在多参数函数中展开可选项

Question

这个问题可能看起来很基础，但我试图在下面的代码中解开middleName 可选参数。我已经在这里成功解开了可选的返回，但我仍然收到Matthew Optional("Matt") Stevenson 记录。如何打开“Matt”并删除可选的返回？

func returnFullName (name: (firstName: String, middleName: String?,
                    lastName: String)) -> String? 
    {
    return ("\(name.firstName) \(name.middleName) \(name.lastName)")
}

var fullName = returnFullName(("Matthew", middleName: "Matt", lastName: "Stevenson"))

if let printFullName = fullName {
    print (printFullName)
}

Answer 1

您可以将nil coalescing operator 用于可选的middleName 属性。如果值为nil，则将打印默认的""，而如果该值包含非零字符串，则将其解包。

return ("\(name.firstName) \(name.middleName ?? "") \(name.lastName)")

但是请注意，您不必让 returnFullName(..) 函数的返回类型是可选的，因为该值永远不会为零。

func returnFullName (name: (firstName: String, middleName: String?,
    lastName: String)) -> String {
    return ("\(name.firstName) \(name.middleName ?? "") \(name.lastName)")
}

print(returnFullName(("Matthew", middleName: "Matt", lastName: "Stevenson"))) 
    // Matthew Matt Stevenson
print(returnFullName(("Matthew", middleName: nil, lastName: "Stevenson")))    
    // Matthew  Stevenson   <-- extra unwanted space 

另外，我没有看到任何明显的理由在函数签名中使用元组，而不是为不同的名称组件单独参数。因此，一个替代方案如下：

func returnFullName (firstName: String, middleName: String?,
    lastName: String) -> String {
    return "\(firstName) \(middleName ?? "") \(lastName)"
}

print(returnFullName("Matthew", middleName: "Matt", lastName: "Stevenson")) 
    // Matthew Matt Stevenson
print(returnFullName("Matthew", middleName: nil, lastName: "Stevenson"))    
    // Matthew  Stevenson   <-- extra unwanted space

最后，为了避免nil-valued middleName 的额外空间，我建议保留单行返回，而是使用带有if let - else 子句的可选绑定：

func returnFullName(firstName: String, middleName: String?, lastName: String) -> String {
    if let middleName = middleName {
        return ("\(firstName) \(middleName) \(lastName)")
    }
    else {
        return ("\(firstName) \(lastName)")
    }
}

print(returnFullName("Matthew", middleName: "Matt", lastName: "Stevenson"))
    // Matthew Matt Stevenson
print(returnFullName("Matthew", middleName: nil, lastName: "Stevenson"))
    // Matthew Stevenson

---

现在，问题涉及解开给定示例的可选变量，上面已经很好地介绍了这一点。不过，我想我会提到，您可以使用闭包和 Swifts 内置函数方法来压缩您的解决方案，如下所示：

// instead of a function: make use of a closure
let returnFullName : (String, String?, String) -> String = {
    [$0, $1, $2].flatMap{ $0 }.joinWithSeparator(" ") }

print(returnFullName("Matthew", "Matt", "Stevenson"))
    // Matthew Matt Stevenson
print(returnFullName("Matthew", nil, "Stevenson"))
    // Matthew Stevenson

Answer 2

您应该只传递 firstName、middleName 和 lastName 的参数，而不是使用元组。在这种特殊情况下，元组似乎是不必要的。

    func returnFullName(firstName: String, middleName: String?, lastName: String) -> String?
    {
        if let middleName = middleName {
            return ("\(firstName) \(middleName) \(lastName)")
        } else {
            return ("\(firstName) \(lastName)")
        }
    }

let fullName = returnFullName("Matthew", middleName: "Matt", lastName: "Stevenson")

if let printFullName = fullName {
    print(printFullName)
}