c# - 异步方法不能返回任何具有可访问的 GetAwaiter 方法的类型

Question

它说 here :

Async methods can have the following return types:

• Task, for an async method that performs an operation but returns novalue.

• Task <TResult>, for an async method that returns a value.

• void, for an event handler.

• Starting with C# 7.0, any type that has an accessible GetAwaitermethod. The object returned by the GetAwaiter method must implementthe System.Runtime.CompilerServices.ICriticalNotifyCompletioninterface.

• Starting with C# 8.0, IAsyncEnumerable<T>, for an async method thatreturns an async stream.

问题。如何理解这句话:“异步方法可以具有以下返回类型:...任何具有可访问的 GetAwaiter 方法的类型”，如果实际不是？

此代码无效:

using System.Runtime.CompilerServices;

async A Method() //Error CS1983 The return type of an async method must be void, Task, Task<T>, a task-like type, IAsyncEnumerable<T>, or IAsyncEnumerator<T>
{
    await new A();
}

class A
{
    public TaskAwaiter GetAwaiter()
    {
        return new TaskAwaiter();
    }
}

Answer 1

摘要文档有点困惑。 GetAwaiter不足以用作返回类型； GetAwaiter更多关于 await比async .

more detailed docs澄清:

In addition, the type returned from the GetAwaiter method must have the System.Runtime.CompilerServices.AsyncMethodBuilderAttribute attribute.

AsyncMethodBuilder属性为编译器提供了足够的信息来实际构建和控制异步状态机的返回类型实例。

TaskAwaiter<T> does not have这个属性。它通常不需要它，因为任务返回类型是“祖父”并且由编译器特殊处理。