C struct完成功能的汇编代码

Question

I was given the following code skeleton:

typedef struct node {
    _______________ x;
    _______________ y;
    struct node *next;
    struct node *prev;
} node_t;

node_t n;
void func() {
    node_t *m;
    m = ______________________;
    m->y /= 16;
    return; 
}

with the following assembly code generated on an IA-32 linux machine:

func:
  pushl %ebp
  movl n+12,%eax
  movl 16(%eax),%eax
  movl %esp,%ebp
  movl %ebp,%esp
  shrw $4,8(%eax)
  popl %ebp
  ret

And I need to fill in the blanks above. I do believe that when I look at he lines:

m->y /= 16; and shrw $4,8(%eax)

this is obviously a right shift on the 8th byte of the node_t *m. So what I think is that there are 8 bytes before y so x is 8 bytes long (double). Also, I think that the total size of the struct 'node' comes from the line

movl n+12, %eax

because that's where the node_t *m is created. I'm just not sure how to determine the size of this from that line.

I was also give this information

It should also be noted that pointers are 4 bytes in size and alignment.

Answer 1

In 32-bit mode, a pointer is 4 bytes, and typically aligned on a 4-byte boundary. So n + 12 is actually loading & n.next to %eax . eg, if y is an unsigned short , there are 2 bytes of padding in the structure before next .

movl 16(%eax),%eax is dereferencing with a 16-byte offset, to give: m = n.next->prev

---

struct node
{
    double x;          /* bytes: 0  .. 7  */
    unsigned short y;  /* bytes: 8  .. 9  */
                       /* bytes: 10 .. 11 */ /* (padding) */
    struct node *next; /* bytes: 12 .. 15 */
    struct node *prev; /* bytes: 16 .. 19 */
};