问题描述
这是关于内存对齐。在下面code，我希望B的结构内部的偏移量为8（32位机）。请参见这里。那里，使 B 总是发生在一个高速缓存行。然而，事实并非如此。在结构测试1 的全局对象的成员 B 似乎一致。我不知道如果是偶然或编译器是这样做的故意。
This is about memory alignment. In code below, I expected that the offset of b inside the structure to be 8 (32-bit machine). See here. There by, making b always occur within a cache line. However, that's not the case. The member b in a global object of struct test1 seems to be aligned. I am not sure if its by chance or compiler is doing this deliberately.
我想明白为什么编译器是不是在填充4个字节的。
I wanted to understand why compiler is not padding 4 bytes after a.
struct test1
{
int a;
double b;
}t1;
int main()
{
struct test1 *p = malloc(sizeof(struct test1));
printf("sizes int %d, float %d, double %d, long double %d\n", sizeof(int), sizeof(float), sizeof(double), sizeof(long double));
printf("offset of b %d\n",(int)&(t1.b)-(int)&(t1));
printf("\naddress of b (on heap) = %p, addr of b (on data seg) = %p\n",&(p->b), &(t1.b));
return 0;
}
输出结果是...
The output is...
sizes int 4, float 4, double 8, long double 12
offset of b 4
address of b (on heap) = 0x804a07c, addr of b (on data seg) = 0x80497e0
我是用标准的gcc编译器在Ubuntu 10.04
I am using standard gcc compiler on ubuntu 10.04
推荐答案
按照系统V ABI i386的，第28页，双只得到4个字节对齐，但是编译器被推荐为8字节提供一个选项，也是如此。看来这就是由GCC在Linux上实现，被称为选项 -malign双。
According to the System V ABI for i386, page 28, double only gets 4 bytes alignment, but compilers are recommended to provide an option for 8 bytes as well. It appears this is what is implemented by GCC on Linux, the option being called -malign-double.
另一种方法是使用 -m64 来获得x86-64的对象code，这已经在一些系统上，包括Mac OS X的缺省值。
Another alternative is to use -m64 to get x86-64 object code, which is already the default on some systems including Mac OS X.
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