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 ","

在Swift 5中，不扩展字符串:

var str = "ABCDEFGH"
for char in str {
if(char == "C") { }
}

以上Swift代码与Java代码相同:

int n = 8;
var str = "ABCDEFGH"
for (int i=0; i<n; i++) {
if (str.charAt(i) == 'C') { }
}

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)!

到目前为止，下标(_:)不可用。就像我们不能这么做一样

str[0] 

与字符串。我们必须提供"String "。指数” 但是，我们如何给出我们自己的索引号呢，我们可以用，

string[str.index(str.startIndex, offsetBy: 0)]

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])
}