Swift - 将UInt8字节转换为位数组

时间:2017-06-28 15:53:49

标签: arrays swift type-conversion byte bit

我正在尝试解码一个protobuff编码的消息,所以我需要将protobuff消息中的第一个字节(密钥)转换成位,这样我就可以找到字段编号。如何将UInt8(字节)转换为位数组?

伪代码

private func findFieldNum(from byte: UInt8) -> Int {
    //Byte is 0001 1010
    var fieldNumBits = byte[1] ++ byte[2] ++ byte[3] ++ byte[4] //concatentates bits to get 0011
    getFieldNum(from: fieldNumBits) //Converts 0011 to field number, 2^1 + 2^0 = 3
}

我看到了this question,它将一个位数组转换为字节数组。

4 个答案:

答案 0 :(得分:10)

这是从字节中获取Bit数组的基本功能:

func bits(fromByte byte: UInt8) -> [Bit] {
    var byte = byte
    var bits = [Bit](repeating: .zero, count: 8)
    for i in 0..<8 {
        let currentBit = byte & 0x01
        if currentBit != 0 {
            bits[i] = .one
        }

        byte >>= 1
    }

    return bits
}

这里,Bit是我定义的自定义枚举类型,如下所示:

enum Bit: UInt8, CustomStringConvertible {
    case zero, one

    var description: String {
        switch self {
        case .one:
            return "1"
        case .zero:
            return "0"
        }
    }
}

使用此设置,输出以下代码:

let byte: UInt8 = 0x1f

print(bits(fromByte: byte))

将是:

[1, 1, 1, 1, 1, 0, 0, 0]

答案 1 :(得分:1)

改善mohak的答案。具有通用功能或扩展功能,可满足的不仅仅是UInt8

enum Bit: UInt8, CustomStringConvertible {
    case zero, one

    var description: String {
        switch self {
        case .one:
            return "1"
        case .zero:
            return "0"
        }
    }
}

func bits<T: FixedWidthInteger>(fromBytes bytes: T) -> [Bit] {
    // Make variable
    var bytes = bytes
    // Fill an array of bits with zeros to the fixed width integer length
    var bits = [Bit](repeating: .zero, count: T.bitWidth)
    // Run through each bit (LSB first)
    for i in 0..<T.bitWidth {
        let currentBit = bytes & 0x01
        if currentBit != 0 {
            bits[i] = .one
        }

        bytes >>= 1
    }

    return bits
}

extension FixedWidthInteger {
    var bits: [Bit] {
        // Make variable
        var bytes = self
        // Fill an array of bits with zeros to the fixed width integer length
        var bits = [Bit](repeating: .zero, count: self.bitWidth)
        // Run through each bit (LSB first)
        for i in 0..<self.bitWidth {
            let currentBit = bytes & 0x01
            if currentBit != 0 {
                bits[i] = .one
            }

            bytes >>= 1
        }

        return bits
    }
}

答案 2 :(得分:0)

您可以执行一些按位运算来获取索引1-4处的位值。

// An example byte
let a: UInt8 = 0b00011010

// Extract the bits (using 0b01111000 as a bitmask) 
// in indices 1 to 4, then shift right by 3 places
// to remove indices 5 to 7
var b = (a & 0b01111000) >> 3 

// In this case, b is now 3 (or 11 in binary)
print(b)

答案 3 :(得分:0)

较短的一个:

enum Bit { case zero, one }

func bit(_ i: Int, of uint8: UInt8) -> Bit {
  let first8PowersOf2 = (0...7).map { return UInt8(1) << $0 }
  return (uint8 & first8PowersOf2[i] != 0) ? Bit.one : Bit.zero
}

func bitsFrom(_ uint8: UInt8) -> [Bit] {
  return Array((0...7)).reversed().map { bit($0, of: uint8) }
}

我们可以将其用作:

let fromUInt8  = bitsFrom(42)
let fromBinary = bitsFrom(0b01011010)
let fromHexa   = bitsFrom(0xFF)