动态参数会导致编译器来想办法回报率是动态 [英] Dynamic parameter causes compiler to think method return is dynamic

问题描述

如果我有一个动态的参数，编译器似乎沟返回类型，并认为这是动态

If I have a dynamic parameter the compiler seems to ditch the return type and think it's dynamic.

例如：

public MethodResult IsValid(object userLogin)
{     
  return new MethodResult();
}

您可能会认为：

var isValidResult = IsValid(someObject());

改为

Should read as

dynamic -> MethodResult 

但它认为它是：

But it thinks that it is:

dynamic -> dynamic

是否加入了动态参数的签名完全知道回报应该是什么，尽管停止编译返回被强类型？

Does adding a dynamic parameter to the signature completely stop the compiler from knowing what the return should be despite the return being strongly typed?

推荐答案

是的，动态的了解上的任何参数，属性或方​​法的返回类型停止编译器类型。添加显式类型转换，如：

Yes, dynamic stops the compiler from knowing the type on any parameters, properties, or method return types. Add an explicit cast like:

(MethodResult)IsValid(someObject));

这里的原因是，一旦你在C＃进入动态的世界，你正在进入后期绑定。编译器无法验证此代码，因为它不能再使用任何静态类型的分析。因此，推迟到以后。您可以通过提供静态类型转换为编译器的导克服这一点。

The reason here is that once you enter the dynamic world in C# you are going into late binding. The compiler can't verify this code because it can no longer use any static type analysis. So it defers until later. You can help overcome this by providing static casts as a guide for the compiler.

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