英文:
Byte order in C bit field?
问题
请看下面的内容：
考虑 uint32_t n = 0x12345678；它存储在大端序（BE）机器或小端序（LE）机器上，就像图片所示；现在我有一个定义如下的结构体：
struct DATA {
uint32_t a : 24;
uint32_t b : 8;
};
int main() {
struct DATA data;
data.a = 0x123456;
data.b = 0x78;
return 0;
}
它在内存中如何存储？
英文:
Consider uint32_t n = 0x12345678; it stores in BE machine or LE machine, like the picture shows; now I have a structure defined like this
struct DATA {
uint32_t a : 24;
uint32_t b : 8;
};
int main() {
struct DATA data;
data.a = 0x123456;
data.b = 0x78;
return 0;
}
How does it store in memory?
答案1
得分: 1
如何存储在内存中？
许多可能性：
- 大端序：无填充：0x12，0x34，0x56，0x78
- 大端序：填充为偶数：0x12，0x34，0x56，...，0x78，...
- 大端序：填充为四字节对齐：0x12，0x34，0x56，...，0x78，...，...，...
- 小端序：无填充：0x56，0x34，0x12，0x78
- 小端序：填充为偶数：0x56，0x34，0x12，...，0x78，...
- 小端序：填充为四字节对齐：0x56，0x34，0x12，...，0x78，...，...，...
- 其他字节序：
- 对于位域，仅类型
int、unsigned、bool有明确定义，因此无效。 - 无：在示例中，经过优化的编译可以消除变量。
- ...
良好的代码不应关心它在内存中的存储方式。
如果代码确实需要特定的顺序，请使用 uint8_t 数组而不是位域。
英文:
> How does it store in memory?
Many possibilities:
- BE: no padding: 0x12, 0x34, 0x56, 0x78
- BE: padding to even: 0x12, 0x34, 0x56, ..., 0x78, ...
- BE: padding to quad: 0x12, 0x34, 0x56, ..., 0x78, ..., ..., ...
- LE: no padding: 0x56, 0x34, 0x12, 0x78
- LE: padding to even: 0x56, 0x34, 0x12, ..., 0x78, ...
- LE: padding to quad: 0x56, 0x34, 0x12, ..., 0x78, ..., ..., ...
- Other Endians:
- Invalid as only types
int,unsigned,boolwell defined for bit-fields. - None: as an optimized compile can eliminate the variable as used in the example.
- ...
Good code should not care how it is stored in memory.
If code really needs a certain order, use an array of uint8_t instead of a bit-field.
Note: many compilers will not store a uint32_t on an odd boundary as in the example.
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