error: cannot convert 'String' to 'const char*' for argument '1' to 'int atoi(const char*)'

解决方法

char floatbuf[32]; // make this at least big enough for the whole string
curLongitude.tochararray(floatbuf,sizeof(floatbuf));
float f = atof(floatbuf);

private static String arabicToDecimal(String number) {
    char[] chars = new char[number.length()];
    for(int i=0;i<number.length();i++) {
        char ch = number.charat(i);
        if (ch >= 0x0660 && ch <= 0x0669)
           ch -= 0x0660 - '0';
        else if (ch >= 0x06f0 && ch <= 0x06F9)
           ch -= 0x06f0 - '0';
        chars[i] = ch;
    }
    return new String(chars);
}

解决方法

let NumberStr: String = "٢٠١٨-٠٦-٠٤"
    let Formatter = NumberFormatter()
    Formatter.locale = NSLocale(localeIdentifier: "EN") as Locale!
    let final = Formatter.number(from: NumberStr)
    if final != 0 {
        print("\(final!)")
    }

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.

let intValue = letters.startIndex.distanceto(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：

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•Swift 3

let letters = "abcdefg"
if let index = letters.characters.index(of: "c") {
    let distance = letters.distance(from: letters.startIndex,to: index)
    print("distance:",distance)
}

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")
}

Xcode 9•Swift 4

let letters = "abcdefg"
if let index = letters.index(of: "c") {
    let distance = letters.distance(from: letters.startIndex,distance)
}

extension String {
    func indexdistance(of character: Character) -> Int? {
        guard let index = index(of: character) else { return nil }
        return distance(from: startIndex,to: index)
    }
}

Swift 4中另一种可能的方法是返回index encodedOffset：

extension String {
    func encodedOffset(of character: Character) -> Int? {
        return index(of: character)?.encodedOffset
    }
    func encodedOffset(of string: String) -> Int? {
        return range(of: string)?.lowerBound.encodedOffset
    }
}

let letters = "abcdefg"
let char: Character = "c"
if let index = letters.encodedOffset(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")
}