带有重写的C#重载 [英] C# overload with override

问题描述

我肯定可以自己编写一个虚拟测试来回答这个问题，但是我想知道人们对这个问题的看法.在这里:

I can surely answer to this question by myself writing a dummy test but I want to know what people think about the question. Here it is:

当我们同时具有重载和重载时，将调用哪个方法?我只考虑类型重载而不考虑Arity重载，并且当类型重载相关时.

Which method will be call when we have both at the same time overloading and overriding? I am only considering Type overloading and not arity overloading and when Type the overload are related.

让我举个例子:

class AA {}
class BB : AA {}

class A {
    public virtual void methodA(AA anAA) { Console.Write("A:methodA(AA) called"); }
    public virtual void methodA(BB aBB) { Console.Write("A:methodA(BB) called"); }
}

class B : A {
    public override void methodA(AA anAA) { Console.Write("B:methodA(AA) called"); }
}

new B().methodA(new BB());     // Case 1
new B().methodA(new AA());     // Case 2
new B().methodA((AA)new BB()); // Case 3

您能说出情况1、2和3会发生什么吗?

Can you tell what will happen in case 1, 2, and 3?

我个人认为重载是邪恶的，没有一致的想法可以得出可预测的答案.这完全基于在compile + vm中实现的约定.

I personally think that overloadaing is evil and that there is no consistent thinking that could lead to a predictable answer. And that is completely base on a convention implemented in the compiler+vm.

如果您对为什么重载是有害的有疑问，可以阅读

If you have some doubt about why overload is evil you can read the blog post from Gilad Brach

谢谢

推荐答案

当编译器确定要调用的方法时，重写的方法将从方法集中排除.请参见成员查找算法.因此，当您在类型B上调用methodA时，将构造类型为B且名称为methodA的成员集，它的基本类型将是:

Overridden methods are excluded from method set when compiler determines which method to call. See member lookup algorithm. So, when you call methodA on type B, set of members with name methodA from type B and it's base type will be constructed:

override B.methodA(AA)
virtual A.methodA(AA)
virtual A.methodA(BB)

然后将具有ovveride修饰符的成员从集合中删除:

Then members with ovveride modifier removed from set:

virtual A.methodA(AA)
virtual A.methodA(BB)

这组方法是查找的结果.之后，过载分辨率用于定义哪个成员调用.

This group of methods is the result of lookup. After that overload resolution applied to define which member to invoke.

1. A.methodA(BB)被调用，因为其参数与参数匹配.
2. A.methodA(AA)将被选择，但是它是虚拟方法，因此实际调用转到B.method(AA)
3. 与选项2相同

1. A.methodA(BB) is invoked, because its argument matches parameter.
2. A.methodA(AA) will be chosen, but it is virtual method, so actually call goes to B.method(AA)
3. Same as option 2

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