How in swift to convert Int16 to two UInt8 Bytes

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

Answer 1

I would just do this:

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

Answer 2

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 3

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]

Answer 4

You could use init(truncatingBitPattern: Int16) initializer:

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