我希望将包含在字符串中的字母的索引转换为整数值。尝试阅读头文件，但是我找不到Index类型，尽管它似乎符合协议ForwardIndexType并具有方法（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)
感激所有的帮助。
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) }
}
Playground testing
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")
}
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
}
函数func distance(from start: String.Index, to end: String.Index) -> String.IndexDistance返回一个IndexDistance，它只是Int的typealias
i 中。然后它计算从字符串起始索引到 i 的距离。这些都是我使用的内置方法，Swift 提供了这些方法。 - Eric33187Swift 4
var str = "abcdefg"
let index = str.index(of: "c")?.encodedOffset // Result: 2
var str = "abcdefgc"
let index = str.index(of: "c")?.encodedOffset // Result: 2
"".index(of: "")?.encodedOffset // 16。 - Leo DabusencodedOffset 自 Swift 4.2 开始已被弃用。
弃用信息：
encodedOffset 已被弃用，因为大多数常见用法是不正确的。使用 utf16Offset(in:) 可以实现相同的行为。
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
let str = ""
let index = str.firstIndex(of: "")?.utf16Offset(in: str) // 结果：8 - Leo Dabus当像这样搜索索引时
⛔️ guard let index = (positions.firstIndex { position <= $0 }) else {
如果您希望将其视为整数，编译器将其视为Array.Index。您需要向编译器提供一个提示，表明您需要一个整数。
✅ guard let index: Int = (positions.firstIndex { position <= $0 }) else {
Swift 5
您可以将其转换为字符数组，然后使用advanced(by:)将其转换为整数。
let myString = "Hello World"
if let i = Array(myString).firstIndex(of: "o") {
let index: Int = i.advanced(by: 0)
print(index) // Prints 4
}
这里有一个扩展程序，它可以让你以Int而不是String.Index值的形式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，这可能会导致一些奇怪的问题，这也是为什么苹果选择让它变得困难的原因。（虽然这是一个有争议的设计决策 - 通过使其变得困难来隐藏危险的事情...）
您可以在Apple的字符串文档中阅读更多信息，但tldr是这源于这些“索引”实际上是特定于实现的。它们代表了字符串在操作系统呈现后的索引，因此根据使用的Unicode规范版本，可以在不同的操作系统之间移动。这意味着通过索引访问值不再是常量时间操作，因为必须对数据运行UTF规范以确定字符串中的正确位置。如果您将其桥接到NSString或底层UTF标量的索引，这些索引也将无法与生成的值对齐。开发者要注意。
func rangeInt<S: StringProtocol>(of aString: S, options: String.CompareOptions = []) -> Range<Int>? {
guard let range = range(of: aString, options: options) else { return nil }
let start = distance(from: startIndex, to: range.lowerBound)
return start..<start+distance(from: range.lowerBound, to: range.upperBound)
} - Leo DabuscountableRange 并返回 CountableRange。 - Leo Dabus要根据索引执行字符串操作，您不能使用传统的索引数字方法。因为swift.index是通过indices函数检索的，并且它不属于Int类型。即使String是字符数组，我们仍然无法按索引读取元素。
这很令人沮丧。
因此，要创建字符串中每个偶数字符的新子字符串，请查看以下代码。
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
var counter = 0let result = mystr.filter { _ in
defer { counter += 1 }
return counter.isMultiple(of: 2)
} - Leo Dabusvar index = mystr.startIndexlet result = mystr.filter { _ in
defer { mystr.formIndex(after: &index) }
return mystr.distance(from: mystr.startIndex, to: index).isMultiple(of: 2)
} - Leo Dabus如果你遇到了“数组越界”错误，你可以尝试这种方法。适用于 Swift 5。
extension String{
func countIndex(_ char:Character) -> Int{
var count = 0
var temp = self
for c in self{
if c == char {
//temp.remove(at: temp.index(temp.startIndex,offsetBy:count))
//temp.insert(".", at: temp.index(temp.startIndex,offsetBy: count))
return count
}
count += 1
}
return -1
}
}
let index = letters.characters.index(of: "c")下一行let int_index = letters.characters.distance(from: letters.startIndex, to: index)翻译： Swift 3：let index = letters.characters.index(of: "c")意思是在字符串letters中找到字符 "c" 的位置。接下来的一行代码：let int_index = letters.characters.distance(from: letters.startIndex, to: index)将字符 "c" 的位置index和字符串letters的起始位置letters.startIndex之间的距离转换为整数型。 - Viktor