如何快速将Int16转换为两个UInt8字节 [英] How in swift to convert Int16 to two UInt8 Bytes

问题描述

我有一些二进制数据，该数据将两个字节的值编码为有符号整数.

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

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

解码

这很容易-我可以使用以下命令从这些字节中提取Int16值:

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

编码

这是我遇到问题的地方.我的大多数数据规范都要求UInt，这很容易，但是我在提取组成Int16

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

问题

如何提取构成Int16值的两个字节

推荐答案

您应使用无符号整数:

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

如果要将UInt16转换为等效的Int16，则可以使用特定的初始值设定项来实现:

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)

反之亦然:

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

在您的情况下:

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

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

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