How exactly does the callstack work?

The call stack could also be called a frame stack.
The things that are stacked after the LIFO principle are not the local variables but the entire stack frames ("calls") of the functions being called. The local variables are pushed and popped together with those frames in the so-called function prologue and epilogue, respectively.

Inside the frame the order of the variables is completely unspecified; Compilers "reorder" the positions of local variables inside a frame appropriately to optimize their alignment so the processor can fetch them as quickly as possible. The crucial fact is that the offset of the variables relative to some fixed address is constant throughout the lifetime of the frame - so it suffices to take an anchor address, say, the address of the frame itself, and work with offsets of that address to the variables. Such an anchor address is actually contained in the so-called base or frame pointer which is stored in the EBP register. The offsets, on the other hand, are clearly known at compile time and are therefore hardcoded into the machine code.

This graphic from Wikipedia shows what the typical call stack is structured like¹:

https://i.stack.imgur.com/uiCRx.png

Add the offset of a variable we want to access to the address contained in the frame pointer and we get the address of our variable. So shortly said, the code just accesses them directly via constant compile-time offsets from the base pointer; It's simple pointer arithmetic.

Example

#include <iostream>

int main()
{
    char c = std::cin.get();
    std::cout << c;
}

gcc.godbolt.org gives us

main:
    pushq   %rbp
    movq    %rsp, %rbp
    subq    $16, %rsp

    movl    std::cin, %edi
    call    std::basic_istream<char, std::char_traits<char> >::get()
    movb    %al, -1(%rbp)
    movsbl  -1(%rbp), %eax
    movl    %eax, %esi
    movl    std::cout, %edi
    call    [... the insertion operator for char, long thing... ]

    movl    $0, %eax
    leave
    ret

.. for main. I divided the code into three subsections. The function prologue consists of the first three operations:

Base pointer is pushed onto the stack.

The stack pointer is saved in the base pointer

The stack pointer is subtracted to make room for local variables.

Then cin is moved into the EDI register² and get is called; The return value is in EAX.

So far so good. Now the interesting thing happens:

The low-order byte of EAX, designated by the 8-bit register AL, is taken and stored in the byte right after the base pointer: That is -1(%rbp), the offset of the base pointer is -1. This byte is our variable c. The offset is negative because the stack grows downwards on x86. The next operation stores c in EAX: EAX is moved to ESI, cout is moved to EDI and then the insertion operator is called with cout and c being the arguments.

Finally,

The return value of main is stored in EAX: 0. That is because of the implicit return statement. You might also see xorl rax rax instead of movl.

leave and return to the call site. leave is abbreviating this epilogue and implicitly Replaces the stack pointer with the base pointer and Pops the base pointer.

Replaces the stack pointer with the base pointer and

Pops the base pointer.

After this operation and ret have been performed, the frame has effectively been popped, although the caller still has to clean up the arguments as we're using the cdecl calling convention. Other conventions, e.g. stdcall, require the callee to tidy up, e.g. by passing the amount of bytes to ret.

Frame Pointer Omission

It is also possible not to use offsets from the base/frame pointer but from the stack pointer (ESB) instead. This makes the EBP-register that would otherwise contain the frame pointer value available for arbitrary use - but it can make debugging impossible on some machines, and will be implicitly turned off for some functions. It is particularly useful when compiling for processors with only few registers, including x86.

This optimization is known as FPO (frame pointer omission) and set by -fomit-frame-pointer in GCC and -Oy in Clang; note that it is implicitly triggered by every optimization level > 0 if and only if debugging is still possible, since it doesn't have any costs apart from that. For further information see here and here.

1 As pointed out in the comments, the frame pointer is presumably meant to point to the address after the return address.

2 Note that the registers that start with R are the 64-bit counterparts of the ones that start with E. EAX designates the four low-order bytes of RAX. I used the names of the 32-bit registers for clarity.

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Because obviously, the next thing we need is to work with a and b but that would mean that the OS/CPU (?) has to pop out d and c first to get back to a and b. But then it would shoot itself in the foot because it needs c and d in the next line.

In short:

There is no need to pop the arguments. The arguments passed by caller foo to function doSomething and the local variables in doSomething can all be referenced as an offset from the base pointer.
So,

When a function call is made, function's arguments are PUSHed on stack. These arguments are further referenced by base pointer. When the function returns to its caller, the arguments of the returning function are POPed from the stack using LIFO method.

In detail:

The rule is that each function call results in a creation of a stack frame (with the minimum being the address to return to). So, if funcA calls funcB and funcB calls funcC, three stack frames are set up one on top of the another. When a function returns, its frame becomes invalid. A well-behaved function acts only on its own stack frame and does not trespass on another's. In another words the POPing is performed to the stack frame on the top (when returning from the function).

https://i.stack.imgur.com/x4A9Q.png

The stack in your question is setup by caller foo. When doSomething and doAnotherThing are called, then they setup their own stack. The figure may help you to understand this:

https://i.stack.imgur.com/w19l1.png

Note that, to access the arguments, the function body will have to traverse down (higher addresses) from the location where the return address is stored, and to access the local variables, the function body will have to traverse up the stack (lower addresses) relative to the location where the return address is stored. In fact, typical compiler generated code for the function will do exactly this. The compiler dedicates a register called EBP for this (Base Pointer). Another name for the same is frame pointer. The compiler typically, as the first thing for the function body, pushes the current EBP value on to the stack and sets the EBP to the current ESP. This means, once this is done, in any part of the function code, argument 1 is EBP+8 away (4 bytes for each of caller's EBP and the return address), argument 2 is EBP+12(decimal) away, local variables are EBP-4n away.

.
.
.
[ebp - 4]  (1st local variable)
[ebp]      (old ebp value)
[ebp + 4]  (return address)
[ebp + 8]  (1st argument)
[ebp + 12] (2nd argument)
[ebp + 16] (3rd function argument) 

Take a look at the following C code for the formation of stack frame of the function:

void MyFunction(int x, int y, int z)
{
     int a, int b, int c;
     ...
}

When caller call it

MyFunction(10, 5, 2);  

the following code will be generated

^
| call _MyFunction  ; Equivalent to: 
|                   ; push eip + 2
|                   ; jmp _MyFunction
| push 2            ; Push first argument  
| push 5            ; Push second argument  
| push 10           ; Push third argument  

and the assembly code for the function will be (set-up by callee before returning)

^
| _MyFunction:
|  sub esp, 12 ; sizeof(a) + sizeof(b) + sizeof(c)
|  ;x = [ebp + 8], y = [ebp + 12], z = [ebp + 16]
|  ;a = [ebp - 4] = [esp + 8], b = [ebp - 8] = [esp + 4], c = [ebp - 12] =   [esp]
|  mov ebp, esp
|  push ebp
 

References:

Function Call Conventions and the Stack.

Frame Pointer and Local Variables.

x86 Disassembly/Functions and Stack Frames.

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Like others noted, there is no need to pop parameters, until they go out of scope.

I will paste some example from "Pointers and Memory" by Nick Parlante. I think the situation is a bit more simple than you envisioned.

Here is code:

void X() 
{
  int a = 1;
  int b = 2;

  // T1
  Y(a);

  // T3
  Y(b);

  // T5
}

void Y(int p) 
{
  int q;
  q = p + 2;
  // T2 (first time through), T4 (second time through)
}

The points in time T1, T2, etc. are marked in the code and the state of memory at that time is shown in the drawing:

https://i.stack.imgur.com/YqZhq.png

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Different processors and languages use a few different stack designs. Two traditional patterns on both the 8x86 and 68000 are called the Pascal calling convention and the C calling convention; each convention is handled the same way in both processors, except for the names of the registers. Each uses two registers to manage the stack and associated variables, called the stack pointer (SP or A7) and the frame pointer (BP or A6).

When calling subroutine using either convention, any parameters are be pushed on the stack before calling the routine. The routine's code then pushes the current value of the frame pointer onto the stack, copies the current value of the stack pointer to the frame pointer, and subtracts from the stack pointer the number of bytes used by local variables [if any]. Once that is done, even if additional data are pushed onto the stack, all local variables will be stored at variables with a constant negative displacement from the stack pointer, and all parameters that were pushed on the stack by the caller may be accessed at a constant positive displacement from the frame pointer.

The difference between the two conventions lies in the way they handle an exit from subroutine. In the C convention, the returning function copies the frame pointer to the stack pointer [restoring it to the value it had just after the old frame pointer was pushed], pops the old frame pointer value, and performs a return. Any parameters the caller had pushed on the stack before the call will remain there. In the Pascal convention, after popping the old frame pointer, the processor pops the function return address, adds to the stack pointer the number of bytes of parameters pushed by the caller, and then goes to the popped return address. On the original 68000 it was necessary to use a 3-instruction sequence to remove the caller's parameters; the 8x86 and all 680x0 processors after the original included a "ret N" [or 680x0 equivalent] instruction which would add N to the stack pointer when performing a return.

The Pascal convention has the advantage of saving a little bit of code on the caller side, since the caller doesn't have to update the stack pointer after a function call. It requires, however, that the called function know exactly how many bytes worth of parameters the caller is going to put on the stack. Failing to push the proper number of parameters onto the stack before calling a function which uses the Pascal convention is almost guaranteed to cause a crash. This is offset, however, by the fact that a little extra code within each called method will save code at the places where the method is called. For that reason, most of the original Macintosh toolbox routines used the Pascal calling convention.

The C calling convention has the advantage of allowing routines to accept a variable number of parameters, and being robust even if a routine doesn't use all the parameters that are passed (the caller will know how many bytes worth of parameters it pushed, and will thus be able to clean them up). Further, it isn't necessary to perform stack cleanup after every function call. If a routine calls four functions in sequence, each of which used four bytes worth of parameters, it may--instead of using an ADD SP,4 after each call, use one ADD SP,16 after the last call to cleanup the parameters from all four calls.

Nowadays the described calling conventions are considered somewhat antiquated. Since compilers have gotten more efficient at register usage, it is common to have methods accept a few parameters in registers rather than requiring that all parameters be pushed on the stack; if a method can use registers to hold all the parameters and local variables, there's no need to use a frame pointer, and thus no need to save and restore the old one. Still, it's sometimes necessary to use the older calling conventions when calling libraries that was linked to use them.

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There are already some really good answers here. However, if you are still concerned about the LIFO behavior of the stack, think of it as a stack of frames, rather than a stack of variables. What I mean to suggest is that, although a function may access variables that are not on the top of the stack, it is still only operating on the item at the top of the stack: a single stack frame.

Of course, there are exceptions to this. The local variables of the entire call chain are still allocated and available. But they won't be accessed directly. Instead, they are passed by reference (or by pointer, which is really only different semantically). In this case a local variable of a stack frame much further down can be accessed. But even in this case, the currently executing function is still only operating on its own local data. It is accessing a reference stored in its own stack frame, which may be a reference to something on the heap, in static memory, or further down the stack.

This is the part of the stack abstraction that makes functions callable in any order, and allows recursion. The top stack frame is the only object that is directly accessed by the code. Anything else is accessed indirectly (through a pointer that lives in the top stack frame).

It might be instructive to look at the assembly of your little program, especially if you compile without optimization. I think you will see that all of the memory access in your function happens through an offset from the stack frame pointer, which is the how the code for the function will be written by the compiler. In the case of a pass by reference, you would see indirect memory access instructions through a pointer that is stored at some offset from the stack frame pointer.

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The call stack is not actually a stack data structure. Behind the scenes, the computers we use are implementations of the random access machine architecture. So, a and b can be directly accessed.

Behind the scenes, the machine does:

get "a" equals reading the value of the fourth element below stack top.

get "b" equals reading the value of the third element below stack top.

http://en.wikipedia.org/wiki/Random-access_machine

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Here is a diagram I created for a call stack for a C++ program on Windows that uses the Windows x64 calling convention. It's more accurate and contemporary than the google image versions:

https://i.stack.imgur.com/45bwQ.png

And corresponding to the exact structure of the above diagram, here is a debug of notepad.exe x64 on windows 7, where the first instruction of a function, 'current function' (because I forgot what function it is), is about to execute.

https://i.stack.imgur.com/vVYgW.png

The low addresses and high addresses are swapped so the stack is climbing upwards in this diagram (it is a vertical flip of the first diagram, also note that the data is formatted to show quadwords and not bytes, so the little endianism cannot be seen). Black is the home space; blue is the return address, which is an offset into the caller function or label in the caller function to the instruction after the call; orange is the alignment; and pink is where rsp is pointing after the prologue of the function, or rather, before the call is made if you are using alloca. The homespace_for_the_next_function+return_address value is the smallest allowed frame on windows, and because the 16 byte rsp alignment right at the start of the called function must be maintained, it includes an 8 byte alignment as well, such that rsp pointing to the first byte after the return address will be aligned to 16 bytes (because rsp was guaranteed to be aligned to 16 bytes when the function was called and homespace+return_address = 40, which is not divisible by 16 so you need an extra 8 bytes to ensure the rsp will be aligned after the function makes a call). Because these functions do not require any stack locals (because they can be optimised into registers) or stack parameters/return values (as they fit in registers) and do not use any of the other fields, the stack frames in green are all alignment+homespace+return_address in size.

The red function lines outline what the callee function logically 'owns' + reads / modifies by value in the calling convention without needing a reference to it (it can modify a parameter passed on the stack that was too big to pass in a register on -Ofast), and is the classic conception of a stack frame. The green frames demarcate what results from the call and the allocation the called function makes: The first green frame shows what the RtlUserThreadStart actually allocates in the duration of the function call (from immediately before the call to executing the next call instruction) and goes from the first byte before the return address to the final byte allocated by the function prologue (or more if using alloca). RtlUserThreadStart allocates the return address itself as null, so you see a sub rsp, 48h and not sub rsp, 40h in the prologue, because there is no call to RtlUserThreadStart, it just begins execution at that rip at the base of the stack.

Stack space that is needed by the function is assigned in the function prologue by decrementing the stack pointer.

For example, take the following C++, and the MASM it compiles to (-O0).

typedef struct _struc {int a;} struc, pstruc;
int func(){return 1;}
int square(_struc num) {
    int a=1;
    int b=2;
    int c=3;
    return func();
}

_DATA SEGMENT
_DATA ENDS

int func(void) PROC ; func
  mov eax, 1
  ret 0
int func(void) ENDP ; func

a$ = 32  //4 bytes from rsp+32 to rsp+35
b$ = 36
c$ = 40 
num$ = 64 

//masm shows stack locals and params relative to the address of rsp; the rsp address
//is the rsp in the main body of the function after the prolog and before the epilog

int square(_struc) PROC ; square
$LN3:
  mov DWORD PTR [rsp+8], ecx
  sub rsp, 56 ; 00000038H
  mov DWORD PTR a$[rsp], 1
  mov DWORD PTR b$[rsp], 2
  mov DWORD PTR c$[rsp], 3
  call int func(void) ; func
  add rsp, 56 ; 00000038H
  ret 0
int square(_struc) ENDP ; square

As can be seen, 56 bytes are reserved, and the green stack frame will be 64 bytes in size when the call instruction allocates the 8 byte return address as well.

The 56 bytes consist of 12 bytes of locals, 32 bytes of home space, and 12 bytes of alignment.

All callee register saving and storing register parameters in the home space happens in the prologue before the prologue reserves (using sub rsp, x instruction) stack space needed by the main body of the function. The alignment is at the highest address of the space reserved by the sub rsp, x instruction, and the final local variable in the function is assigned at the next lower address after that (and within the assignment for that primitive data type itself it starts at the lowest address of that assignment and works towards the higher addresses, bytewise, because it is little endian), such that the first primitive type (array cell, variable etc.) in the function is at the top of the stack, although the locals can be allocated in any order. This is shown in the following diagram for a different random example code to the above, that does not call any functions (still using x64 Windows cc):

https://i.stack.imgur.com/GQcWi.png

If you remove the call to func(), it only reserves 24 bytes, i.e. 12 bytes of of locals and 12 bytes of alignment. The alignment is at the start of the frame. When a function pushes something to the stack or reserves space on the stack by decrementing the rsp, rsp needs to be aligned, regardless of whether it is going to call another function or not. If the allocation of stack space can be optimised out and no homespace+return_addreess is required because the function does not make a call, then there will be no alignment requirement as rsp does not change. It also does not need to align if the stack will be aligned by 16 with just the locals (+ homespace+return_address if it makes a call) that it needs to allocate, essentially it rounds up the space it needs to allocate to a 16 byte boundary.

rbp is not used on the x64 Windows calling convention unless alloca is used.

On gcc 32 bit cdecl and 64 bit system V calling conventions, rbp is used, and the new rbp points to the first byte after the old rbp (only if compiling using -O0, because it is saved to the stack on -O0, otherwise, rbp will point to the first byte after the return address). On these calling conventions, if compiling using -O0, it will, after callee saved registers, store register parameters to the stack, and this will be relative to rbp and part of the stack reservation done by the rsp decrement. Data within the stack reservation done by the rsp decrement is accessed relative rbp rather than rsp, unlike Windows x64 cc. On the Windows x64 calling convention, it stores parameters that were passed to it in registers to the homespace that was assigned for it if it is a varargs function or compiling using -O0. If it is not a varargs function then on -O1, it will not write them to the homespace but the homespace will still be provided to it by the calling function, this means that it actually accesses those variables from the register rather from the homespace location on the stack after it stores it there, unlike O0 (which saves them to the homespace and then accesses them through the stack and not the registers).

If a function call is placed in the function represented by the previous diagram, the stack will now look like this before the callee function's prologue starts (Windows x64 cc):

https://i.stack.imgur.com/cL8yi.png

Orange indicates the part that the callee can freely arrange (arrays and structs remain contiguous of course, and work their way towards higher addresses, each element being little endian), so it can put the variables and the return value allocation in any order, and it passes a pointer for the return value allocation in rcx for the callee to write to when the return type of the function it is calling cannot be passed in rax. On -O0, if the return value cannot be passed in rax, there is also an anonymous variable created (as well as the return value space and as well as any variable it is assigned to, so there can be 3 copies of the struct). -Ofast cant optimise out the return value space because it is return by value, but it optimises out the anonymous return variable if the return value is not used, or assigns it straight to the variable the return value is being assigned to without creating an anonymous variable, so -Ofast has 2 / 1 copies and -O0 has 3 / 2 copies (return value assigned to a variable / return value not assigned to a variable). Blue indicates the part the callee must provide in exact order for the calling convention of the callee (the parameters must be in that order, such that the first stack parameter from left to right in the function signature is at the top of the stack, which is the same as how cdecl (which is a 32 bit cc) orders its stack parameters. The alignment for the callee can however be in any location, although I've only ever seen it to be between the locals and callee pushed registers.

If the function calls multiple functions, the call is in the same place on the stack for all the different possible callsites in the function, this is because the prologue caters for the whole function, including all calls it makes, and the parameters and homespace for any called function is always at the end of the allocation made in the prologue.

It turns out that C/C++ Microsoft calling convention only passes a struct in the registers if it fits into one register, otherwise it copies the local / anonymous variable and passes a pointer to it in the first available register. On gcc C/C++, if the struct does not fit in the first 2 parameter registers then it's passed on the stack and a pointer to it is not passed because the callee knows where it is due to the calling convention.

Arrays are passed by reference regardless of their size. So if you need to use rcx as the pointer to the return value allocation then if the first parameter is an array, the pointer will be passed in rdx, which will be a pointer to the local variable that is being passed. In this case, it does not need to copy it to the stack as a parameter because it's not passed by value. The pointer however is passed on the stack when passing by reference if there are no registers available to pass the pointer in.

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