Mov instructions and registers - confusion!

Question

I read Computer Systems: A Programmer's Perspective, Chapter 3 explains the mov instruction, and the explanation given in the book confuses me.

give function (p. 142 1 issue)

 int exchange( int *xp, int y) { int x = *xp; *xp = y; return x; } 

Function body assembly code

 movl 8(%ebp), %eax //Get xp movl 12(%ebp), %edx //Get y movl (%eax), %ecx //Get x at *xp movl %edx, (%eax) //Store y at *xp movl %ecx, %eax //Set x as return value 

What bothers me is what will be stored, and where Here's how I understand it:

 movl 8(%ebp), %eax //Get xp 

The CPU moves +8 bytes to the stack (from the frame pointer %ebp ), takes the value stored in this place, and stores this value in the %eax register (for allocation, it stores the value, not the address)

I'm right? Thanks!

Answer 1

Yes, it looks like everything is fine with you. IMHO, the syntax of AT & T 8(%ebp) less intuitive than Intel [ebp+8] , which is clearer. The brackets indicate that you are using the value at the address in the register, and the number is the offset from the address you really want.

Answer 2

Yes, this is the AT & T syntax, which takes the form:

 instruction source, dest 

Intel assembly is the opposite order .

You are also right at 8(%ebp) moving 8 bytes from the frame pointer. The reason that it moves 8 bytes, in particular, is that the parameters are pushed onto the stack in the reverse order ("right" to "left" when viewing a typical function call). Thus, y was pressed first, then xp and finally the return address of the calling function (so you are moving 8 bytes, not 4).

Answer 3

You need to understand what a stack stack is. Find out exactly what push and pop instructions do. Before this code was

  push y_val push xp_ptr call exchange .cont ... .exchange push ebp mov ebp, esp // .. rest of code // stack frame: old_ebp_val ; [ebp] points here .cont ; [ebp + 4] xp_ptr ; [ebp + 8] y_val