Find out function arguments value from stack pointer

Question

Given stack pointer value, is it possible to determine the value of the passed arguments to the function? Where are the arguments stored in the stack frame.

Lets say, executing gcc compiled ELF binary on x86 architecture on Linux platform:

int foo(int a, int b)
{
...
}

foo(a,b) is called from main() and I know the stack pointer(SP) value which is pointing to foo() now. How can I retrive the value of arguments a and b?

EDIT: If stack grows from smaller address to larger address, and arguments are passed right to left using cdecl, can I obtain args value like this:

b = *(SP + 1);
a = *(SP + 2);

EDIT: The following program prints the value of functions args a, b using above arch and specifications.

void foo(int a, int b)
{
        int i;
        register int stackptr asm("sp");
        int *sp = (int *)stackptr;
        printf("\n\ta=%d b=%d\n", a, b);
        for (i=0; i<16; i++) {
                printf("*(sp + %d) = %d\n", i, *(sp +i));
        }
}

int main()
{
        foo(3, 8);
        foo(9, 2);
        foo(1, 4);
        return 0;
}

The output of above code is:

        a=3 b=8
*(sp + 0) = 134514016
*(sp + 1) = 0
*(sp + 2) = 0
*(sp + 3) = 134513373
*(sp + 4) = 8239384
*(sp + 5) = 134513228
*(sp + 6) = 6
*(sp + 7) = -1076716032
*(sp + 8) = 134513456
*(sp + 9) = 0
*(sp + 10) = -1076715960
*(sp + 11) = 134513759
*(sp + 12) = 3  //value of arg a
*(sp + 13) = 8  //value of arg b
*(sp + 14) = 134513817
*(sp + 15) = 10612724

        a=9 b=2
*(sp + 0) = 134514016
*(sp + 1) = 0
*(sp + 2) = 0
*(sp + 3) = 134513373
*(sp + 4) = 8239384
*(sp + 5) = 134513228
*(sp + 6) = 6
*(sp + 7) = -1076716032
*(sp + 8) = 134513456
*(sp + 9) = 0
*(sp + 10) = -1076715960
*(sp + 11) = 134513779
*(sp + 12) = 9  //value of arg a
*(sp + 13) = 2  //value of arg b
*(sp + 14) = 134513817
*(sp + 15) = 10612724

        a=1 b=4
*(sp + 0) = 134514016
*(sp + 1) = 0
*(sp + 2) = 0
*(sp + 3) = 134513373
*(sp + 4) = 8239384
*(sp + 5) = 134513228
*(sp + 6) = 6
*(sp + 7) = -1076716032
*(sp + 8) = 134513456
*(sp + 9) = 0
*(sp + 10) = -1076715960
*(sp + 11) = 134513799
*(sp + 12) = 1  //value of arg a
*(sp + 13) = 4  //value of arg b 
*(sp + 14) = 134513817
*(sp + 15) = 10612724

Why function arguments are stored from offset 12 of SP? Also notice values at offset 0 to 10 are always same, and value at offset 11 increases by 20 on each invocation of function foo().

UPDATE: I found that gcc has in-build function to retrieve frame pointer address

void * __builtin_frame_address (unsigned int level)

When I print values at offsets starting from __builtin_frame_address(0) the function arguments start from offset 2. How can I confirm that this behavior is always consistent?

Answer 1

You must know the calling convention to know what order the arguments are pushed onto the stack, or even if they are on the stack. Many pass the first few arguments in registers. Even on x86, you have fastcall, pascal, register, stdcall and cdecl, just to name a few.

EDIT: Don't forget that printf is also a function, and local variables also go on the stack. So, in your sample app, you have your parameters (since it's cdecl), then your locals, then your function saved state and return address, then parameters to printf (maybe, not sure if it's cdecl or fastcall), then printf's locals by the time anything actually makes it to the screen.