[英]Using Task.FromResult to implicitly convert a Task<T> to a Task<X> where T : X?
Task<IDictionary<double, double>> GetTaskDict()
{
return Task.FromResult(new Dictionary<double, double> () );
}
This code does not compile because we can not convert between Task<Dictionary<double, double>> to Task<IDictionary<double, double>>.此代码无法编译,因为我们无法将 Task<Dictionary<double, double>> 转换为 Task<IDictionary<double, double>>。 Why does this not work, and is it possible to make this type of call work?为什么这不起作用,是否可以使这种类型的呼叫起作用? This opposed to a method like this which compiles这与像这样编译的方法相反
IDictionary<double, double> GetTaskDict()
{
return new Dictionary<double, double> ();
}
return Task.FromResult<IDictionary<double, double>>(new Dictionary<double, double>());
This is because Task<T>
is not covariant, so a Task<Foo>
is not interchangeable with a Task<IFoo>
even if Foo: IFoo
.这是因为Task<T>
不是协变的,因此即使Foo: IFoo
, Task<Foo>
也不能与Task<IFoo>
互换。
You may also wish to consider using ValueTask<T>
if you're going to have lots of results which are available synchronously - it is significantly cheaper in that scenario.如果您要获得大量同步可用的结果,您可能还希望考虑使用ValueTask<T>
- 在这种情况下它要便宜得多。 You can even use the implicit new()
usage in that scenario:您甚至可以在该场景中使用隐式new()
用法:
ValueTask<IDictionary<double, double>> GetTaskDict()
{
return new(new Dictionary<double, double>());
}
(here, new(...)
is interpreted as new ValueTask<IDictionary<double, double>>(...)
from the declared return-type of the method) (这里, new(...)
被解释为new ValueTask<IDictionary<double, double>>(...)
从方法的声明返回类型)
Suppose it were possible.假设这是可能的。
Then someone could do this:然后有人可以这样做:
Task<IDictionary<double, double>> myTaskDict = new Task<Dictionary<double, double>>();
Task.SetResult(myIDictionaryWhichIsNotActuallyADictionary);
In other words, as @MarcGravell has said, Task
is not covariant, and for very good reason.换句话说,正如@MarcGravell 所说, Task
不是协变的,并且有很好的理由。
You can specify the type explicitly like this though:您可以像这样显式指定类型:
Task<IDictionary<double, double>> GetTaskDict()
{
return Task.FromResult<IDictionary<double, double>>(new Dictionary<double, double>());
}
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