我的解决方案是在一行中，假设cadena是字符串，4是你想要的第n个位置:

let character = cadena[advance(cadena.startIndex, 4)]

简单的…我想Swift在未来的版本中会包含更多关于子字符串的内容。

我刚想出了一个巧妙的变通办法

var firstChar = Array(string)[0]

Swift 4.2或更高版本

使用String的indexes属性进行范围和部分范围下标

作为@LeoDabus nice answer的变体，我们可以为DefaultIndices添加一个额外的扩展，目的是允许我们在为后者实现自定义下标(通过Int专用范围和部分范围)时使用String的indexes属性。

extension DefaultIndices {
    subscript(at: Int) -> Elements.Index { index(startIndex, offsetBy: at) }
}

// Moving the index(_:offsetBy:) to an extension yields slightly
// briefer implementations for these String extensions.
extension String {
    subscript(range: Range<Int>) -> SubSequence {
        let start = indices[range.lowerBound]
        return self[start..<indices[start...][range.count]]
    }
    subscript(range: ClosedRange<Int>) -> SubSequence {
        let start = indices[range.lowerBound]
        return self[start...indices[start...][range.count]]
    }
    subscript(range: PartialRangeFrom<Int>) -> SubSequence {
        self[indices[range.lowerBound]...]
    }
    subscript(range: PartialRangeThrough<Int>) -> SubSequence {
        self[...indices[range.upperBound]]
    }
    subscript(range: PartialRangeUpTo<Int>) -> SubSequence {
        self[..<indices[range.upperBound]]
    }
}

let str = "foo bar baz bax"
print(str[4..<6]) // "ba"
print(str[4...6]) // "bar"
print(str[4...])  // "bar baz bax"
print(str[...6])  // "foo bar"
print(str[..<6])  // "foo ba"

感谢@LeoDabus指出我在使用索引属性作为字符串下标的(其他)替代方案的方向!

允许负指数

它总是有用的，不必总是写string[string]。长度- 1]用于在使用下标扩展名时获取最后一个字符。这(Swift 3)扩展允许负索引，范围和CountableClosedRange。

extension String {
    var count: Int { return self.characters.count }

    subscript (i: Int) -> Character {
        // wraps out of bounds indices
        let j = i % self.count
        // wraps negative indices
        let x = j < 0 ? j + self.count : j

        // quick exit for first
        guard x != 0 else {
            return self.characters.first!
        }

        // quick exit for last
        guard x != count - 1 else {
            return self.characters.last!
        }

        return self[self.index(self.startIndex, offsetBy: x)]
    }

    subscript (r: Range<Int>) -> String {
        let lb = r.lowerBound
        let ub = r.upperBound

        // quick exit for one character
        guard lb != ub else { return String(self[lb]) }

        return self[self.index(self.startIndex, offsetBy: lb)..<self.index(self.startIndex, offsetBy: ub)]
    }

    subscript (r: CountableClosedRange<Int>) -> String {
        return self[r.lowerBound..<r.upperBound + 1]
    }
}

如何使用:

var text = "Hello World"

text[-1]    // d
text[2]     // l
text[12]    // e
text[0...4] // Hello
text[0..<4] // Hell

对于更彻底的程序员:在这个扩展中包括一个防止空字符串的保护

subscript (i: Int) -> Character {
    guard self.count != 0 else { return '' }
    ...
}

subscript (r: Range<Int>) -> String {
    guard self.count != 0 else { return "" }
    ...
}

斯威夫特4.2

这个答案是理想的，因为它在一个扩展中扩展了String及其所有子序列(Substring)

public extension StringProtocol {
    
    public subscript (i: Int) -> Element {
        return self[index(startIndex, offsetBy: i)]
    }

    public subscript (bounds: CountableClosedRange<Int>) -> SubSequence {
        let start = index(startIndex, offsetBy: bounds.lowerBound)
        let end = index(startIndex, offsetBy: bounds.upperBound)
        return self[start...end]
    }
    
    public subscript (bounds: CountableRange<Int>) -> SubSequence {
        let start = index(startIndex, offsetBy: bounds.lowerBound)
        let end = index(startIndex, offsetBy: bounds.upperBound)
        return self[start..<end]
    }
    
    public subscript (bounds: PartialRangeUpTo<Int>) -> SubSequence {
        let end = index(startIndex, offsetBy: bounds.upperBound)
        return self[startIndex..<end]
    }
    
    public subscript (bounds: PartialRangeThrough<Int>) -> SubSequence {
        let end = index(startIndex, offsetBy: bounds.upperBound)
        return self[startIndex...end]
    }
    
    public subscript (bounds: CountablePartialRangeFrom<Int>) -> SubSequence {
        let start = index(startIndex, offsetBy: bounds.lowerBound)
        return self[start..<endIndex]
    }
}

使用

var str = "Hello, playground"

print(str[5...][...5][0])
// Prints ","

我想指出的是，如果你有一个很大的字符串，并且需要从中随机访问许多字符，你可能想要支付额外的内存成本，并将字符串转换为一个数组以获得更好的性能:

// Pay up front for O(N) memory
let chars = Array(veryLargeString.characters)

for i in 0...veryLargeNumber {
    // Benefit from O(1) access
    print(chars[i])
}