位字段用C含结构的工会结构 [英] Bitfields in C with struct containing union of structs

问题描述

唔...那为什么，当我打印的sizeof（结构MYSTRUCT），它输出（而不是2）3本code <？ / p>

Hm... why is it that, when I print sizeof(struct MyStruct), it outputs 3 (instead of 2) for this code?

#pragma pack(push, 1)
    struct MyStruct
    {
        unsigned char a : 6;
        union
        {
            struct
            {
                unsigned int b : 9;
            };
        };
    };
#pragma pack(pop)

在这重要的情况下，我在Windows 7上运行x64的GCC的MinGW 4.5.0，但说实话，结果是怪异够我说，我不认为编译器和OS太大的关系在这里。 ：\\

In case it matters, I'm running MinGW GCC 4.5.0 on Windows 7 x64, but honestly, the result is weird enough for me that I don't think the compiler and the OS matter too much here. :\

推荐答案

您不能开始在未字节对齐地址领域。
你期待：

You can't have the field starting at an address that is not byte aligned. You're expecting:

6 bits + 9 bits -> 15 bits -> 2 bytes

但你得到的是：

6 bits -> 1 byte
9 bits -> 2 bytes
total ->  3 bytes

的数据被存储为：

The data is being stored as:

| 1 byte | 2 byte |3 byte | 
 aaaaaaXX bbbbbbbb bXXXXX  

当你期待：

| 1 byte | 2 byte |
 aaaaaabb bbbbbbbX  

编辑：
为了澄清基于以下意见：

edit: To clarify based on the comments below:

工会（和含结构）必须字节对齐。这不要紧，内容只有9位，工会/结构本身是一个完整的16位。请注意，你不能做到以下几点：

The union (and the containing struct) must be byte aligned. It doesn't matter that the contents are only 9 bits, the union/struct itself is a full 16 bits. Notice that you cannot do the following:

struct MyStruct
{
    unsigned char a : 6;
    union
    {
        struct
        {
            unsigned int b : 9;
        } c:9;
    } d:9;
};

以C不会让您指定的整个结构的比特大小。

As C won't let you specify the entire struct's bit-size.

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