Assembly Language for Beginners

1.10 Accessing passed arguments

1.10 Accessing passed arguments

Now we figured out that thecallerfunction is passing arguments to thecalleevia the stack. But how does

thecalleeaccess them?

Listing 1.86: simple example

#include <stdio.h>

int f (int a, int b, int c)

{

return a*b+c;

};

int main()

{

printf ("%d\n", f(1, 2, 3));

return 0;

};

1.10.1 x86

MSVC

Here is what we get after compilation (MSVC 2010 Express):

Listing 1.87: MSVC 2010 Express

_TEXT SEGMENT

_a$ = 8 ; size = 4

_b$ = 12 ; size = 4

_c$ = 16 ; size = 4

_f PROC

push ebp

mov ebp, esp

mov eax, DWORD PTR _a$[ebp]

imul eax, DWORD PTR _b$[ebp]

add eax, DWORD PTR _c$[ebp]

pop ebp

ret 0

_f ENDP

_main PROC

push ebp

mov ebp, esp

push 3 ; 3rd argument

push 2 ; 2nd argument

push 1 ; 1st argument

call _f

add esp, 12

push eax

push OFFSET $SG2463 ; '%d', 0aH, 00H

call _printf

add esp, 8

; return 0

xor eax, eax

pop ebp

ret 0

_main ENDP

What we see is that themain()function pushes 3 numbers onto the stack and callsf(int,int,int).

Argument access insidef()is organized with the help of macros like:
_a$ = 8, in the same way as local variables, but with positive offsets (addressed withplus). So, we are
addressing theouterside of thestack frameby adding the_a$macro to the value in theEBPregister.

Then the value ofais stored intoEAX. AfterIMULinstruction execution, the value inEAXis aproductof the

value inEAXand the content of_b.