Understanding stack and base pointer Y86 - linked List sum

Question

Note : This is for a class. I'm just trying to understand how the rsp and rbp work so I can understand what part of my code is messed up. Sorry I'm new to this . Thanks for any help.

So I'm writing Y86 code to find sum of nodes of linked list

        .pos 0
    init:   irmovq  Stack, %rsp     # Set up stack pointer
    rrmovq  %rsp,%rbp       # Set up base pointer
    irmovq  ele1,%rdi
    irmovq  $18, %r8        #Note 1 : unsure of what this should be
    call    rsum_list        # Execute main program
    halt                    # Terminate program


  # Sample linked list
    .align 8
  ele1:   .quad 0x00
    .quad ele2
  ele2:   .quad 0x0b0
    .quad ele3
  ele3:   .quad 0xc00
    .quad 0

        # int sum_list(list_ptr ls)
  rsum_list:  pushq   %rbp
        rrmovq  %rsp,%rbp
        xorq    %rax,%rax       # val = 0
        subq %r8, %rsp              #subtract constant from rsp , not sure why we need to do this -> saw this in x86 code
        rmmovq %rdi, -18(%rbp)      #move ls to -18(%rbp)
        andq %rdi, %rdi             #check if 0   
        je End
        mrmovq -18(%rbp), %rax
        mrmovq (%rax), %rax            #rax gets ls->val
        rmmovq %rax, -10(%rbp)         #the val obtained is stored in -10(%rbp)
        mrmovq -18(%rbp), %rax         #rax becomes ls again
        mrmovq 8(%rax), %rax          # rax gets ls-> next
        rmmovq %rax, %rdi             #this is copied to rdi for next recursive call
        call rsum_list
        rmmovq %rax, -8(%rbp)          #rax will contain ans from recursive call
        mrmovq -10(%rbp), %rdx       #retrieve ans. of current node from -10(%rbbp) , where we stored it before recursion
        mrmovq -8(%rbp), %rax      
        addq %rdx, %rax           #adding

  End :
       rrmovq %rbp, %rsp
       popq %rbp
       nop                     # makes sure stop in 31 steps
       ret

I suspect I'm making a mistake in storing values on the stack somewhere that is being messed up due to recursion. Sorry but I really don't undderstand this and want to. Any help is appreciated. Thanks!

What it should give me in rax finally : 0x0000cba What I'm getting is 0x0000040

Answer 1

You seem to partially overwrite your stored value from the current node with the result of your recursive call. I've added asterisks to the relevant lines below:

*rmmovq %rax, -10(%rbp)         #the val obtained is stored in -10(%rbp)
 mrmovq -18(%rbp), %rax         #rax becomes ls again
 mrmovq 8(%rax), %rax          # rax gets ls-> next
 rmmovq %rax, %rdi             #this is copied to rdi for next recursive call
 call rsum_list
*rmmovq %rax, -8(%rbp)          #rax will contain ans from recursive call
 mrmovq -10(%rbp), %rdx       #retrieve ans. of current node from -10(%rbbp) , where we stored it before recursion
 mrmovq -8(%rbp), %rax

Note that the two storage locations are only 2 bytes apart, though the values written are 8-bytes long. Given Intel's little endian storage convention, writes will align as follows, where the columns give the placement of the three writes that occur at a given recursion depth (with ans denoting the answer from the recursive call)

RBP: 
 -1:                         ans(MSB)     
 -2:                         ans     
 -3:           ls->val(MSB)  ans     
 -4:           ls->val       ans     
 -5:           ls->val       ans     
 -6:           ls->val       ans     
 -7:           ls->val       ans     
 -8:           ls->val       ans(LSB)
 -9:           ls->val     
-10:           ls->val(LSB)     
-11: ls(MSB)                
-12: ls                
-13: ls                
-14: ls                
-15: ls                
-16: ls                
-17: ls                
-18: ls(LSB)                

Unfortunately, this alone shouldn't have produced your 0x40, so there's something else wrong.