如何在Swift中将"Index"转换为"Int"类型？

Question

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,但我找不到类型Index,虽然它似乎符合协议ForwardIndexType与方法(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.

Answer 1

Xcode 9•Swift 4或更高版本

extension StringProtocol {
    func indexDistance(of element: Element) -> Int? { firstIndex(of: element)?.distance(in: self) }
    func indexDistance<S: StringProtocol>(of string: S) -> Int? { range(of: string)?.lowerBound.distance(in: self) }
}
extension String.Index {
    func distance<S: StringProtocol>(in string: S) -> Int { string.distance(from: string.startIndex, to: self) }
}

extension Collection where Element: Equatable {
    func indexDistance(of element: Element) -> Int? {
        guard let index = firstIndex(of: element) else { return nil }
        return distance(from: startIndex, to: index)
    }
}
extension StringProtocol {
    func indexDistance<S: StringProtocol>(of string: S) -> Int? {
        guard let index = range(of: string)?.lowerBound else { return nil }
        return distance(from: startIndex, to: index)
    }
}

let letters = "abcdefg"

let char: Character = "c"
if let distance = letters.indexDistance(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")
}

Swift 4中另一种可能的方法是返回索引__CODE__:

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

extension StringProtocol {
    func indexDistance(of element: Element) -> Int? { firstIndex(of: element)?.distance(in: self) }
    func indexDistance<S: StringProtocol>(of string: S) -> Int? { range(of: string)?.lowerBound.distance(in: self) }
}
extension String.Index {
    func distance<S: StringProtocol>(in string: S) -> Int { string.distance(from: string.startIndex, to: self) }
}

extension Collection where Element: Equatable {
    func indexDistance(of element: Element) -> Int? {
        guard let index = firstIndex(of: element) else { return nil }
        return distance(from: startIndex, to: index)
    }
}
extension StringProtocol {
    func indexDistance<S: StringProtocol>(of string: S) -> Int? {
        guard let index = range(of: string)?.lowerBound else { return nil }
        return distance(from: startIndex, to: index)
    }
}

Xcode 8•Swift 3

let letters = "abcdefg"

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

extension StringProtocol {
    func indexDistance(of element: Element) -> Int? { firstIndex(of: element)?.distance(in: self) }
    func indexDistance<S: StringProtocol>(of string: S) -> Int? { range(of: string)?.lowerBound.distance(in: self) }
}
extension String.Index {
    func distance<S: StringProtocol>(in string: S) -> Int { string.distance(from: string.startIndex, to: self) }
}

老答案

您需要使用与原始字符串起始索引相关的distanceTo(index)方法:

extension Collection where Element: Equatable {
    func indexDistance(of element: Element) -> Int? {
        guard let index = firstIndex(of: element) else { return nil }
        return distance(from: startIndex, to: index)
    }
}
extension StringProtocol {
    func indexDistance<S: StringProtocol>(of string: S) -> Int? {
        guard let index = range(of: string)?.lowerBound else { return nil }
        return distance(from: startIndex, to: index)
    }
}

您还可以使用方法扩展String以返回字符串中第一次出现的字符,如下所示:

let letters = "abcdefg"

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

Answer 2

适用于Xcode 13和Swift 5

let myString = "Hello World"

if let i = myString.firstIndex(of: "o") {
  let index: Int = myString.distance(from: myString.startIndex, to: i)
  print(index) // Prints 4
}

该函数func distance(from start: String.Index, to end: String.Index) -> String.IndexDistance返回一个IndexDistance只是一个typealiasforInt

Answer 3

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，则只会从左侧获得最接近的一个字符

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