如何使用C++构造函数初始化位域？

Question

首先,我不关心可移植性,并且可以安全地假设字节顺序不会改变.假设我读取了硬件寄存器值,我想将该寄存器值覆盖在位域上,这样我就可以参考寄存器中的各个字段而不使用位掩码.

编辑:修复了GMan指出的问题,并调整了代码,以便将来读者更清楚.

请参阅: Anders K.和Michael J的答案,以获得更有说服力的解决方案.

#include <iostream>

/// \class HardwareRegister
/// Abstracts out bitfields in a hardware register.
/// \warning  This is non-portable code.
class HardwareRegister
{
   public:
      /// Constructor.
      /// \param[in]  registerValue - the value of the entire register. The
      ///                             value will be overlayed onto the bitfields
      ///                             defined in this class.
      HardwareRegister(unsigned long registerValue = 0)
      {
         /// Lots of casting to get registerValue to overlay on top of the
         /// bitfields
         *this = *(reinterpret_cast<HardwareRegister*>(&registerValue));
      }


      /// Bitfields of this register.
      /// The data type of this field should be the same size as the register
      /// unsigned short for 16 bit register
      /// unsigned long for 32 bit register.
      ///
      /// \warning Remember endianess! Order of the following bitfields are
      ///          important.
      ///          Big Endian    - Start with the most signifcant bits first.
      ///          Little Endian - Start with the least signifcant bits first.
      unsigned long field1: 8;
      unsigned long field2:16;
      unsigned long field3: 8;
}; //end class Hardware


int main()
{

   unsigned long registerValue = 0xFFFFFF00;
   HardwareRegister  testRegister(registerValue);

   // Prints out for little endianess machine
   // Field 1 = 0
   // Field 2 = 65535
   // Field 3 = 255
   std::cout << "Field 1 = " << testRegister.field1 << std::endl;
   std::cout << "Field 2 = " << testRegister.field2 << std::endl;
   std::cout << "Field 3 = " << testRegister.field3 << std::endl;
}

Answer 1

Bitfields无法正常工作.您不能将标量值分配给struct完整的位域.看起来你已经知道了,因为你使用过reinterpret_cast,但由于reinterpret_cast不能保证做得非常多,它只是滚动骰子.

如果要在位域结构和标量之间进行转换,则需要对值进行编码和解码.

    HW_Register(unsigned char value)
      : field1( value & 3 ),
        field2( value >> 2 & 3 ),
        field3( value >> 4 & 7 )
        {}

编辑:您没有得到任何输出的原因是与字段中的数字对应的ASCII字符是非打印的.试试这个:

std::cout << "Field 1 = " << (int) testRegister.field1 << std::endl;
std::cout << "Field 2 = " << (int) testRegister.field2 << std::endl;
std::cout << "Field 3 = " << (int) testRegister.field3 << std::endl;

Answer 2

不要这样做

 *this = *(reinterpret_cast<HW_Register*>(&registerValue));

'this'指针不应该以这种方式摆弄:

HW_Register reg(val)
HW_Register *reg = new HW_Register(val)

这里'这'在记忆中有两个不同的地方

相反,有一个内部联合/结构来保存值,这样很容易来回转换(因为你对可移植性不感兴趣)

例如

union
{
   struct {
     unsigned short field1:2;
     unsigned short field2:4;
     unsigned short field3:2;
    ...
   } bits;
   unsigned short value;
} reg

编辑:足够真实,名称为'register'