Question

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.

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;
}

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

I am using standard gcc compiler on ubuntu 10.04

Answer 1

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.

Another alternative is to use -m64 to get x86-64 object code, which is already the default on some systems including Mac OS X.