15 Aug 2014, 00:00

Say hello to x86_64 Assembly [part 3]


The stack is special region in memory, which operates on the principle lifo (Last Input, First Output).

We have 16 general-purpose registers for temporary data storage. They are RAX, RBX, RCX, RDX, RDI, RSI, RBP, RSP and R8-R15. It’s too few for serious applications. So we can store data in the stack. Yet another usage of stack is following: When we call a function, return address copied in stack. After end of function execution, address copied in commands counter (RIP) and application continue to executes from next place after function.

For example:

global _start

section .text

		mov rax, 1
		call incRax
		cmp rax, 2
		jne exit
		;; Do something

		inc rax

Here we can see that after application runnning, rax is equal to 1. Then we call a function incRax, which increases rax value to 1, and now rax value must be 2. After this execution continues from 8 line, where we compare rax value with 2. Also as we can read in System V AMD64 ABI, the first six function arguments passed in registers. They are:

  • rdi - first argument
  • rsi - second argument
  • rdx - third argument
  • rcx - fourth argument
  • r8 - fifth argument
  • r9 - sixth

Next arguments will be passed in stack. So if we have function like this:

int foo(int a1, int a2, int a3, int a4, int a5, int a6, int a7)
    return (a1 + a2 - a3 - a4 + a5 - a6) * a7;

Then first six arguments will be passed in registers, but 7 argument will be passed in stack.

Stack pointer

As i wroute about we have 16 general-purpose registers, and there are two interesting registers - RSP and RBP. RBP is the base pointer register. It points to the base of the current stack frame. RSP is the stack pointer, which points to the top of current stack frame.


We have two commands for work with stack:

  • push argument - increments stack pointer (RSP) and stores argument in location pointed by stack pointer
  • pop argument - copied data to argument from location pointed by stack pointer

Let’s look on one simple example:

global _start

section .text

		mov rax, 1
		mov rdx, 2
		push rax
		push rdx

		mov rax, [rsp + 8]

		;; Do something

Here we can see that we put 1 to rax register and 2 to rdx register. After it we push to stack values of these registers. Stack works as LIFO (Last In First Out). So after this stack or our application will have following structure:

stack diagram

Then we copy value from stack which has address rsp + 8. It means we get address of top of stack, add 8 to it and copy data by this address to rax. After it rax value will be 1.


Let’s see one example. We will write simple program, which will get two command line arguments. Will get sum of this arguments and print result.

section .data
		SYS_WRITE equ 1
		STD_IN    equ 1
		SYS_EXIT  equ 60
		EXIT_CODE equ 0

		NEW_LINE   db 0xa
		WRONG_ARGC db "Must be two command line argument", 0xa

First of all we define .data section with some values. Here we have four constants for linux syscalls, for sys_write, sys_exit and etc… And also we have two strings: First is just new line symbol and second is error message.

Let’s look on the .text section, which consists from code of program:

section .text
        global _start

		pop rcx
		cmp rcx, 3
		jne argcError

		add rsp, 8
		pop rsi
		call str_to_int

		mov r10, rax
		pop rsi
		call str_to_int
		mov r11, rax

		add r10, r11

Let’s try to understand, what is happening here: After _start label first instruction get first value from stack and puts it to rcx register. If we run application with command line arguments, all of their will be in stack after running in following order:

    [rsp] - top of stack will contain arguments count.
    [rsp + 8] - will contain argv[0]
    [rsp + 16] - will contain argv[1]
    and so on...

So we get command line arguments count and put it to rcx. After it we compare rcx with 3. And if they are not equal we jump to argcError label which just prints error message:

    ;; sys_write syscall
    mov     rax, 1
    ;; file descritor, standard output
	mov     rdi, 1
    ;; message address
    mov     rsi, WRONG_ARGC
    ;; length of message
    mov     rdx, 34
    ;; call write syscall
    ;; exit from program
	jmp exit

Why we compare with 3 when we have two arguments. It’s simple. First argument is a program name, and all after it are command line arguments which we passed to program. Ok, if we passed two command line arguments we go next to 10 line. Here we shift rsp to 8 and thereby missing the first argument - the name of the program. Now rsp points to first command line argument which we passed. We get it with pop command and put it to rsi register and call function for converting it to integer. Next we read about str_to_int implementation. After our function ends to work we have integer value in rax register and we save it in r10 register. After this we do the same operation but with r11. In the end we have two integer values in r10 and r11 registers, now we can get sum of it with add command. Now we must convert result to string and print it. Let’s see how to do it:

mov rax, r10
;; number counter
xor r12, r12
;; convert to string
jmp int_to_str

Here we put sum of command line arguments to rax register, set r12 to zero and jump to int_to_str. Ok now we have base of our program. We already know how to print string and we have what to print. Let’s see at str_to_int and int_to_str implementation.

            xor rax, rax
            mov rcx,  10
	    cmp [rsi], byte 0
	    je return_str
	    mov bl, [rsi]
            sub bl, 48
	    mul rcx
	    add rax, rbx
	    inc rsi
	    jmp next


At the start of str_to_int, we set up rax to 0 and rcx to 10. Then we go to next label. As you can see in above example (first line before first call of str_to_int) we put argv[1] in rsi from stack. Now we compare first byte of rsi with 0, because every string ends with NULL symbol and if it is we return. If it is not 0 we copy it’s value to one byte bl register and substract 48 from it. Why 48? All numbers from 0 to 9 have 48 to 57 codes in asci table. So if we substract from number symbol 48 (for example from 57) we get number. Then we multiply rax on rcx (which has value - 10). After this we increment rsi for getting next byte and loop again. Algorthm is simple. For example if rsi points to ‘5’ ‘7’ ‘6’ ‘\000’ sequence, then will be following steps:

    rax = 0
    get first byte - 5 and put it to rbx
    rax * 10 --> rax = 0 * 10
    rax = rax + rbx = 0 + 5
    Get second byte - 7 and put it to rbx
    rax * 10 --> rax = 5 * 10 = 50
    rax = rax + rbx = 50 + 7 = 57
    and loop it while rsi is not \000

After str_to_int we will have number in rax. Now let’s look at int_to_str:

		mov rdx, 0
		mov rbx, 10
		div rbx
		add rdx, 48
		add rdx, 0x0
		push rdx
		inc r12
		cmp rax, 0x0
		jne int_to_str
		jmp print

Here we put 0 to rdx and 10 to rbx. Than we exeute div rbx. If we look above at code before str_to_int call. We will see that rax contains integer number - sum of two command line arguments. With this instruction we devide rax value on rbx value and get reminder in rdx and whole part in rax. Next we add to rdx 48 and 0x0. After adding 48 we’ll get asci symbol of this number and all strings much be ended with 0x0. After this we save symbol to stack, increment r12 (it’s 0 at first iteration, we set it to 0 at the _start) and compare rax with 0, if it is 0 it means that we ended to convert integer to string. Algorithm step by step is following: For example we have number 23

    123 / 10. rax = 12; rdx = 3
    rdx + 48 = "3"
    push "3" to stack
    compare rax with 0 if no go again
    12 / 10. rax = 1; rdx = 2
    rdx + 48 = "2"
    push "2" to stack
    compare rax with 0, if yes we can finish function execution and we will have "2" "3" ... in stack

We implemented two useful function int_to_str and str_to_int for converting integer number to string and vice versa. Now we have sum of two integers which was converted into string and saved in the stack. We can print result:

	;;;; calculate number length
	mov rax, 1
	mul r12
	mov r12, 8
	mul r12
	mov rdx, rax

	;;;; print sum
	mov rax, SYS_WRITE
	mov rdi, STD_IN
	mov rsi, rsp
	;; call sys_write

    jmp exit

We already know how to print string with sys_write syscall, but here is one interesting part. We must to calculate length of string. If you will look on the int_to_str, you will see that we increment r12 register every iteration, so it contains amount of digits in our number. We must multiple it to 8 (because we pushed every symbol to stack) and it will be length of our string which need to print. After this we as everytime put 1 to rax (sys_write number), 1 to rdi (stdin), string length to rdx and pointer to the top of stack to rsi (start of string). And finish our program:

	mov rax, SYS_EXIT
	exit code
	mov rdi, EXIT_CODE

That’s All.