How are the function local variables accessed from the stack?

Question

From http://www.learncpp.com/cpp-tutorial/79-the-stack-and-the-heap/

Here is the sequence of steps that takes place when a function is called:

Answer 1

The stack is a metaphoric stack. Remember it is still a RAM, so you can access each address without popping the rest, if you know what you are looking for.

Since the automatic variable's size is known at compile time - the compiler marks offset for each variable, the offset is determined from where the automatic variables section on stack start [or the stack's head, both are valid and the specific implementation depends might depend on architecture], and it access them by merely: start + offset for each variable's offset.

Answer 2

No they are not. The stack pointer (typically the esp registry) points to a, esp+8h points to b, esp+16h points to c and so on. There's no need for a to be popped.

Note that this is an implementation detail. You shouldn't worry about these. The number I've given is purely theoretical, on some architectures descending addresses are given to latter parameters, on others the other way around. There's no guarantee this happens.

EDIT: It seems to me like that's not a very reliable source of information. It speaks of stack and heap, but these are implementation details, and might not even be there.

There's no constraint in the standard for anything to be implemented via a stack either. For example, I have the following code generated:

void foo(int x, int y, int z)
{
01241380  push        ebp  
01241381  mov         ebp,esp 
01241383  sub         esp,0CCh 
01241389  push        ebx  
0124138A  push        esi  
0124138B  push        edi  
0124138C  lea         edi,[ebp-0CCh] 
01241392  mov         ecx,33h 
01241397  mov         eax,0CCCCCCCCh 
0124139C  rep stos    dword ptr es:[edi] 
    int c = x;
0124139E  mov         eax,dword ptr [x] 
012413A1  mov         dword ptr [c],eax 
    c = y;
012413A4  mov         eax,dword ptr [y] 
012413A7  mov         dword ptr [c],eax 
    c = z;
012413AA  mov         eax,dword ptr [z] 
012413AD  mov         dword ptr [c],eax 
}
012413B0  pop         edi  
012413B1  pop         esi  
012413B2  pop         ebx  
012413B3  mov         esp,ebp 
012413B5  pop         ebp  

So you see, there's no stack there. The runtime has direct access to the elements: dword ptr [x], etc.

Answer 3

It uses the stack pointer and a relative adress to point out c.