After the sentence struct test _ s * t = NULL is executed, t->; The address of pad0 is 0x00000000. Because all members in the structure are int-type, t->; The address of pad 1 is 0x00000004, and so on t->; The address of pad9 is 0x00000020.

& This is the address operator. (t->； Pad9) means to take the address of t->pad9, then convert it into int type and output 32.

If t is not defined as NULL, the printed address is relative to sss, as shown in the following example.

int main(int argc，char *argv[])

{

Structural test _ s sss

Structural test = & ampsss

printf("%d\n "，(int)& amp； (t->； pad 2))；

Returns 0;

}