X =＆gt;将＆QUOT Lamdba; {抛出..}＆QUOT;推断的匹配函数功能＆LT; T，任务＆GT;在重载的方法？ [英] Lamdba of "x => { throw .. }" inferred to match Func<T,Task> in overloaded method?

问题描述

我不明白，为什么C＃结束了在以下LINQPad code执行一个不正确的扩展方法：

I don't understand why C# ends up executing an incorrect extension method in the following LINQPad code:

void Main()
{
    // Actual: Sync Action
    "Expected: Sync Action".Run(x => { x.Dump(); });

    // Actual: Async Task
    "Expected: Async Task".Run(async x => { await System.Threading.Tasks.Task.Run(() => x.Dump()); });

    // Actual: Async Task!!
    "Expected: Sync Action".Run(x => { throw new Exception("Meh"); });
}

static class Extensions
{
    public static void Run<T>(this T instance, Action<T> action)
    {
        "Actual: Sync Action".Dump();
        action(instance);
    }

    public static void Run<T>(this T instance, Func<T, System.Threading.Tasks.Task> func)
    {
        "Actual: Async Task".Dump();
        func(instance).Wait();
    }
}

有人可以帮助我了解为什么编译器认为拉姆达返回任务吗？

Can someone help me understand why does the compiler think that the lambda returns a Task here?

我期望看到的实际：同步行动，在第三调用运行（）因为没有在拉姆达表示这是一个Func键返回任务

I expected to see "Actual: Sync Action" in the third call to Run() since nothing in the lambda indicates that this is a Func returning Task.

推荐答案

这是一个简单的重载问题。显然，在lambda X =＆GT; {抛出新的异常（咩）; } 可以转换到任何一个动作＆LT; T＆GT; 或一个 Func键＆LT; T，SomeNonVoidType＆GT; （以及许多其它代表类型无关这个问题）。它只是说美元的这种字母p $ PFER后者C＃的重载决策规则。

This is simply an overload resolution issue. Clearly, the lambda x => { throw new Exception("Meh"); } can be converted to either an Action<T> or to a Func<T, SomeNonVoidType> (as well as to many other delegate types irrelevant to this question). It's simply C#'s overload resolution rules that prefer the latter in this case.

下面是一个更重presentative例如：

Here's a more representative example:

void Main()
{
    // Output: Func<T, int>
    "Test".WhatsThis(x => { throw new Exception("Meh"); });
}

static class Extensions
{
    public static void WhatsThis<T>(this T dummy, Action<T> action)
    {
       "Action<T>".Dump();
    }
    public static void WhatsThis<T>(this T dummy, Func<T, int> func)
    {
       "Func<T, int>".Dump();
    }
}

至于为什么的是这样的话，我不是100％肯定，但随便看的规范显示我的下面可能的解释（重点煤矿）：

As for why this is the case, I'm not 100% sure, but a casual look at the spec shows me the below likely explanation (emphasis mine):

7.5.3重载

[...]

7.5.3.3从EX pression更好的转换

7.5.3.3 Better conversion from expression

由于一个隐式转换C1，它把来自一名前pression E至一个类型T1和隐式转换C2，它把来自一名前pression E至一个类型T2，C1是一个比C2更好的转换如果发生以下中的至少一个成立：

Given an implicit conversion C1 that converts from an expression E to a type T1, and an implicit conversion C2 that converts from an expression E to a type T2, C1 is a better conversion than C2 if at least one of the following holds:

[...]

•E为匿名函数，T1或者是委托类型D1或前pression树型   防爆pression＆LT; D​​1＆GT; ，T2要么是委托类型D2或前pression树   键入防爆pression＆LT; D​​2＆GT; 键，下列情况之一成立：

• E is an anonymous function, T1 is either a delegate type D1 or an expression tree type Expression<D1>, T2 is either a delegate type D2 or an expression tree type Expression<D2> and one of the following holds:

[...]

•D1具有返回类型Y和D2是无效的返回

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