如何快速将Int16转换为两个UInt8字节

Question

I have some binary data that encodes a two byte value as a signed integer.

bytes[1] = 255  // 0xFF
bytes[2] = 251  // 0xF1

Decoding

This is fairly easy - I can extract an Int16 value from these bytes with:

Int16(bytes[1]) << 8 | Int16(bytes[2])

Encoding

This is where I'm running into issues. Most of my data spec called for UInt and that is easy but I'm having trouble extracting the two bytes that make up an Int16

let nv : Int16 = -15
UInt8(nv >> 8) // fail
UInt8(nv)      // fail

Question

How would I extract the two bytes that make up an Int16 value

Answer 1

You should work with unsigned integers:

let bytes: [UInt8] = [255, 251]
let uInt16Value = UInt16(bytes[0]) << 8 | UInt16(bytes[1])
let uInt8Value0 = uInt16Value >> 8
let uInt8Value1 = UInt8(uInt16Value & 0x00ff)

If you want to convert UInt16 to bit equivalent Int16 then you can do it with specific initializer:

let int16Value: Int16 = -15
let uInt16Value = UInt16(bitPattern: int16Value)

And vice versa:

let uInt16Value: UInt16 = 65000
let int16Value = Int16(bitPattern: uInt16Value)

In your case:

let nv: Int16 = -15
let uNv = UInt16(bitPattern: nv)

UInt8(uNv >> 8)
UInt8(uNv & 0x00ff)

Answer 2

You could use init(truncatingBitPattern: Int16) initializer:

let nv: Int16 = -15
UInt8(truncatingBitPattern: nv >> 8) // -> 255
UInt8(truncatingBitPattern: nv) // -> 241

Answer 3

I would just do this:

let a = UInt8(nv >> 8 & 0x00ff)  // 255
let b = UInt8(nv & 0x00ff)       // 241

Answer 4

extension Int16 {
    var twoBytes : [UInt8] {
        let unsignedSelf = UInt16(bitPattern: self)
        return [UInt8(truncatingIfNeeded: unsignedSelf >> 8),
                UInt8(truncatingIfNeeded: unsignedSelf)]
    }
}

var test : Int16 = -15
test.twoBytes // [255, 241]