string – 如何将“Index”转换为Swift中的“Int”类型？

概述参见英文答案 > Finding index of character in Swift String                                    2 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，但是我找不到Index的类型，尽管它似乎符合使用方法(例如distanceTo)的ForwardIndexType协议。 var letters = "abcde 参见英文答案 > Finding index of character in Swift String2
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，但是我找不到Index的类型，尽管它似乎符合使用方法(例如distanceto)的ForwardindexType协议。

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)

任何帮助是赞赏。

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

let intValue = letters.startIndex.distanceto(index)

您还可以使用一种方法扩展字符串，以返回字符串中第一个出现的字符串，如下所示：

extension String {    func indexdistanceOfFirst(character character: Character) -> Int? {        guard let index = characters.indexOf(character) else { return nil }        return startIndex.distanceto(index)    }}let letters = "abcdefg"let char: Character = "c"if let index = letters.indexdistanceOfFirst(character: char) {    print("character \(char) was found at position #\(index)")   // "character c was found at position #2\n"} else {    print("character \(char) was not found")}

Xcode 8 beta 3•Swift 3

extension String {    func indexdistance(of character: Character) -> Int? {        guard let index = characters.index(of: character) else { return nil }        return distance(from: startIndex,to: index)    }}let letters = "abcdefg"let char: Character = "c"if let index = letters.indexdistance(of: char) {    print("character \(char) was found at position #\(index)")   // "character c was found at position #2\n"} else {    print("character \(char) was not found")}

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