在linux 系统中的应用程序,struct 并非是采用自然对齐,而是gcc强制对齐
何为自然对齐:在数据项大小的整数倍(例如:8字节数据项存入8的整数倍的地址)的地址处存储
数据项。
#include
#include
#include
/*
* Define several data structures, all of them start with a lone char
* in order to present an unaligned offset for the next field
*/
struct c {char c; char t;} c;
struct s {char c; short t;} s;
struct i {char c; int t;} i;
struct l {char c; long t;} l;
struct ll {char c; long long t;} ll;
struct p {char c; void * t;} p;
struct u1b {char c; __u8 t;} u1b;
struct u2b {char c; __u16 t;} u2b;
struct u4b {char c; __u32 t;} u4b;
struct u8b {char c; __u64 t;} u8b;
int main(int argc, char **argv)
{
struct utsname name;
uname(&name); /* never fails :) */
printf("arch Align: char short int long ptr long-long "
" u8 u16 u32 u64\n");
printf( "%-12s %3i %3i %3i %3i %3i %3i "
"%3i %3i %3i %3i\n",
name.machine,
/* note that gcc can subtract void * values, but it's not ansi */
(int)((void *)(&c.t) - (void *)&c),
(int)((void *)(&s.t) - (void *)&s),
(int)((void *)(&i.t) - (void *)&i),
(int)((void *)(&l.t) - (void *)&l),
(int)((void *)(&p.t) - (void *)&p),
(int)((void *)(&ll.t) - (void *)&ll),
(int)((void *)(&u1b.t) - (void *)&u1b),
(int)((void *)(&u2b.t) - (void *)&u2b),
(int)((void *)(&u4b.t) - (void *)&u4b),
(int)((void *)(&u8b.t) - (void *)&u8b));
return 0;
}
[root@hujunlinux driver_pratice]# ./dataalign
arch Align: char short int long ptr long-long u8 u16 u32 u64
i686 1 2 4 4 4 4 1 2 4 4
可见最大的对齐单位是4
int main()
{
struct A{ char c;
long long b;
}one;
printf("sizeof(one)=%d,sizeof(c)=%d\n",sizeof(one),
(int)((void*)(&(one.b))-(void*)(&one))
);
return 0;
}
./charSize
sizeof(one)=12,sizeof(c)=4
int main()
{
struct A{ char c;
short b;
}one;
printf("sizeof(one)=%d,sizeof(c)=%d\n",sizeof(one),
(int)((void*)(&(one.b))-(void*)(&one))
);
return 0;
}
./charSize
sizeof(one)=4,sizeof(c)=2
int main()
{
struct A{ char c;
int b;
}one;
printf("sizeof(one)=%d,sizeof(c)=%d\n",sizeof(one),
(int)((void*)(&(one.b))-(void*)(&one))
);
return 0;
}
./charSize
sizeof(one)=8,sizeof(c)=4
阅读(3090) | 评论(0) | 转发(0) |