[英]How can Ok() be both Task<IActionResult> and IActionResult?
In a Controller
in .NET Core you can return Ok()
as an IActionResult
. 在.NET Core中的
Controller
中,您可以将Ok()
作为IActionResult
。 But I do not understand how it can also return a Task<IActionResult>
. 但是我不明白它怎么也可以返回
Task<IActionResult>
。
Example: 例:
public class FooController : Controller
{
[HttpGet]
public async Task<IActionResult> OkResultAsync()
{
// This is ok. But I don't get why since OkResult != Task<IActionResult>
OkResult result = Ok();
return result;
}
[HttpGet]
public IActionResult OkResult()
{
// This is ok, and it seems logical since OkResult implements IActionResult.
OkResult result = Ok();
return result;
}
[HttpGet]
public FooResult Bar()
{
// This is ok.
return new FooResult();
}
[HttpGet]
public async Task<FooResult> BarAsync()
{
// This is not ok since FooResult != Task<FooResult>
return new FooResult();
}
}
Ok()
returns a OkResult
, which in turn implements IActionResult
. Ok()
返回一个OkResult
,后者进而实现IActionResult
。 How does .NET know how to handle it (without awaiting) if the method signature is returning a Task<IActionResult>
? 如果方法签名返回
Task<IActionResult>
,.NET如何知道如何处理(不等待)?
The async
keyword causes the compiler to take care of this automatically. async
关键字使编译器自动执行此操作。 Async methods implicitly "wrap" the return value in a Task. 异步方法隐式“包装”任务中的返回值。
async Task<int> GetNumber()
{
return 42;
}
vs 与
Task<int> GetNumber()
{
return Task.FromResult(42);
}
The async keyword is a shorthand that wraps the contents of the method in a Task. async关键字是将方法的内容包装在Task中的简写形式。 When you return inside an async method the compiler wraps it up into a Task for you.
当您在异步方法内部返回时,编译器会为您包装成一个Task。 For example these two methods are essentially the same:
例如,这两种方法本质上是相同的:
private static Task<string> Hello()
{
return new Task<string>(() => "hello");
}
private static async Task<string> AsyncHello()
{
return "hello";
}
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