在Swift中将两个字节的UInt8数组转换为UInt16

  • Z时代
  • 2024-01-10
  • 分类：问答

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

快速尝试：

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，并且正在寻找不使用临时数组的解决方案。

回答：

如果您想通过NSData，它将像这样工作：

let bytes:[UInt8] = [0x01, 0x02]
println("bytes: \(bytes)") // bytes: [1, 2]
let data = NSData(bytes: bytes, 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按主机字节顺序排列。

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

以上是 在Swift中将两个字节的UInt8数组转换为UInt16 的全部内容， 来源链接： utcz.com/qa/429880.html

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