如何在Swift中将字节数组[UInt8]转换为hexa字符串

Question

我发现如何将六进制字符串转换为字节[UInt8],但我还没有找到如何将字节[UInt8]转换为Swift中的六进制字符串

这hexstring转换为string代码:

static func bytesConvertToHexstring(byte : [UInt8]) -> String {
    var string = ""

    for val in byte {
        //getBytes(&byte, range: NSMakeRange(i, 1))
        string = string + String(format: "%02X", val)
    }

    return string
}

像这样的结果:

"F063C52A6FF7C8904D3F6E379EB85714ECA9C1CB1E8DFD6CA5D3B4A991269D60F607C565C327BD0ECC0985F74E5007E0D276499E1ADB4E0C92D8BDBB46E57705B2D5390FF5CBD4ED1B850C537301CA7E"

UInt8 数组: [0, 11, 8, 15, 6, 6, 5, 8, 8, 4, 14, 14, 0, 0, 9, 12, 6, 4, 10, 6, 4, 8, 6, 2, 14, 2, 6, 13, 3, 3, 12, 4, 3, 12, 8, 13, 14, 4, 10, 1, 12, 15, 4, 0, 14, 14, 0, 8, 8, 14, 6, 15, 2, 2, 9, 15, 13, 6, 2, 6, 8, 15, 4, 2, 12, 1, 0, 13, 13, 4, 6, 0, 9, 6, 8, 2, 7, 0, 6, 1, 3, 3, 9, 15, 5, 7, 12, 8, 7, 5, 13, 14, 15, 6, 7, 6, 12, 6, 7, 7, 11, 9, 6, 0, 14, 5, 6, 14, 1, 5, 13, 10, 12, 13, 14, 2, 13, 14, 4, 7, 13, 0, 3, 10, 6, 11, 9, 12, 7, 11, 5, 3, 5, 11, 4, 9, 6, 10, 14, 0, 11, 7, 15, 9, 3, 14, 5, 1, 10, 14, 5, 6, 12, 4, 12, 14, 4, 3, 9, 8, 0]

Answer 1

Xcode 9或更高版本•Swift 4或更高版本

extension StringProtocol {
    var hexa: [UInt8] {
        var startIndex = self.startIndex
        return (0..<count/2).compactMap { _ in
            let endIndex = index(after: startIndex)
            defer { startIndex = index(after: endIndex) }
            return UInt8(self[startIndex...endIndex], radix: 16)
        }
    }
}

extension Sequence where Element == UInt8 {
    var data: Data { .init(self) }
    var hexa: String { map { .init(format: "%02x", $0) }.joined() }
}