在Swift中将两个字节的UInt8数组转换为UInt16

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使用Swift我想将字节从uint8_t数组转换为整数。

“C”示例:

  1. char bytes[2] = {0x01,0x02};
  2. NSData *data = [NSData dataWithBytes:bytes length:2];
  3. NSLog(@"data: %@",data); // data: <0102>
  4.  
  5. uint16_t value2 = *(uint16_t *)data.bytes;
  6. NSLog(@"value2: %i",value2); // value2: 513

Swift尝试:

  1. let bytes:[UInt8] = [0x01,0x02]
  2. println("bytes: \(bytes)") // bytes: [1,2]
  3. let data = NSData(bytes: bytes,length: 2)
  4. println("data: \(data)") // data: <0102>
  5.  
  6. let integer1 = *data.bytes // This fails
  7. let integer2 = *data.bytes as UInt16 // This fails
  8.  
  9. let dataBytePointer = UnsafePointer<UInt16>(data.bytes)
  10. let integer3 = dataBytePointer as UInt16 // This fails
  11. let integer4 = *dataBytePointer as UInt16 // This fails
  12. 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:
  1. let bytes:[UInt8] = [0x01,length: 2)
  2. print("data: \(data)") // data: <0102>
  3.  
  4. var u16 : UInt16 = 0 ; data.getBytes(&u16)
  5. // Or:
  6. let u16 = UnsafePointer<UInt16>(data.bytes).memory
  7.  
  8. println("u16: \(u16)") // u16: 513

或者:

  1. let bytes:[UInt8] = [0x01,0x02]
  2. let u16 = UnsafePointer<UInt16>(bytes).memory
  3. 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):

  1. let bytes: [UInt8] = [0x01,0x02]
  2. let u16 = UnsafePointer(bytes).withMemoryRebound(to: UInt16.self,capacity: 1) {
  3. $0.pointee
  4. }
  5. print("u16: \(u16)") // u16: 513

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