结构和对齐 [英] Structs and Alignment
问题描述
如果我有
struct foo
{
char bar;
long baz ;
};
在许多平台上，填充和对齐正在进行以获得baz到
从偶数地址开始或4分的地址。所以baz不是
必然是一个字节（char的大小）到struct foo。
我怎么能准确知道struct foo baz中有多少字节？
/ David
If I have
struct foo
{
char bar;
long baz;
};
on many platforms, padding and alignment is taking place to get baz to
start at an even address or an address that 4 divides. So baz is not
necessarily one byte (the size of char) into struct foo.
How can I know exactly how many bytes into struct foo baz is?
/David
推荐答案
David Rasmussen写道：
David Rasmussen wrote:
如果我在许多平台上，填充和填充
结构foo
{
char bar;
long baz;
};
正在进行调整以使baz从偶数地址或4分频的地址开始。因此baz不一定是struct foo中的一个字节（char的大小）。
我怎样才能准确知道struct foo baz中有多少字节？
/ David
If I have
struct foo
{
char bar;
long baz;
};
on many platforms, padding and alignment is taking place to get baz to
start at an even address or an address that 4 divides. So baz is not
necessarily one byte (the size of char) into struct foo.
How can I know exactly how many bytes into struct foo baz is?
/David
使用< stddef.h>中的offsetof宏：
#include< stddef.h> ;
struct foo
{
char bar;
long baz;
};
int main（无效）
{
size_t baz_offset = offsetof（struct foo，baz） ;
返回0;
}
-Michael
-
电子邮件：我的是/ at / gmx / dot / de地址。
Use the offsetof macro from <stddef.h>:
#include <stddef.h>
struct foo
{
char bar;
long baz;
};
int main (void)
{
size_t baz_offset = offsetof(struct foo, baz);
return 0;
}
-Michael
--
E-Mail: Mine is an /at/ gmx /dot/ de address.
Pierre Maurette写道：
Pierre Maurette wrote:
David Rasmussenécrit：
David Rasmussen a écrit :
如果我有
结构foo
{
char bar;
long baz;
};
在很多平台上，填充和对齐都是为了让baz从偶数地址开始n地址为4分。所以baz不一定是一个字节（char的大小）到struct foo。
在我的平台上，你可以试着打赌：
- 没有头部填充，& struct ==& struct.firstfield（rule）。
- 每个字段根据编译器选项对齐，可能是它自己的大小。
- 结束填充，计算结果：
struct.firstfield正确对齐。
If I have
struct foo
{
char bar;
long baz;
};
on many platforms, padding and alignment is taking place to get baz to
start at an even address or an address that 4 divides. So baz is not
necessarily one byte (the size of char) into struct foo.
"on my platform", you can try to bet on:
- no head padding, &struct == &struct.firstfield (rule).
- each field aligned according the compiler options, maybe its own size.
- end padding, calculated for:
struct.firstfield correctly aligned.
不：这是计算得出的具有
严格对齐要求的结构成员在
中正确对齐了一个类型为struct foo的> = 2元素的数组。没有填充的结构数组（规则）。
是的。
注意：在C99中，具有灵活阵列成员的结构无法构建阵列。
所有这些都是递归的。当然。
Nope: it is calculated s.t. the structure member with the
strictest alignment requirements is correctly aligned in
an array of >=2 elements of type struct foo. array of structs without padding (rule).
Yep.
Note: In C99, structures with flexible array members cannot
build up arrays.
All this recursively, of course.
我怎样才能准确知道struct foo baz中有多少字节？
也许：
size_t pad =（ char *）& foo.baz - & foo;
How can I know exactly how many bytes into struct foo baz is?
maybe:
size_t pad = (char*)&foo.baz - &foo;
非便携式。可以使用
offsetof宏来计算偏移量。
结果应该是不同的，例如你改变
long baz;
到
char baz;
Non-portable. The offset can be calculated using the
offsetof macro.
The result should be different if for example you change
long baz;
to
char baz;
s /应该/可能是/
干杯
迈克尔
-
电子邮件：我的是/ at / gmx / dot / de地址。
s/should be/is probably/
Cheers
Michael
--
E-Mail: Mine is an /at/ gmx /dot/ de address.
（取代< mn *********************** @ YOURBRAnoos.fr>）
David Rasmussen写道01/01/05：
(supersedes <mn***********************@YOURBRAnoos.fr>)
David Rasmussen wrote on 01/01/05 :
如果我有
结构foo
{
char bar;
long baz;
在许多平台上，正在进行填充和对齐以使baz从偶数地址或4分割的地址开始。所以baz不一定是struct foo中的一个字节（char的大小）。
我怎样才能准确知道struct foo baz中有多少字节？
If I have
struct foo
{
char bar;
long baz;
};
on many platforms, padding and alignment is taking place to get baz to
start at an even address or an address that 4 divides. So baz is not
necessarily one byte (the size of char) into struct foo.
How can I know exactly how many bytes into struct foo baz is?
大小：sizeof运算符是你的朋友（你的C-book也是如此）。
Offset：使用offsetof（）宏。
-
Emmanuel
C-FAQ： http://www.eskimo.com/~scs/C-faq/faq.html
C -library： http://www.dinkumware.com/refxc.html
显然你的代码不符合原始规格。
你用湿面条被判30鞭。
- alc ++中的Jerry Coffin
Size : The sizeof operator is your friend (so is your C-book).
Offset : Use the offsetof() macro.
--
Emmanuel
The C-FAQ: http://www.eskimo.com/~scs/C-faq/faq.html
The C-library: http://www.dinkumware.com/refxc.html
"Clearly your code does not meet the original spec."
"You are sentenced to 30 lashes with a wet noodle."
-- Jerry Coffin in a.l.c.c++
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