.GetGrain<IHashGeneratorGrain>(0);
Console.WriteLine("Client making a call");
hashGenerator.Call_A_ToTemp();
await Task.Delay(10000);
hashGenerator.Call_B_ToTemp();
// HashGeneratorGrain
public async Task Call_A_ToTemp()
{
Console.WriteLine("Making call A to a fellow grain");
ITempGrain grain = this.GrainFactory.GetGrain<ITempGrain>(1);
// CallA() returns successfully in 20s that is less than the timeout.
grain.CallA().ContinueWith(async (t)=>
{
if (t.IsCompletedSuccessfully)
{
Console.WriteLine("Task success");
Console.WriteLine("Call_A_ToTemp: Counter value now is: " + counter);
var selfGrain = this.GrainFactory.GetGrain<IHashGeneratorGrain>(0);
await selfGrain.IncCounter();
Console.WriteLine("Call_A_ToTemp: after increment: " + this.counter);
}
}
);
}
public async Task IncCounter()
{
this.counter += 1;
}
public async Task Call_B_ToTemp()
{
Console.WriteLine("Call_B_ToTemp called");
Console.WriteLine("Call_B_ToTemp: Counter value now is: " + counter);
Console.WriteLine("Call_B_ToTemp: now sleep for 20s");
await Task.Delay(20000);
Console.WriteLine("Call_B_ToTemp: after sleep value: " + counter);
this.counter += 1;
Console.WriteLine("Call_B_ToTemp: after increment: " + this.counter);
}
// TempGrain
public async Task CallA()
{
Console.WriteLine("Call A came to TempGrain");
await Task.Delay(20000);
}
输出为:
Client making a call
Making call A to a fellow grain
Call A came to TempGrain
Call_B_ToTemp called
Call_B_ToTemp: Counter value now is: 0
Call_B_ToTemp: now sleep for 20s
Task success
Call_A_ToTemp: Counter value now is: 0
Call_A_ToTemp: after increment: 1
Call_B_ToTemp: after sleep value: 1
Call_B_ToTemp: after increment: 2
如我们所见,Call_B_ToTemp()
尚未完成,但是IncCounter()
在退出之间和退出之后被调用,Call_B_ToTemp()
被执行并完成。
这不违反奥尔良谷物的单螺纹性吗?