MDK 字节对齐 __packed 与 #pragma packed区别

这2种字节对齐我都在用，感觉没什么区别，这次又碰到了，通过查看MDK帮助文档，终于弄清楚了，这2个只有细微的差别。

Comparison of a __packed `struct` and a #pragma packed `struct`

第1种实现方法

__packed struct foobar
{char x;short y[10];
};
short get_y0(struct foobar *s)
{// Unaligned-capable loadreturn *s->y;
}
short *get_y(struct foobar *s)
{return s->y;    // Compile error
}

第二种实现Compiler User Guide: Comparisons of an unpacked struct, a __packed struct, and a struct with individually __packed fields, and of a __packed struct and a #pragma packed struct

#pragma push
#pragma pack(1)
struct foobar
{char x;short y[10];
};
#pragma pop
short get_y0(struct foobar *s)
{// Unaligned-capable loadreturn *s->y;
}
short *get_y(struct foobar *s)
{return s->y;    // No error// Potentially illegal unaligned load,// depending on use of result
}

In the first implementation, taking the address of a field in a __packed struct or a __packed field in a struct yields a __packed pointer, and the compiler generates a type error if you try to implicitly cast this to a non-__packed pointer. In the second implementation, in contrast, taking the address of a field in a #pragma packed struct does not yield a __packed-qualified pointer. However, the field might not be properly aligned for its type, and dereferencing such an unaligned pointer results in Undefined behavior.

如果使用此指针，则Potentially illegal unaligned load, depending on use of result

在第一个实现中，获取字节对齐结构中的字段或字节对齐结构中的成员字段的地址会生成一个packed指针，如果试图将其隐式转换为非packed指针，编译器会生成一个类型错误。

相比之下，在第二种实现中，在#pragma packed struct中获取字段的地址不会生成_packed-qualified指针。但是字段可能没有正确对齐其类型，间接引用此未对齐的指针会导致未定义的行为。

潜在的非法未对齐加载，取决于结果的使用

AC5与AC6用法不太一样，详细如下：

AC5

Compiler User Guide: Comparisons of an unpacked struct, a __packed struct, and a struct with individually __packed fields, and of a __packed struct and a #pragma packed struct

#pragma push
#pragma pack(1)
struct foobar
{char x;short y[10];
};
#pragma pop

AC6

#pragma pack(push)

#pragma pack([n])

#pragma pack(pop).