Question

我正在尝试将传统的Objective-C函数转换为Swift并使用它运行到一些障碍中。这是Objective-C函数 -

UIImageView

我无法找到Swift等效内容的几行

- (SHCardType)cardType:(NSString *)cardNum {
    if (cardNum.length <= 0) {
        return SHCTUnknown;
    }
    const char *s = [cardNum UTF8String];

    if (sizeof(cardTypes[0]) == 0) {
        return SHCTUnknown;
    }
    int cardsCount = sizeof(cardTypes) / sizeof(cardTypes[0]);
    for (int i = 0; i < cardsCount; i++) {
        int dig = log10(cardTypes[i].max) + 1;
        int n = 0;

        for (int j = 0; j < dig; j++) {
            n = n * 10 + (s[j] - '0');
        }

        if ((cardTypes[i].min <= n) && (n <= cardTypes[i].max)) {
            SHCardType cardType = cardTypes[i].cardType;
            int numLength = 16;
            if (cardType == SHCTAmericanExpress) {
                // Length is 15. Should confirm its type before all card number is entered.
                return cardType;
            } else if (cardType == SHCTDinersClubCarteBlanche || cardType == SHCTDinersClubenRoute || cardType == SHCTDinersClubInternational || cardType == SHCTDinersClubUnitedStatesCanada) {
                numLength = 14;
            }

            // For DinersClubCarteBlanche card, the length may be 16.
            if (numLength == cardNum.length || cardType == SHCTDinersClubCarteBlanche || cardType == SHCTDinersClubInternational) {
                return cardType;
            }
        }
    }

    return SHCTUnknown;
}

我是否可以使用Swift函数将String转换为UTF-8

const char *s = [cardNum UTF8String];

我如何代表角色＆＃39; 0＆＃39;在迅速。我可以使用＆＃39; - ＆＃39;操作数呢？

任何指针都会受到赞赏。

Answer 1

试试这个：

let s = cardNum.cStringUsingEncoding(NSUTF8StringEncoding)! // s is an Array<Int8>
...
let CHAR_ZERO : Int8 = 48
n = n * 10 + (s[j] - CHAR_ZERO);

将字符转换为Swift中ASCII表中的整数值很麻烦。所以你也可以把它定义为一个整数常量。

修改：如果您想要一种更安全的方式进行展开：

let s : Array<Int8>
if cardNum.canBeConvertedToEncoding(NSUTF8StringEncoding) {
    s = cardNum.cStringUsingEncoding(NSUTF8StringEncoding)!
} else {
    s = []
    // Handle error
}

Answer 2

来自How can I get the Unicode code point(s) of a Character?

let ch: Character = "A"
let s = String(ch).unicodeScalars
s[s.startIndex].value // returns 65

所以也许试试：

let z = "0".unicodeScalars
s[j] - z[z.startIndex].value

让我知道它是怎么回事。

Answer 3

我不确定，但你可以试试这个：

func cardType(cardNum: String) -> SHCardType {
    if cardNum.length <= 0 {
        return SHCTUnknown
    }
    let s: Character = cardNum.UTF8String()
    if sizeof([0]) == 0 {
        return SHCTUnknown
    }
    var cardsCount: Int = sizeof() / sizeof([0])
    for var i = 0; i < cardsCount; i++ {
        var dig: Int = log10(cardTypes[i].max) + 1
        var n: Int = 0
        for var j = 0; j < dig; j++ {
            n = n * 10 + (s[j] - "0")
        }
        if (cardTypes[i].min <= n) && (n <= cardTypes[i].max) {
            var cardType: SHCardType = cardTypes[i].cardType
            var numLength: Int = 16
            if cardType == SHCTAmericanExpress {
                return cardType
            }
            else {
                if cardType == SHCTDinersClubCarteBlanche || cardType == SHCTDinersClubenRoute || cardType == SHCTDinersClubInternational || cardType == SHCTDinersClubUnitedStatesCanada {
                    numLength = 14
                }
            }
            if numLength == cardNum.length || cardType == SHCTDinersClubCarteBlanche || cardType == SHCTDinersClubInternational {
                return cardType
            }
        }
    }
    return SHCTUnknown
}

将Swift String转换为UTF-8并使用char的ASCII值进行算术运算

3 个答案: