Assembly operator AND

Question

In order to continue this: Debugging C program (int declaration) I decided to test more code and see how compiler reacts to it. So I decided to try this one to test local variables:

#include <stdio.h>
main()
{
  int a,b,c,d,e,f,g;
  a=0xbeef;
  b=0xdead;
  c=0x12;
  d=0x65;
  e=0xfed;
  f=0xaa;
  g=0xfaceb00c;
  a=a+b;
  printf("%d",a);
}

Ok I did that int a,b,c... just to test the main's frame size and see the sub $0x10,%esp growing up, (I'm under linux so that is why maybe is sub), now to sub $0x30,%esp so here is the the gdb output with "disas main" command:

   0x0804841c <+0>:  push   %ebp
   0x0804841d <+1>:  mov    %esp,%ebp
   0x0804841f <+3>:  and    $0xfffffff0,%esp
   0x08048422 <+6>:  sub    $0x30,%esp ;7 int vars 4-byte is 7*4=28. 30 is enough
   0x08048425 <+9>:  movl   $0xbeef,0x14(%esp)
   0x0804842d <+17>: movl   $0xdead,0x18(%esp)
   0x08048435 <+25>: movl   $0x12,0x1c(%esp)
   0x0804843d <+33>: movl   $0x65,0x20(%esp)
   0x08048445 <+41>: movl   $0xfed,0x24(%esp)
   0x0804844d <+49>: movl   $0xaa,0x28(%esp)
   0x08048455 <+57>: movl   $0xfaceb00c,0x2c(%esp)
   0x0804845d <+65>: mov    0x18(%esp),%eax
   0x08048461 <+69>: add    %eax,0x14(%esp)
   0x08048465 <+73>: mov    0x14(%esp),%eax
   0x08048469 <+77>: mov    %eax,0x4(%esp)
   0x0804846d <+81>: movl   $0x8048510,(%esp)
   0x08048474 <+88>: call   0x80482f0 <printf@plt>
   0x08048479 <+93>: leave  
   0x0804847a <+94>: ret    

This line: 0x0804841f <+3>:and $0xfffffff0,%esp what is and operator and why is there a large number?

And why the offset in movl commands isn't negative like: movl $0xa,-0x4(%ebp) So far I know is the AND is a logical operator like 1 and 1 is 1, 0 and 0 is 0, 1 and 0 is 0 etc... If it is the case, %esp has the ebp value that was the base frame address of who called the main function.

can any of you explain why this is compiled like this?

I think I'm missing something. Edit: I saw some "topics" on stackoverflow talking about this. Going to share: link1 link2 link3

Answer 1

• Why is the offset in movl $0xbeef,0x14(%esp) not negative?

Because unlike in the other example, addressing is relative to esp, not ebp. esp is on one end of the stack, esp on the other one. So in order to get an address inside the current stack frame, you need to add to esp or subtract from ebp.

• Why and $0xfffffff0,%esp?

For alignment. @BlackBear explains this in the answer to your previous question: Debugging C program (int declaration)