如何使局部变量的地址是8字节对齐？ [英] How to make local variable's address is 8 byte alignment?
问题描述
HP-UX 11.23，HP aC ++ B3910B A.03.63
以下是样本。
bjhp1 / nfs / users / pjia> ; cat tt.c
#include< stdlib.h>
int main（）
{
int i = 10;
char p [1024] ="" ;
printf（"％p，％p \ n"，& i，p）;
返回0;
}
bjhp1 / nfs / users / pjia> aCC + DD64 tt.c
bjhp1 / nfs / users / pjia> ./a.out
800003ffbfff0f78,8000000ffbfff0f7c
我希望指针p'的地址是800003ffbfff0f80，而不是
800003ffbfff0f7c
有没有选择来实现这个目标？
HP-UX 11.23, HP aC++ B3910B A.03.63
Here is a samples.
bjhp1 /nfs/users/pjia>cat tt.c
#include <stdlib.h>
int main()
{
int i =10 ;
char p[1024] ="" ;
printf("%p, %p\n", &i, p) ;
return 0 ;
}
bjhp1 /nfs/users/pjia>aCC +DD64 tt.c
bjhp1 /nfs/users/pjia>./a.out
800003ffbfff0f78, 800003ffbfff0f7c
I hope the pointer p''s address is 800003ffbfff0f80, not
800003ffbfff0f7c
Is there any option to realize this?
推荐答案
Pengjun Jia写道：
Pengjun Jia wrote:
HP-UX 11.23，HP aC ++ B3910B A.03.63
以下是样本。
bjhp1 / nfs / users / pjia> cat tt.c
#include< stdlib.h>
int main（）
{
int i = 10;
char p [1024] ="" ;
printf（"％p，％p \ n"，& i，p）;
返回0;
}
bjhp1 / nfs / users / pjia> aCC + DD64 tt.c
bjhp1 / nfs / users / pjia> ./a.out
800003ffbfff0f78,8000000ffbfff0f7c
我希望指针p'的地址是800003ffbfff0f80，而不是
800003ffbfff0f7c
HP-UX 11.23, HP aC++ B3910B A.03.63
Here is a samples.
bjhp1 /nfs/users/pjia>cat tt.c
#include <stdlib.h>
int main()
{
int i =10 ;
char p[1024] ="" ;
printf("%p, %p\n", &i, p) ;
return 0 ;
}
bjhp1 /nfs/users/pjia>aCC +DD64 tt.c
bjhp1 /nfs/users/pjia>./a.out
800003ffbfff0f78, 800003ffbfff0f7c
I hope the pointer p''s address is 800003ffbfff0f80, not
800003ffbfff0f7c
为什么这对你很重要？
Why is this important to you?
是否有任何选项可以实现这一点？
Is there any option to realize this?
这是特定于您的平台的，我想，所以在
平台特定组中询问会更合适。
一般情况下，我认为你需要尝试在联盟（或可能是结构）中嵌入p
来玩弄技巧强制对齐。
例如，你可能会发现类似的东西： -
union {
double for_alignment ;
char p [1024];
}软糖。
将以8字节边界对齐。 />
但我重申这是特定于平台的，所以YMMV ......
This is specific to your platform, I''d think, so asking in a
platform-specific group would be more appropriate.
In general, I think you''d need to try playing tricks with embedding "p"
in a union (or possibly a structure) to force alignment.
For example, you may find that something like :-
union {
double for_alignment;
char p[1024];
} fudge.
Would get aligned at an 8-byte boundary.
But I repeat this is platform-specific, so YMMV...
Pengjun Jia写道：
Pengjun Jia wrote:
>
HP-UX 11.23，HP aC ++ B3910B A.03.63
以下是样本。
bjhp1 / nfs / users / pjia> cat tt.c
#include< stdlib.h>
int main（）
{
int i = 10;
char p [1024] ="" ;
printf（"％p，％p \ n"，& i，p）;
返回0;
}
bjhp1 / nfs / users / pjia> aCC + DD64 tt.c
bjhp1 / nfs / users / pjia> ./a.out
800003ffbfff0f78,8000000ffbfff0f7c
我希望指针p'的地址是800003ffbfff0f80，而不是
800003ffbfff0f7c
有没有选择来实现这个目标？
>
HP-UX 11.23, HP aC++ B3910B A.03.63
Here is a samples.
bjhp1 /nfs/users/pjia>cat tt.c
#include <stdlib.h>
int main()
{
int i =10 ;
char p[1024] ="" ;
printf("%p, %p\n", &i, p) ;
return 0 ;
}
bjhp1 /nfs/users/pjia>aCC +DD64 tt.c
bjhp1 /nfs/users/pjia>./a.out
800003ffbfff0f78, 800003ffbfff0f7c
I hope the pointer p''s address is 800003ffbfff0f80, not
800003ffbfff0f7c
Is there any option to realize this?
/ * BEGIN eight_bytes.c * /
#include< stdio.h>
#define BYTES 8
#define NUM 256
int main（无效）
{
int i [NUM + BYTES / sizeof（int）] = {10};
char * p = BYTES +（char *）i;
printf（"％p，％p \ n"，（void *）i，（void *）p）;
返回0;
}
/ * END eight_bytes.c * /
-
pete
/* BEGIN eight_bytes.c */
#include <stdio.h>
#define BYTES 8
#define NUM 256
int main(void)
{
int i[NUM + BYTES / sizeof(int)] = {10};
char *p = BYTES + (char *)i ;
printf("%p, %p\n", (void *)i, (void *)p) ;
return 0 ;
}
/* END eight_bytes.c */
--
pete
pete写道，On 10/01/08 11:42：
pete wrote, On 10/01/08 11:42:
Pengjun Jia写道：
Pengjun Jia wrote:
> HP-UX 11.23，HP aC ++ B3910B A.03.63
这是一个示例。
bjhp1 / nfs / users / pjia> ; cat tt.c
#include< stdlib.h>
int main（）
{i / 10;
char p [ 1024] ="" ;
printf（"％p，％p \ n"，& i，p）;
返回0;
}
bjhp1 / nfs / users / pjia> aCC + DD64 tt.c
bjhp1 /nfs/users/pjia>./a.out
800003ffbfff0f78,8000000ffbfff0f7c
我希望指针p'的地址是800003ffbfff0f80，而不是
800003ffbfff0f7c
是否有任何选项可以实现这一点？
>HP-UX 11.23, HP aC++ B3910B A.03.63
Here is a samples.
bjhp1 /nfs/users/pjia>cat tt.c
#include <stdlib.h>
int main()
{
int i =10 ;
char p[1024] ="" ;
printf("%p, %p\n", &i, p) ;
return 0 ;
}
bjhp1 /nfs/users/pjia>aCC +DD64 tt.c
bjhp1 /nfs/users/pjia>./a.out
800003ffbfff0f78, 800003ffbfff0f7c
I hope the pointer p''s address is 800003ffbfff0f80, not
800003ffbfff0f7c
Is there any option to realize this?
/ * BEGIN eight_bytes.c * /
#include< stdio.h>
#define BYTES 8
#define NUM 256
int main（无效）
{
int i [NUM + BYTES / sizeof（int）] = {10};
/* BEGIN eight_bytes.c */
#include <stdio.h>
#define BYTES 8
#define NUM 256
int main(void)
{
int i[NUM + BYTES / sizeof(int)] = {10};
您已经更改了可能存在问题的i类型。你还没有注意到这很可能*不能*做你想做的事情，因为C
语言并不保证在内存中如何安排事情。对于
实例，一些编译器开始玩巧妙的技巧，其中数组是放置在堆栈上的数组和/或放置保护值。数组的任一侧要检测缓冲区溢出和/或尝试检测它们何时发生。
You have change the type of i which is likely to be a problem. You have
also failed to note that this may well *not* do what is wanted as the C
language does not guarantee how things will be arranged in memory. For
instance some compiler start playing clever tricks with where arrays are
placed on the stack and/or put "guard values" either side of arrays to
detect buffer overflows and/or try and detect when they have occurred.
char * p = BYTES +（ char *）i;
printf（"％p，％p \ n"，（void *）i，（void *）p）;
char *p = BYTES + (char *)i ;
printf("%p, %p\n", (void *)i, (void *)p) ;
如果你告诉OP为什么你添加了
演员，那会很有帮助。原因是％p特别需要一个void *指针
和其他指针类型可以有不同的大小，表示或
传递机制。
It would have been helpful if you had told the OP why you added the
casts. The reason being that %p specifically requires a void* pointer
and other pointer types can have different sizes, representations or
passing mechanisms.
返回0;
}
/ * END eight_bytes.c * /
return 0 ;
}
/* END eight_bytes.c */
-
Flash Gordon
--
Flash Gordon
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