转换字节到UINT8（SWIFT 3）

Question

我已经写在Objective-C以下代码：

int port1 = SERVER_DEVICE_PORT; 
int port2 = SERVER_DEVICE_PORT>>8; 

Byte port1Byte[1] = {port1}; 
Byte port2Byte[1] = {port2}; 

NSData *port1Data = [[NSData alloc]initWithBytes: port1Byte length: sizeof(port1Byte)]; 
NSData *port2Data = [[NSData alloc]initWithBytes: port2Byte length: sizeof(port2Byte)]; 

我已转换它为夫特3像这样：

let port1: Int = Int(SERVER_DEVICE_PORT) 
let port2: Int = Int(SERVER_DEVICE_PORT) >> 8 

let port1Bytes: [UInt8] = [UInt8(port1)] 
let port2Bytes: [UInt8] = [UInt8(port2)] 

let port1Data = NSData(bytes: port1Bytes, length: port1) 
let port2Data = NSData(bytes: port2Bytes, length: port2) 

然而，用这种代码我收到以下错误：

如何这个问题能解决？

Answer 1

斯威夫特3，最简单的方式来获得一个32位的值最低的两个字节是

var SERVER_DEVICE_PORT : Int32 = 55056 

let data = Data(buffer: UnsafeBufferPointer(start: &SERVER_DEVICE_PORT, count: 1)) 
// or let data = Data(bytes: &SERVER_DEVICE_PORT, count: 2) 
let port1Data = data[0] 
let port2Data = data[1] 

print(port1Data, port2Data) 

这导致UInt8值，得到Data使用

let port1Data = Data([data[0]]) 
let port2Data = Data([data[1]]) 

如果 - 因某种原因 - 32位值是大端（最小地址中的最高有效字节），然后是port1Data = data[3]和port2Data = data[2]。