在Swift中将两个字节的UInt8数组转换为UInt16

发布时间：2020-12-14 05:58:28 所属栏目：百科来源：网络整理

导读：使用Swift我想将字节从uint8_t数组转换为整数。 “C”示例： char bytes[2] = {0x01,0x02};NSData *data = [NSData dataWithBytes:bytes length:2];NSLog(@"data: %@",data); // data: 0102uint16_t value2 = *(uint16_t *)data.bytes;NSLog(@"value2: %i",v

使用Swift我想将字节从uint8_t数组转换为整数。

“C”示例：

char bytes[2] = {0x01,0x02};
NSData *data = [NSData dataWithBytes:bytes length:2];
NSLog(@"data: %@",data); // data: <0102>

uint16_t value2 = *(uint16_t *)data.bytes;
NSLog(@"value2: %i",value2); // value2: 513

Swift尝试：

let bytes:[UInt8] = [0x01,0x02]
println("bytes: (bytes)") // bytes: [1,2]
let data = NSData(bytes: bytes,length: 2)
println("data: (data)") // data: <0102>

let integer1 = *data.bytes // This fails
let integer2 = *data.bytes as UInt16 // This fails

let dataBytePointer = UnsafePointer<UInt16>(data.bytes)
let integer3 = dataBytePointer as UInt16 // This fails
let integer4 = *dataBytePointer as UInt16 // This fails
let integer5 = *dataBytePointer // This fails

从Swift中的UInt8数组创建UInt16值的正确语法或代码是什么？

我对NSData版本感兴趣，并且正在寻找一个不使用临时数组的解决方案。

如果你想通过NSData去，那么它将像这样工作：

let bytes:[UInt8] = [0x01,length: 2)
print("data: (data)") // data: <0102>

var u16 : UInt16 = 0 ; data.getBytes(&u16)
// Or:
let u16 = UnsafePointer<UInt16>(data.bytes).memory

println("u16: (u16)") // u16: 513

或者：

let bytes:[UInt8] = [0x01,0x02]
let u16 = UnsafePointer<UInt16>(bytes).memory
print("u16: (u16)") // u16: 513

两种变体都假定字节是主机字节顺序。

更新Swift 3(Xcode 8)：

let bytes: [UInt8] = [0x01,0x02]
let u16 = UnsafePointer(bytes).withMemoryRebound(to: UInt16.self,capacity: 1) {
    $0.pointee
}
print("u16: (u16)") // u16: 513

（编辑：李大同）

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