给定Swift中的十六进制字符串，转换为十六进制值

Suppose I am given a string like this:

假设我得到一个这样的字符串:

D7C17A4F

How do I convert each individual character to a hex value?

如何将每个字符转换为十六进制值?

So D should be 0xD, 7 should be 0x7…

所以D应该是0xD,7应该是0x7 ......

Right now, I have each individual character represented as it's ASCII value. D is 68, 7 is 55. I'm trying to pack those two values into one byte. For example: D7 becomes 0xD7 and C1 becomes 0xC1. I can't do that using the ASCII decimal values though.

现在,我将每个单独的字符表示为ASCII值。 D是68,7是55.我试图将这两个值打包成一个字节。例如:D7变为0xD7,C1变为0xC1。我不能使用ASCII十进制值来做到这一点。

4 个解决方案

#1

A possible solution:

可能的解决方案:

let string = "D7C17A4F"

let chars = Array(string)
let numbers = map (stride(from: 0, to: chars.count, by: 2)) {
    strtoul(String(chars[$0 ..< $0+2]), nil, 16)
}

Using the approach from https://*.com/a/29306523/1187415, the string is split into substrings of two characters. Each substring is interpreted as a sequence of digits in base 16, and converted to a number with strtoul().

使用https://*.com/a/29306523/1187415中的方法,字符串将拆分为两个字符的子字符串。每个子字符串被解释为基数16中的数字序列,并转换为带有strtoul()的数字。

Verify the result:

验证结果:

println(numbers)
// [215, 193, 122, 79]

println(map(numbers, { String(format: "%02X", $0) } ))
// [D7, C1, 7A, 4F]

Update for Swift 2 (Xcode 7):

更新Swift 2(Xcode 7):

let string = "D7C17A4F"
let chars = Array(string.characters)

let numbers = 0.stride(to: chars.count, by: 2).map {
    UInt8(String(chars[$0 ..< $0+2]), radix: 16) ?? 0
}

print(numbers)

let string = "D7C17A4F"

var numbers = [UInt8]()
var from = string.startIndex
while from != string.endIndex {
    let to = from.advancedBy(2, limit: string.endIndex)
    numbers.append(UInt8(string[from ..< to], radix: 16) ?? 0)
    from = to
}

print(numbers)

The second solution looks a bit more complicated but has the small advantage that no additional chars array is needed.

第二种解决方案看起来有点复杂,但具有不需要额外的字符串阵列的小优势。

#2

Swift 3 version, modified from @Martin R's answer. This variant also accepts incoming string with odd length.

Swift 3版本,改编自@Martin R的答案。此变体还接受奇数长度的传入字符串。

let string = "D7C17A4F"

let chars = Array(string.characters)
let numbers = stride(from: 0, to: chars.count, by: 2).map() {
    strtoul(String(chars[$0 ..< min($0 + 2, chars.count)]), nil, 16)
}

#3

My variation of @martin-r answer:

我对@ martin-r的变化回答:

extension String {

    func hexToByteArray() -> [UInt8] {
        let byteCount = self.utf8.count / 2
        var array = [UInt8](count: byteCount, repeatedValue: 0)
        var from = self.startIndex
        for i in 0..<byteCount {
            let to = from.successor()
            let sub = self.substringWithRange(from...to)
            array[i] = UInt8(sub, radix: 16) ?? 0
            from = to.successor()
        }
        return array
    }

}

#4

here is the more generic, "pure swift" approach (no Foundation required :-))

这里是更通用的,“纯粹的快速”方法(没有基金会要求:-))

extension UnsignedInteger {
    var hex: String {
        var str = String(self, radix: 16, uppercase: true)
        while str.characters.count < 2 * MemoryLayout<Self>.size {
            str.insert("0", at: str.startIndex)
        }
        return str
    }
}

extension Array where Element: UnsignedInteger {
    var hex: String {
        var str = ""
        self.forEach { (u) in
            str.append(u.hex)
        }
        return str
    }
}

let str = [UInt8(1),22,63,41].hex  // "01163F29"
let str2 = [UInt(1),22,63,41].hex  // "00000000000000010000000000000016000000000000003F0000000000000029"

extension String {
    func toUnsignedInteger<T:UnsignedInteger>()->[T]? {
        var ret = [T]()
        let nibles = MemoryLayout<T>.size * 2
        for i in stride(from: 0, to: characters.count, by: nibles) {
            let start = self.index(startIndex, offsetBy: i)
            guard let end = self.index(start, offsetBy: nibles, limitedBy: endIndex),
                let ui = UIntMax(self[start..<end], radix: 16) else { return nil }
            ret.append(T(ui))
        }
        return ret
    }
}

let u0:[UInt8]? = str.toUnsignedInteger()                   // [1, 22, 63, 41]
let u1 = "F2345f".toUnsignedInteger() as [UInt8]?           // [18, 52, 95]
let u2 = "12345f".toUnsignedInteger() as [UInt16]?          // nil
let u3 = "12345g".toUnsignedInteger() as [UInt8]?           // nil
let u4 = "12345f".toUnsignedInteger() as [UInt]?            // nil
let u5 = "12345678".toUnsignedInteger() as [UInt8]?         // [18, 52, 86, 120]
let u6 = "12345678".toUnsignedInteger() as [UInt16]?        // [4660, 22136]
let u7 = "1234567812345678".toUnsignedInteger() as [UInt]?  // [1311768465173141112]

It is very easily to do the same for SignedInteger as well, but better approach will be to map results to signed type

对于SignedInteger也很容易做同样的事情,但更好的方法是将结果映射到有符号类型

let u8 = u1?.map { Int8(bitPattern: $0) }                    // [-14, 52, 95]

#1