编译器如何将返回值转换为return Task< value>在异步方法？ [英] How does compiler converts return value into return Task<value> in async methods?

问题描述

我设计了以下创建记录的方法。

I've designed the following method to create records.

public Task<Guid> NotAwaited()
{
  Account account = new Account();
  Context.Accounts.Add(account);
  Context.SaveChangesAsync();
  return new Task<Guid>(() => account.Id);
}

然后，我意识到目前存在储蓄未完成的风险归还GUID。因此，我添加了 await ，这要求我用 async 装饰方法签名。在此之后，我遇到了一个错误，要求返回的内容使用更简单的语法，就像这样。

Then, I realized that there's risk of the saving not being finished at the moment of returning the guid. So I've added await, which required me to decorate the method signature with async. Upon that, I got an error demanding a simpler syntax of what's being returned, like this.

public async Task<Guid> Awaited()
{
  Account account = new Account();
  Context.Accounts.Add(account);
  await Context.SaveChangesAsync();
  return account.Id;
}

我知道 account.Id 部分以某种方式转换为任务。我只是不确定如何。感觉好像是黑魔法（据我所知不是）。

I understand that the account.Id part gets converted to a task somehow. I'm just uncertain how. It feel like if it's black magic (which I understand it isn't).

是否存在隐式转换？还是我仍然不正确地执行异步调用？

Is there an implicit conversion? Or am I still performing the asynchronous call improperly?

推荐答案

您可以想到 async 作为将结果（返回值和异常）包装到 Task< T> 中。

You can think of async as wrapping results (both return values and exceptions) into a Task<T>.

等待解开结果（提取返回值或引发异常）。

Likewise, await unwraps the results (extracting the return value or raising an exception).

我有一个 async 简介进一步详细介绍，我建议 async 最佳做法作为后续措施。附带说明一下，永远不要使用 Task 构造函数。

I have an async intro that goes into more detail, and I recommend async best practices as followup to that. On a side note, you should never use the Task constructor.

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