Struct 为什么这个typedef给出的sizeof（）值大于预期值？

我使用这种形式的typedef来简化对微处理器寄存器及其位字段的访问

  typedef union
  {
     uint8_t        u8Byte;           ///< REG_8 as unsigned byte
     int8_t         i8Byte;           ///< REG_8 as signed byte
     struct
     {
        unsigned b0:1;                ///< Bit 0 of REG_8 type
        unsigned b1:1;                ///< Bit 1 of REG_8 type
        unsigned b2:1;                ///< Bit 2 of REG_8 type
        unsigned b3:1;                ///< Bit 3 of REG_8 type
        unsigned b4:1;                ///< Bit 4 of REG_8 type
        unsigned b5:1;                ///< Bit 5 of REG_8 type
        unsigned b6:1;                ///< Bit 6 of REG_8 type
        unsigned b7:1;                ///< Bit 7 of REG_8 type
     };
  } REG_8;

将

a.b2

的值设为1，因此不存在对齐问题

删除

struct

会使

sizeof

的值为1，因此看起来似乎存在填充问题，但如果是，原因是什么

有人能解释一下吗？我正在使用Microchip XC16编译器（基于GCC），目标是一个16位处理器。

在您的计算机上可能是sizeof（unsigned）=2，因此任何“unsigned”位字段都至少占用2个字节。将unsigned替换为uint8应使sizeof（REG_8）变为1

另见这个问题：

---

看来@twin的想法是正确的，尽管我也找到了另一个解决方案。给出预期

sizeof（REG_8）==1

的两个备选方案是：

  typedef union
  {
     uint8_t        u8Byte;           ///< REG_8 as unsigned byte
     int8_t         i8Byte;           ///< REG_8 as signed byte
     struct
     {
        unsigned b0:1;                ///< Bit 0 of REG_8 type
        unsigned b1:1;                ///< Bit 1 of REG_8 type
        unsigned b2:1;                ///< Bit 2 of REG_8 type
        unsigned b3:1;                ///< Bit 3 of REG_8 type
        unsigned b4:1;                ///< Bit 4 of REG_8 type
        unsigned b5:1;                ///< Bit 5 of REG_8 type
        unsigned b6:1;                ///< Bit 6 of REG_8 type
        unsigned b7:1;                ///< Bit 7 of REG_8 type
     } __attribute__((packed));
  } REG_8;

typedef联合
{
uint8\u t u8Byte；//

……或者

  typedef union
  {
     uint8_t        u8Byte;           ///< REG_8 as unsigned byte
     int8_t         i8Byte;           ///< REG_8 as signed byte
     struct
     {
        uint8_t b0:1;                ///< Bit 0 of REG_8 type
        uint8_t b1:1;                ///< Bit 1 of REG_8 type
        uint8_t b2:1;                ///< Bit 2 of REG_8 type
        uint8_t b3:1;                ///< Bit 3 of REG_8 type
        uint8_t b4:1;                ///< Bit 4 of REG_8 type
        uint8_t b5:1;                ///< Bit 5 of REG_8 type
        uint8_t b6:1;                ///< Bit 6 of REG_8 type
        uint8_t b7:1;                ///< Bit 7 of REG_8 type
     };
  } REG_8;

typedef联合
{
uint8\u t u8Byte；//

我猜，由于严格的数据对齐寻址，16位机器的单字节联合将有1字节的填充。这是一个猜测，因为我不熟悉机器/编译器。@I在这台16位机器上，字确实需要在偶数地址上对齐，但机器可以在任何地址（即，无对齐要求）访问字节。sizeof（unsigned）为2，但：1限定符不是覆盖了结构成员的字节吗？此外，如果b7成员得到2个字节，那么union不应该增长到3个字节（2+7/8四舍五入）？结构成员将各自占用1位无符号字节。（我不理解你评论的第二部分-你将如何声明b7？）如果你声明未签名的b7:16，结构需要两个未签名的b7:16，因此我应该有大小为4Silliness的部分-我假设未签名的b7的LSB将代表它的值。请忽略！

  typedef union
  {
     uint8_t        u8Byte;           ///< REG_8 as unsigned byte
     int8_t         i8Byte;           ///< REG_8 as signed byte
     struct
     {
        uint8_t b0:1;                ///< Bit 0 of REG_8 type
        uint8_t b1:1;                ///< Bit 1 of REG_8 type
        uint8_t b2:1;                ///< Bit 2 of REG_8 type
        uint8_t b3:1;                ///< Bit 3 of REG_8 type
        uint8_t b4:1;                ///< Bit 4 of REG_8 type
        uint8_t b5:1;                ///< Bit 5 of REG_8 type
        uint8_t b6:1;                ///< Bit 6 of REG_8 type
        uint8_t b7:1;                ///< Bit 7 of REG_8 type
     };
  } REG_8;