Stack in Assembly

I need help with a code in Assembly.

I'm coworking on this and my group just did this Assembly code where it supposed to do the same as what I did in #c.

Could someone help me to understand what happens with de stack at least on the first steps so I can go on and conclude the rest?

I'm a begginer in Assembly, but I know that these lines just save the value for the calling function, takes a frame for the called function and save space for local variables, but I can't figure out the next first steps.

mov ebp
mov ebp,esp
sub esp, 16


Here is what I did in #c:

void mult_integer(int X[A_Linhas][A_Colunas], int number)
{
int c, l;

for (l = 0; l < A_Linhas; l++)
{
for (c = 0; c < A_Colunas; c++)
{
X[l][c] = number * X[l][c];
}
}
}


And here is the code in Assembly:

mul_integer:
push ebp
mov ebp, esp
sub esp, 16
mov dword [ebp-4H], 0
jmp L_020

L_017: mov dword [ebp-8H], 0
jmp L_019

L_018: mov edx, dword [ebp-4H]
mov eax, edx
add eax, eax
add eax, edx
shl eax, 2
mov edx, eax
mov eax, dword [ebp+8H]
lea ecx, [edx+eax]
mov edx, dword [ebp-4H]
mov eax, edx
add eax, eax
add eax, edx
shl eax, 2
mov edx, eax
mov eax, dword [ebp+8H]
add edx, eax
mov eax, dword [ebp-8H]
mov eax, dword [edx+eax*4]
imul eax, dword [ebp+0CH]
mov edx, eax
mov eax, dword [ebp-8H]
mov dword [ecx+eax*4], edx
add dword [ebp-8H], 1
L_019: cmp dword [ebp-8H], 2
jle L_018
add dword [ebp-4H], 1
L_020: cmp dword [ebp-4H], 3
jle L_017
nop
leave
ret


[ebp-4]

l

[ebp-8]

c

l = 0

ebp

esp

call, ret, push, pop, ...

esp

call function

call

ebp

1 Answer
1

When I write mov edx, [ebp-4] or +4, or -4, +8, -8 what am I doing with the stack, exactly?

mov edx, [ebp-4]

Take a look at the stack (diagram from Wikipedia's Call Stack article). Note that low memory addresses are at the top of this diagram, while higher memory address are at the bottom.

The frame pointer is stored in register ebp (on x86). It contains the address of the Return Address.

ebp

Your local variables are stored before the Return Address. Variable c is 4 bytes in size. By subtracting 4 from the ebp address, you are now pointing to your first local variable, c. Subtract another 4 (making it -8), and you're now pointing to your second local variable l.

c

ebp

c

l

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.