Question

Can someone explain to me why in the second question proco2, that x-> b.i -> f[3]?

Answer 1

Actually on the stack you only have a pointer to the struct s2 (in 8(%ebp)). Therefore after

movl 8(%ebp), %eax

in %eax you have an address of struct s2.

The 8th-11th bytes of struct s2 constitute f[0] and the 12th-15th bytes constitute f[1] and therefore you have

return x->f[1]

In the second case after

movl 8(%ebp), %eax

in %eax you have an address of struct s1.

The 4th-7th bytes of struct s1 constitute the b field of type union u1. Therefore after

movl 4(%eax), %eax

in %eax you have union u1. Since it’s a union the %eax contains all field values at the same time (h, i and j). So

movl 20(%eax), %eax

is actually getting the 20th-23th byte of whatever pointer is in %eax (it can’t be the j field since it’s not a pointer). It can’t be h field since it points to struct s1 and sizeof (struct s1) is 12 which is <20. Therefore it must be the i field. The 20th-23th byte of struct s2 is f[3] and therefore you have:

return x->b.i->f[3]