Stack

rdi - first argument

rsi - second argument

rdx - third argument

rcx - fourth argument

r8 - fifth argument

r9 - sixth

Stack pointer

Commands

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

Example

[rsp] - top of stack will contain arguments count.

[rsp + 8] - will contain argv[0]

[rsp + 16] - will contain argv[1]

and so on...

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:





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.





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

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

Conclusion

Some time ago i started to write a series of posts about assembly x64 programming. It is third part and it will be about stack. The stack is special region in memory (built into the CPU), 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:Here we can see that after application runnning,is equal to 1. Then we call a function, which increasesvalue to 1, and nowvalue must be 2. After this execution continues from 8 line, where we comparevalue with 2. Also as we can read in System V AMD64 ABI , the first six function arguments passed in registers. They are:Next arguments will be passed in stack. So if we have function like this:Then first six arguments will be passed in registers, but 7 argument will be passed in stack.As i wroute about we have 16 general-purpose registers, and there are two interesting registers -andis the base pointer register. It points to the base of the current stack frame.is the stack pointer, which points to the top of current stack frame.We have two commands for work with stack:Let's look on one simple example:Here we can see that we put 1 toregister and 2 toregister. 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:Then we copy value from stack which has address. It means we get address of top of stack, add 8 to it and copy data by this address to. After itvalue 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.First of all we definesection 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 at .text section, which consists from code of program:Let's try to understand, what is happening here: Afterlabel first instruction get first value from stack and puts it toregister. If we run application with command line arguments, all of their will be in stack after running in following order:So we get command line arguments count and put it to. After it we comparewith 3. And if they are not equal we jump tolabel which just prints error message:Why we compareAfterwe will have number in. Now let's look atHere we put 0 toand 10 to. Than we exeute. If we look above at code before str_to_int call. We will see thatcontains integer number - sum of two command line arguments. With this instruction we devidevalue onvalue and get reminder inand whole part in. Next we add to48 and. 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 23We implemented to useful functionandfor 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:We already know how to print string withsyscall, but here is one interesting part. We must to calculate length of string. If you will look at, you will see that we incrementregister 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(sys_write number), 1 to(stdin), string length toand pointer to the top of stack to(start of string). And finish our program:That's all.It was third part of series of posts about x64 assembly programming in Linux. If you will have questions/suggestions write me a comment. Sour code of all examples you can find - here You can find all parts: