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

问题描述

我设计了以下方法来创建记录.

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，我意识到在返回 guid 的那一刻，存在保存未完成的风险.所以我添加了await，它要求我用async 来装饰方法签名.在那之后，我收到了一个错误，要求对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 部分会以某种方式转换为任务.我只是不确定如何.感觉就像是黑魔法(我理解它不是).

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?还是我执行的异步调用仍然不正确?

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

推荐答案

您可以将 async 视为将结果(包括返回值和异常)包装到一个 Task 中代码>.

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

同样，await 解包结果(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 介绍 更详细，我推荐 async最佳实践作为后续.附带说明一下，你永远不应该使用 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.

这篇关于编译器如何将返回值转换为返回值 Task<value>在异步方法中?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！