如何在Swift中将“索引”转换为“ Int”类型？

一尘不染

如何在Swift中将“索引”转换为“ Int”类型？

swift

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)

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阅读 375

2020-07-07

共1个答案

一尘不染

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")
}

2020-07-07