c# - 这是斯巴达，还是？

Question

下面是一道面试题。我想出了一个解决方案，但我不确定它为什么有效。

---

问题:

在不修改 Sparta 类的情况下，编写一些代码使 MakeItReturnFalse 返回 false。

public class Sparta : Place
{
    public bool MakeItReturnFalse()
    {
        return this is Sparta;
    }
}

---

我的解决方案:(剧透)

公共(public)类地点
{
公共(public)接口(interface)斯巴达{}
}

但为什么 MakeItReturnFalse() 中的 Sparta 引用 {namespace}.Place.Sparta 而不是 {namespace}。斯巴达?

Answer 1

But why does Sparta in MakeItReturnFalse() refer to {namespace}.Place.Sparta instead of {namespace}.Sparta?

基本上，因为名称查找规则就是这么说的。在 C# 5 规范中，相关的命名规则在第 3.8 节(“命名空间和类型名称”)中。

第一对项目符号 - 截断和注释 - 阅读:

• If the namespace-or-type-name is of the form I or of the form I<A1, ..., AK> [so K = 0 in our case]:
  • If K is zero and the namespace-or-type-name appears within a generic method declaration [nope, no generic methods]
  • Otherwise, if the namespace-or-type-name appears within a type declaration, then for each instance type T (§10.3.1), starting with the instance type of that type declaration and continuing with the instance type of each enclosing class or struct declaration (if any):
    • If K is zero and the declaration of T includes a type parameter with name I, then the namespace-or-type-name refers to that type parameter. [Nope]
    • Otherwise, if the namespace-or-type-name appears within the body of the type declaration, and T or any of its base types contain a nested accessible type having name I and K type parameters, then the namespace-or-type-name refers to that type constructed with the given type arguments. [Bingo!]
• If the previous steps were unsuccessful then, for each namespace N, starting with the namespace in which the namespace-or-type-name occurs, continuing with each enclosing namespace (if any), and ending with the global namespace, the following steps are evaluated until an entity is located:
  • If K is zero and I is the name of a namespace in N, then... [Yes, that would succeed]

所以最后的要点是选择 Sparta 的原因class 如果第一个项目符号没有找到任何东西...但是当基类 Place定义接口(interface) Sparta ，它在我们考虑 Sparta 之前被发现类。

请注意，如果您创建嵌套类型 Place.Sparta一个类而不是一个接口(interface)，它仍然编译并返回 false - 但编译器发出警告，因为它知道 Sparta 的一个实例永远不会是类 Place.Sparta 的实例.同样，如果您保留 Place.Sparta一个接口(interface)，但使 Sparta类 sealed ，你会收到警告，因为没有 Sparta实例可以实现接口(interface)。