Question

我试图在Swift中创建一个函数，它将字符串字典作为参数，并返回一个字符串元组。我希望字典中的键值对是选项，因为如果它返回的元组中的一个值是“nil”，我不希望我的程序崩溃。

func songBreakdown(song songfacts: [String?: String?]) -> (String, String, String) {
return (songfacts["title"], songfacts["artist"], songfacts["album"])
}
if let song = songBreakdown(song: ["title": "The Miracle", 
                              "artist": "U2", 
                              "album": "Songs of Innocence"]) {
println(song);
}

第一行上有一条错误消息：“类型'字符串？不符合协议'Hashable'。

我尝试为参数而不是选项...

创建键值对

func songBreak(song songfacts: [String: String]) -> (String?, String?, String?) {
    return (songfacts["title"], songfacts["artist"], songfacts["album"])
}
if let song = songBreak(song: ["title": "The Miracle", "artist": "U2", "album": "Songs of Innocence"]) {
    println(song);
}

但是有一条错误消息说：“条件绑定中的绑定值必须是可选类型。我该如何解决这个问题？

Answer 1

正如您已经想到的那样，您不能使用选项作为字典的键。所以，从你的第二次尝试开始，你得到错误，因为你的函数返回的值是一个选项元组，而不是一个可选的元组。而不是尝试解包函数返回的值，将其分配给变量，然后打开每个组件：

func songBreak(song songfacts: [String: String]) -> (String?, String?, String?) {
    return (songfacts["title"], songfacts["artist"], songfacts["album"])
}

let song = songBreak(song: ["title": "The Miracle", "artist": "U2", "album": "Songs of Innocence"])

if let title = song.0 {
    println("Title: \(title)")
}
if let artist = song.1 {
    println("Artist: \(artist)")
}
if let album = song.2 {
    println("Album: \(album)")
}

正如Rob Mayoff在评论中所说，命名元组元素是更好的风格：

func songBreak(song songfacts: [String: String]) -> (title: String?, artist: String?, album: String?) {
    return (title: songfacts["title"], artist: songfacts["artist"], album: songfacts["album"])
}

let song = songBreak(song: ["title": "The Miracle", "artist": "U2", "album": "Songs of Innocence"])

if let title = song.title {
    println("Title: \(title)")
}
if let artist = song.artist {
    println("Artist: \(artist)")
}
if let album = song.album {
    println("Album: \(album)")
}

你如何在swift“hashable”中创建字符串选项？

1 个答案: