c# - 为什么从接口(interface)继承的属性会变成虚拟的？

Question

假设我有一个接口(interface)和两个类，其中一个类实现了这个接口(interface):

interface IAAA
{
    int F1 { get; set; }
}

class AAA1
{
    public int F1 { get; set; }
    public int F2 { get; set; }
}

class AAA2 : IAAA
{
    public int F1 { get; set; }
    public int F2 { get; set; }
}

在 AAA2 类中，属性 F1 是从接口(interface) IAAA“继承”(我不确定)，然后我使用反射来检查一个属性是否是虚拟的:

Console.WriteLine("AAA1 which does not implement IAAA");
foreach (var prop in typeof(AAA1).GetProperties())
{
    var virtualOrNot = prop.GetGetMethod().IsVirtual ? "" : " not";
    Console.WriteLine($@"{prop.Name} is{virtualOrNot} virtual");
}

Console.WriteLine("AAA2 which implements IAAA");
foreach (var prop in typeof(AAA2).GetProperties())
{
    var virtualOrNot = prop.GetGetMethod().IsVirtual ? "" : " not";
    Console.WriteLine($"{prop.Name} is{virtualOrNot} virtual");
}

输出是:

AAA1 which does not implement IAAA
F1 is not virtual
F2 is not virtual
AAA2 which implements IAAA
F1 is virtual
F2 is not virtual

有什么原因吗？

Answer 1

从 remarks section of MS docs 开始:

A virtual member may reference instance data in a class and must be referenced through an instance of the class... The common language runtime requires that all methods that implement interface members must be marked as virtual; therefore, the compiler marks the method virtual final

如果您需要确定此方法是否可重写，那么检查 IsVirtual 是不够的，您还需要检查 IsFinal 是否为 false。

这是执行此检查的扩展方法:

public static bool IsOverridable(this MethodInfo method)
    => method.IsVirtual && !method.IsFinal;