swift - Swift 3.1 中自定义初始化器的访问级别

Question

引自 The Swift Programming Language (Swift 3.1):

Custom initializers can be assigned an access level less than or equal to the type that they initialize. The only exception is for required initializers (as defined in Required Initializers). A required initializer must have the same access level as the class it belongs to.

如果是这样，为什么这段代码可以编译和工作？

private class GoofyClass {
    public init(mood: String) {}
    public required init(isCrazy: Bool) {}
}

private let shock = GoofyClass(mood: "shocked")
private let crazy = GoofyClass(isCrazy: true)

Answer 1

在 Swift 中，访问级别限制低于类/结构本身的类或结构的成员会自动降级到与类/结构相同的级别。我相信这是语言设计者经过深思熟虑的设计决定。

在你的例子中，假设类是在文件的顶层声明的(即它没有嵌套在另一个类型中)，你声明的 init 是 public 实际上是 fileprivate。

The only exception is for required initializers (as defined in Required Initializers). A required initializer must have the same access level as the class it belongs to.

这是指您不能使所需初始化程序的访问级别比其类更严格，例如

open class Foo 
{
    internal required init() // error
}