64位 内存对齐问题
MSDN中的一段话:
On 64-bit Windows, if a data structure is misaligned, routines that manipulate the structure, such as RtlCopyMemory and memcpy, will not fault. Instead, they will raise an exception. For example:
Copy
#pragma pack (1) /* also set by /Zp switch */
struct Buffer {
ULONG size;
void *ptr;
};
void SetPointer(void *p) {
struct Buffer s;
s.ptr = p; /* will cause alignment fault */ //为什么这里有问题?
...
}
You could use the UNALIGNED macro to fix this:
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void SetPointer(void *p) {
struct Buffer s;
*(UNALIGNED void *)&s.ptr = p;
}
Unfortunately, using the UNALIGNED macro is very expensive on Itanium-based processors. A better solution is to put 64-bit values and pointers at the beginning of the structure.
问题:
s.ptr = p;有什么问题?
我的理解是:原来64位是8字节对齐,现在#pragma pack (1)后就不会在size后填充4个字节,s.ptr = p;中“.”操作仍然会从s地址的第8个字节开始读,造成错误。不知道MSDN是不是这个意思?求教~