C/C++ 结构偏移
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C/C++ Structure offset
提问by davenpcj
I'm looking for a piece of code that can tell me the offset of a field within a structure without allocating an instance of the structure.
我正在寻找一段代码，它可以告诉我结构中字段的偏移量，而无需分配结构的实例。
IE: given
IE：给定
struct mstct {
int myfield;
int myfield2;
};
I could write:
我可以写：
mstct thing;
printf("offset %lu\n", (unsigned long)(&thing.myfield2 - &thing));
And get offset 4for the output. How can I do it without that mstct thingdeclaration/allocating one?
并获取offset 4输出。如果没有那个mstct thing声明/分配一个，我怎么能做到这一点？
I know that &<struct>does not always point at the first byte of the first field of the structure, I can account for that later.
我知道&<struct>并不总是指向结构的第一个字段的第一个字节，我可以稍后解释。
回答by Michael Burr
How about the standard offsetof() macro (in stddef.h)?
标准的 offsetof() 宏（在 stddef.h 中）怎么样？
Edit: for people who might not have the offsetof() macro available for some reason, you can get the effect using something like:
编辑：对于由于某种原因可能没有 offsetof() 宏可用的人，您可以使用以下内容获得效果：
#define OFFSETOF(type, field) ((unsigned long) &(((type *) 0)->field))
回答by Dan Lenski
Right, use the offsetofmacro, which (at least with GNU CC) is available to both C and C++ code:
是的，使用offsetof宏，它（至少对于 GNU CC）可用于 C 和 C++ 代码：
offsetof(struct mstct, myfield2)
回答by Frank Krueger
printf("offset: %d\n", &((mstct*)0)->myfield2);
printf("偏移量：%d\n", &((mstct*)0)->myfield2);
