如何在Swift中将十六进制字符串转换为UInt8字节数组？

Question

我有以下代码：

var encryptedByteArray: Array<UInt8>?
do {
    let aes = try AES(key: "passwordpassword", iv: "drowssapdrowssap")
    encryptedByteArray = try aes.encrypt(Array("ThisIsAnExample".utf8))
} catch {
    fatalError("Failed to initiate aes!")
}

print(encryptedByteArray!) // Prints [224, 105, 99, 73, 119, 70, 6, 241, 181, 96, 47, 250, 108, 45, 149, 63]

let hexString = encryptedByteArray?.toHexString()

print(hexString!) // Prints e0696349774606f1b5602ffa6c2d953f

如何转换hexString回相同的UInt8字节数组？

我问的原因是因为我想通过加密的十六进制字符串与服务器通信，所以我需要将其转换回UInt8字节数组以将字符串解码为其原始格式。

Answer 1

您可以将六进制字符串转换回UInt8数组，该数组每两个六个字符重复一次，并使用UInt8基数16初始化程序从中初始化UInt8：

编辑/更新： Xcode 14•Swift 5.1

extension StringProtocol {
    var hexaData: Data { .init(hexa) }
    var hexaBytes: [UInt8] { .init(hexa) }
    private var hexa: UnfoldSequence<UInt8, Index> {
        sequence(state: startIndex) { startIndex in
            guard startIndex < self.endIndex else { return nil }
            let endIndex = self.index(startIndex, offsetBy: 2, limitedBy: self.endIndex) ?? self.endIndex
            defer { startIndex = endIndex }
            return UInt8(self[startIndex..<endIndex], radix: 16)
        }
    }
}

let string = "e0696349774606f1b5602ffa6c2d953f"
let data = string.hexaData    // 16 bytes
let bytes = string.hexaBytes  // [224, 105, 99, 73, 119, 70, 6, 241, 181, 96, 47, 250, 108, 45, 149, 63]

操场：

let hexaString = "e0696349774606f1b5602ffa6c2d953f"

let bytes = hexaString.hexa   // [224, 105, 99, 73, 119, 70, 6, 241, 181, 96, 47, 250, 108, 45, 149, 63]