c++ - 为什么 "unsigned"映射到 MSVC C++ 中的 32 位？

Question

我在 MSVC++ 2008 中定义了以下结构:

struct{
 Uint16  XDD;
 unsigned XDD_UI:8;
 unsigned XDD_CR:8;
}byte;

当对上述结构执行 sizeof 时，它表明 unsigned 使用 32 位来存储数据。

• 这是为什么，它与内存对齐有关吗？
• 我怎么能强制编译器只使用定义的 8 位而不用 触及结构定义？

Answer 1

unsigned 是 unsigned int 的缩写，在 Windows 上是 32 位宽。这意味着它具有 4 的对齐方式。由于您的结构是对齐的，因此在 XDD 和 XDD_UI 之间有两个字节填充，并且在结构末尾有两个字节填充。

你的结构是这样布局的:

0-1  XDD
2-3  <padding>
4-4  XDD_UI
5-5  XDD_CR
6-7  <padding>

If you want the struct to be packed then you need to pack it. Use #pragma pack to achieve that. However, even if you do that, the compiler produces a struct with size 6. That's because the bitfields are packed into an unsigned int and so your two bitfields will always consume 4 bytes.

If you made sure that your bitfields were declared as being of a type whose size was no larger than 2 bytes, then your struct would be 4 bytes. And that would be true for an aligned struct. For instance, this struct has size 4.

struct s {
    unsigned short XDD;
    unsigned short XDD_UI:8;
    unsigned short XDD_CR:8;
};

但是，在我看来，声明没有位域的结构更明智:

struct s {
    Uint16 XDD;
    unsigned char XDD_UI;
    unsigned char XDD_CR;
};

您可以声明该结构对齐并具有您想要的布局。