swift 2.0子字符串更新

public extension String {
    public subscript (i: Int) -> String {
        return self.substringWithRange(self.startIndex..<self.startIndex.advancedBy(i + 1))
    }

    public subscript (r: Range<Int>) -> String {
        get {
            return self.substringWithRange(self.startIndex.advancedBy(r.startIndex)..<self.startIndex.advancedBy(r.endIndex))
        }
    }

}

我刚想出了一个巧妙的变通办法

var firstChar = Array(string)[0]

Swift 2.0的Xcode 7通用种子

var text = "Hello, world!"

let firstChar = text[text.startIndex.advancedBy(0)] // "H"

对于第n个字符，将0替换为n-1。

编辑:Swift 3.0

text[text.index(text.startIndex, offsetBy: 0)]

注意，有更简单的方法来获取字符串中的某些字符

例如:let firstChar = text.characters.first

swift字符串类不提供在特定索引处获取字符的能力，因为它本身支持UTF字符。UTF字符在内存中的可变长度使得直接跳转到字符是不可能的。这意味着每次都必须手动遍历字符串。

您可以扩展String以提供一个方法，该方法将遍历字符，直到您需要的索引

extension String {
    func characterAtIndex(index: Int) -> Character? {
        var cur = 0
        for char in self {
            if cur == index {
                return char
            }
            cur++
        }
        return nil
    }
}

myString.characterAtIndex(0)!

不使用整数进行索引，只使用String.Index。主要是线性复杂度。您还可以从String中创建范围。索引并使用它们获取子字符串。

斯威夫特3.0

let firstChar = someString[someString.startIndex]
let lastChar = someString[someString.index(before: someString.endIndex)]
let charAtIndex = someString[someString.index(someString.startIndex, offsetBy: 10)]

let range = someString.startIndex..<someString.index(someString.startIndex, offsetBy: 10)
let substring = someString[range]

快2.倍

let firstChar = someString[someString.startIndex]
let lastChar = someString[someString.endIndex.predecessor()]
let charAtIndex = someString[someString.startIndex.advanceBy(10)]

let range = someString.startIndex..<someString.startIndex.advanceBy(10)
let subtring = someString[range]

请注意，不能使用从一个字符串到另一个字符串创建的索引(或范围)

let index10 = someString.startIndex.advanceBy(10)

//will compile
//sometimes it will work but sometimes it will crash or result in undefined behaviour
let charFromAnotherString = anotherString[index10]

swift 2.0子字符串更新

public extension String {
    public subscript (i: Int) -> String {
        return self.substringWithRange(self.startIndex..<self.startIndex.advancedBy(i + 1))
    }

    public subscript (r: Range<Int>) -> String {
        get {
            return self.substringWithRange(self.startIndex.advancedBy(r.startIndex)..<self.startIndex.advancedBy(r.endIndex))
        }
    }

}