带有可选参数的类初始化 - 不可能？

Question

谁能帮我理解为什么我无法将 middleName 参数设置为一个值？我已经将它声明为一个可选的，但是每当我尝试初始化它时，我总是在 getFullName() 函数中的保护语句中运行 nil。

var firstName: String
var middleName: String?
var lastName: String

init(firstName: String, middleName: String, lastName: String)
{
    self.firstName = firstName
    self.middleName? = middleName
    self.lastName = lastName
}

func getFullName() -> String
{        
    guard let midName = self.middleName else
    {
        return "\(firstName) \(lastName)"
    }

    return "\(firstName) \(midName) \(lastName)"
}

let person = Person(firstName: "Foo", middleName: "Bar", lastName: "FooBar")

打印时的返回值为“Foo Foobar”。这是为什么？

Answer 1

在分配属性时，您不需要（或不想）解开该属性。你直接引用就行了；

init(firstName: String, middleName: String, lastName: String)
{
    self.firstName = firstName
    self.middleName = middleName
    self.lastName = lastName
}

当middleName 为nil（最初是这样）时，条件解包self.middleName? 失败，因此不执行分配。

有条件地展开 self.middleName 实际上是一种简写方式

if self.middleName != nil {
    self.middleName = middleName
}

由于middleName 是可选的，那么您可能希望初始化参数也可选：

init(firstName: String, middleName: String?=nil, lastName: String)
{
    self.firstName = firstName
    self.middleName = middleName
    self.lastName = lastName
}

Answer 2

如果一个属性是可选的，你必须让init方法中的相应参数也是可选的，否则将属性声明为可选是没有意义的。

init(firstName: String, middleName: String?, lastName: String)
{
    self.firstName = firstName
    self.middleName = middleName
    self.lastName = lastName
}

您甚至可以将nil 指定为默认值 以便能够省略参数

init(firstName: String, middleName: String? = nil, lastName: String)
{
    self.firstName = firstName
    self.middleName = middleName
    self.lastName = lastName
}

let person = Person(firstName: "Foo", lastName: "FooBar")