Question

这个问题看起来很基本，但我试图在下面的代码中展开public class TeleportReference : MonoBehaviour { private GameObject reference; void Awake () { reference = null; } public void SetReference(GameObject other){ reference = other; } public GameObject GetReference(){ return reference; }可选参数。我已经成功解开了可选返回，但我仍然记录了middleName。我如何打开＆＃34;马特＆＃34;并删除可选的退货？

Matthew Optional("Matt") Stevenson

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，则会将其解包。

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

但请注意，您不需要将returnFullName(..)函数的返回类型设置为可选，因为该值永远不会为nil。

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

此外，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分别使用参数。因此，另一种选择如下：

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，为避免nil - 值middleName的额外空间，我建议保留单行返回，而是使用带if let - else子句的可选绑定：< / p>

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，问题涉及展开给定示例的可选变量，现在已经很好地解决了这个问题。但是，我想我会提到你可以使用闭包和Swifts内置功能方法来压缩你的解决方案，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)
}

在多参数功能中展开可选项

2 个答案: