'String'不能转换为'DictionaryIndex <String,[(String)]？>'编译错误,具有非常基本的字典用法

Question

我正在尝试创建一个非常简单的字典,将字符串映射到Swift中的字符串数组.代码如下所示:

class FirstViewController: UIViewController {

    var characters:[String] = []
    var adjacency = [String : [String]?]()

    override func viewDidLoad() {
        super.viewDidLoad()

        characters = loadCharacters()
        adjacency = loadAdjacency()

        var character:String = characters[0]
        var adj:[String] = adjacency[character] // This line gives the first compiler error
        adj = adjacency["a"] // This line gives the second compiler error

        println(adj)
    }

    func loadCharacters() -> [String] {
        return ["a", "b", "c"]
    }

    func loadAdjacency() -> [String : [String]?] {
        return ["a": ["a", "b", "c"], "b": ["b", "c", "d"], "c": ["c", "d", "e"]]
    }
}

第一个编译器错误是: 'String' is not convertible to 'DictionaryIndex<String, [(String)]?>'

第二个编译器错误是: '(String, [(String)]?)' is not convertible to '[String]'

据我所知,这两行应该是等效和正确的 - 我从字典查找中使用字符串作为键来获取字符串数组.我哪里错了？

如果我按如下方式编写它,代码将编译并正确运行:

class SecondViewController: UIViewController {

    var keyArray:[String] = []
    var dict = [String : [String]?]()

    override func viewDidLoad() {
        super.viewDidLoad()
        // Do any additional setup after loading the view, typically from a nib.

        keyArray = ["a", "b", "c"]
        dict = ["a": ["a", "b", "c"], "b": ["b", "c", "d"], "c": ["c", "d", "e"]]

        println(dict[keyArray[1]])

    }

}

顺便说一下,我正在使用Xcode Version 6.1(6A1052d),而我正在运行OSX 10.9.4.我即将升级到优胜美地,以防万一我的设置发生了一些奇怪的事情,但我想我只是遗漏了一些明显的东西.

Answer 1

错误消息在这里不是很有用,它根本不涉及实际问题....

解决方案1:

var adj/*:[String]??*/ = adjacency[character] // adj is Optional<Optional<[String]>>
adj = adjacency["a"]
println(adj) // Optional(Optional(["a", "b", "c"]))

解决方案2:

var adj/*:[String]*/ = adjacency[character]!! // adj is [String]
adj = adjacency["a"]!!
println(adj) // [a, b, c]

解决方案3 - 安全的方式:

if let adj = adjacency[character] { // adj is Optional<String[]>
    if let adj2 = adj {  // adj2 is String[]
        println(adj2)
    }
}