如何在 Swift 中将字节数组 [UInt8] 转换为十六进制字符串

Question

我找到了如何将六进制字符串转换为字节 [UInt8]，但我还没有找到如何在 Swift 中将字节 [UInt8] 转换为六进制字符串

此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
}

类似这样的结果：

"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 11 • Swift 5.1 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 DataProtocol {
    var data: Data { .init(self) }
    var hexa: String { map { .init(format: "%02x", $0) }.joined() }
}

---

"0f00ff".hexa                 // [15, 0, 255]
"0f00ff".hexa.data            // 3 bytes
"0f00ff".hexa.data.hexa       // "0f00ff"
"0f00ff".hexa.data as NSData  // <0f00ff>

---

注意：Swift 4 或 5 语法点击here

Answer 2

感谢布赖恩的例行公事。它可以方便地添加为 Swift 扩展，如下所示。

 extension Array where Element == UInt8 {
  func bytesToHex(spacing: String) -> String {
    var hexString: String = ""
    var count = self.count
    for byte in self
    {
        hexString.append(String(format:"%02X", byte))
        count = count - 1
        if count > 0
        {
            hexString.append(spacing)
        }
    }
    return hexString
}

}

调用示例：

let testData: [UInt8] = [15, 0, 255]
print(testData.bytesToHex(spacing: " "))          // 0F 00 FF

Answer 3

XCode 12-beta 6.

我知道它很晚，但我使用这个简单的例程，它给出了一个任意间距。我在 Android 上从 Java 转换它。我也有 C 和其他语言版本 - 当您不记得所有特定语言的库时，很容易从一种语言转换到另一种语言。

public static func bytesToHex(bytes: [UInt8], spacing: String) -> String
{
    var hexString: String = ""
    var count = bytes.count
    for byte in bytes
    {
        hexString.append(String(format:"%02X", byte))
        count = count - 1
        if count > 0
        {
            hexString.append(spacing)
        }
    }
    return hexString
}

它创建一个两位数的十六进制字符串，元素之间有任意间距字符串。例如，我像这样使用它来显示特征读取的结果

let charValue = [UInt8](characteristic.value ?? Data())
    print("Characteristic \(characteristic.uuid) read with value: \(Btle.bytesToHex(bytes: charValue, spacing: " "))")

输出如下所示：

Characteristic System ID read with value: 00 1C 05 FF FE FF E8 74

我在 Swift 和 iOS 方面的经验非常有限；也许经验丰富的 Swift 人可以“迅速化”它。