C# 我必须做什么才能使我的方法可等待？
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What must I do to make my methods awaitable?
提问by B. Clay Shannon
How can I roll my own async awaitable methods?
如何推出自己的异步等待方法？
I see that writing an async method is easy as pie in some cases:
我发现在某些情况下编写异步方法很容易：
private async Task TestGeo()
{
Geolocator geo = new Geolocator();
Geoposition pos = await geo.GetGeopositionAsync();
double dLat = pos.Coordinate.Latitude;
double dLong = pos.Coordinate.Latitude;
}
...but sadly also see that not just any method can be made async willy-nilly, though; to wit: this doesn't work:
...但遗憾的是，也不是任何方法都可以随意异步；即：这不起作用：
private async Task TestAsyncAwait()
{
int i = await TaSLs_Classes.TASLsUtils.GetZoomSettingForDistance(5);
}
...it stops me with the compile error, "Cannot await int"; a hint at design time similarly tells me, "type 'int' is not awaitable"
...它阻止我编译错误，“不能等待 int”；设计时的提示同样告诉我，“类型 'int' 不可等待”
I also tried this with the same results:
我也尝试过，结果相同：
Task<int> i = await TaSLs_Classes.TASLsUtils.GetZoomSettingForDistance(5);
What must I do to make my methods awaitable?
我必须做什么才能使我的方法可等待？
UPDATE
更新
As Linebacker and S. Cleary indicated (any relation to that cat who used to be on KNBR?), this works:
正如 Linebacker 和 S. Cleary 指出的（与曾经在 KNBR 上的那只猫有任何关系？），这是有效的：
int i = await Task.Run(() => TaSLs_Classes.TASLsUtils.GetZoomSettingForDistance(5));
...that is to say, it compiles-- but it never "moves."
...也就是说，它会编译——但它永远不会“移动”。
At runtime, it tells me I should "await" the CALL to TestAsyncAwait(), but if I do that, it doesn't compile at all...
在运行时，它告诉我应该“等待”对 TestAsyncAwait() 的 CALL，但如果我这样做，它根本不会编译......
采纳答案by Stephen Cleary
You only needto return an awaitable. Task/Task<TResult>is a common choice; Tasks can be created using Task.Run(to execute code on a background thread) or TaskCompletionSource<T>(to wrap an asynchronous event).
你只需要返回一个 awaitable。Task/Task<TResult>是一个常见的选择；Tasks 可以使用Task.Run（在后台线程上执行代码）或TaskCompletionSource<T>（包装异步事件）来创建。
Read the Task-Based Asynchronous Patternfor more information.
阅读基于任务的异步模式了解更多信息。
回答by Mayank
You need to either consume the awaited return or return Task<Type of await call>
您需要消耗等待的返回或返回 Task<Type of await call>
private async Task<Geoposition> TestGeo() { Geolocator geo = new Geolocator(); return await geo.GetGeopositionAsync(); }private async Task<int> TestAsyncAwait() { return await TaSLs_Classes.TASLsUtils.GetZoomSettingForDistance(5); }
private async Task<Geoposition> TestGeo() { Geolocator geo = new Geolocator(); return await geo.GetGeopositionAsync(); }private async Task<int> TestAsyncAwait() { return await TaSLs_Classes.TASLsUtils.GetZoomSettingForDistance(5); }
回答by p3n
var something = Task<int>.Factory.StartNew(() => 0);
something.Wait();
int number = something.Result;
回答by Jeson Martajaya
Your method
你的方法
private async Task TestAsyncAwait()
{
int i = await TaSLs_Classes.TASLsUtils.GetZoomSettingForDistance(5);
}
should be written like this
应该这样写
private async Task TestAsyncAwait()
{
Task<int> t = new Task<int>(() =>
{
return TaSLs_Classes.TASLsUtils.GetZoomSettingForDistance(5);
});
t.Start();
await t;
}
If you need to return the int, replace the Task type:
如果需要返回 int，请替换 Task 类型：
private async Task<int> TestAsyncAwait()
{
Task<int> t = new Task<int>(() =>
{
return TaSLs_Classes.TASLsUtils.GetZoomSettingForDistance(5);
});
t.Start();
return await t;
}
More explanation here.
更多解释在这里。
