如何在Swift中将“索引”转换为“ Int”类型？

我想将字符串中包含的字母的索引转换为整数值。尝试读取头文件，但找不到的类型Index，尽管它似乎符合ForwardIndexType使用方法的协议（例如distanceTo）。

var letters = "abcdefg"
let index = letters.characters.indexOf("c")!

// ERROR: Cannot invoke initializer for type 'Int' with an argument list of type '(String.CharacterView.Index)'
let intValue = Int(index)  // I want the integer value of the index (e.g. 2)

任何帮助表示赞赏。

编辑/更新：

Xcode 11•Swift 5.1或更高版本

extension StringProtocol {
    func distance(of element: Element) -> Int? { firstIndex(of: element)?.distance(in: self) }
    func distance<S: StringProtocol>(of string: S) -> Int? { range(of: string)?.lowerBound.distance(in: self) }
}

extension Collection {
    func distance(to index: Index) -> Int { distance(from: startIndex, to: index) }
}

extension String.Index {
    func distance<S: StringProtocol>(in string: S) -> Int { string.distance(to: self) }
}

游乐场测试

let letters = "abcdefg"

let char: Character = "c"
if let distance = letters.distance(of: char) {
    print("character \(char) was found at position #\(distance)")   // "character c was found at position #2\n"
} else {
    print("character \(char) was not found")
}

let string = "cde"
if let distance = letters.distance(of: string) {
    print("string \(string) was found at position #\(distance)")   // "string cde was found at position #2\n"
} else {
    print("string \(string) was not found")
}

斯威夫特4

var str = "abcdefg"
let index = str.index(of: "c")?.encodedOffset // Result: 2

注意：如果String包含相同的多个字符，则只会从左边获得最接近的一个字符

var str = "abcdefgc"
let index = str.index(of: "c")?.encodedOffset // Result: 2

encodedOffset已从Swift 4.2中弃用。

弃用消息： encodedOffset已被弃用，因为大多数常用用法都不正确。使用utf16Offset(in:)以达到相同的行为。

所以我们可以这样使用utf16Offset(in:)：

var str = "abcdefgc"
let index = str.index(of: "c")?.utf16Offset(in: str) // Result: 2

要基于index执行字符串操作，您不能使用传统的索引数字方法来执行。因为swift.index是由indexs函数检索的，并且不是Int类型。即使String是字符数组，我们仍然无法按索引读取元素。

这真令人沮丧。

因此，要创建字符串的每个偶数字符的新子字符串，请检查以下代码。

let mystr = "abcdefghijklmnopqrstuvwxyz"
let mystrArray = Array(mystr)
let strLength = mystrArray.count
var resultStrArray : [Character] = []
var i = 0
while i < strLength {
    if i % 2 == 0 {
        resultStrArray.append(mystrArray[i])
      }
    i += 1
}
let resultString = String(resultStrArray)
print(resultString)

输出：acegikmoqsuwy

提前致谢

这是一个扩展，使您可以使用Ints而不是String.Index值来访问子字符串的边界：

import Foundation

/// This extension is available at
/// https://gist.github.com/zackdotcomputer/9d83f4d48af7127cd0bea427b4d6d61b
extension StringProtocol {
    /// Access the range of the search string as integer indices
    /// in the rendered string.
    /// - NOTE: This is "unsafe" because it may not return what you expect if
    ///     your string contains single symbols formed from multiple scalars.
    /// - Returns: A `CountableRange<Int>` that will align with the Swift String.Index
    ///     from the result of the standard function range(of:).
    func countableRange<SearchType: StringProtocol>(
        of search: SearchType,
        options: String.CompareOptions = [],
        range: Range<String.Index>? = nil,
        locale: Locale? = nil
    ) -> CountableRange<Int>? {
        guard let trueRange = self.range(of: search, options: options, range: range, locale: locale) else {
            return nil
        }

        let intStart = self.distance(from: startIndex, to: trueRange.lowerBound)
        let intEnd = self.distance(from: trueRange.lowerBound, to: trueRange.upperBound) + intStart

        return Range(uncheckedBounds: (lower: intStart, upper: intEnd))
    }
}

请注意，这可能会导致怪异，这就是Apple选择加倍努力的原因。（尽管这是一个值得商design的设计决策-通过使其变得困难来隐藏危险的东西...）

您可以从Apple的String文档中了解更多信息，但tldr的原因是这些“索引”实际上是特定于实现的。它们代表OS渲染后到字符串中的索引，因此可以根据所使用的Unicode规范版本在OS与OS之间切换。这意味着按索引访问值不再是固定时间的操作，因为必须对数据运行UTF规范才能确定字符串中的正确位置。这些索引也不会与NSString生成的值对齐（如果您将其桥接），也不会与基础UTF标量中的索引对齐。警告开发人员。