Basic Assembly Tutorial, part 3

February 2016, Cremona, Italy

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GOTO lesson 4

Here we are to part III. Things are starting to get interesting! This time we see macros and basic operations.

Constants with EQU

This is a pretty simple topic. You can assign a constant with the EQU directive. The assembler will substitute the label you choose with the given value.

my_length equ 13

One useful way to use equ is defining the length of a data section, such as a string.
section .data msg db 'This is a message.' len equ $ - msg
The meaning is simple: remember that the name 'msg' is just a label for a memory address, right? The assembler stores data in order. One piece of data after the other, consecutively. The '$' symbol means 'here' and you subtract the position labeled 'msg', thus we obtain the length of our stored message.


Macros are groups of instruction that can be very handy. In NASM, you specify the start and end of a macro with the directives "%macro" and "%endmacro". Parameters to the macro are passed just like you do for the mov directive, separed by commas. Here's the syntax:
%macro name-of-macro number-of-parameters [instructions] %endmacro
Note that an extensive use of macros will result in a pretty ugly code. For sets of instructions of more than a few lines long, one should better use subroutines, that can be stored in another file if you like, making everything more clean. "Yes, but one can include an extern file containing macros" you're saying.. Macros are ugly and subroutines are good. Remember this :-)

Example: macros for printing and for exiting
;;;;;; Macro print_msg ;;;;;;;;;;;;;;;;;;;;;;;;;;;;; %macro print_msg 2 ; This macro has 2 parameters: the message address ; and its length. ; The macro fills the registers and calls sys_write mov eax, 4 ; sys_write mov ebx, 1 ; stdout mov ecx, %1 ; passing first parameter to ecx mov edx, %2 ; passing second parameter to edx int 0x80 ; perform system call %endmacro ;;;;;; Macro exit ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; %macro exit 1 ; This macro exits with a specified error status mov eax, 1 ; sys_exit mov ebx, %1 ; parameter, error status int 0x80 %endmacro ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; section .data msg db 'Here is my message.', 0xA ; 0xA is newline len equ $ - msg ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; section .text global _start _start: ; Printing message print_msg msg, len ; 'msg' and 'len' are the parameters ; Now exit exit 0 ; 0 is the error status

Note that:
  1. I have defined 'len' with the equ directive, it is thus a constant.
  2. I have defined the macro *before* calling it in the .text section.
  3. Into the macro, you refer to parameters with the '%' symbol

Basic Instructions

Time to introduce some basic instructions.

Instruction Meaning 1st Operand 2nd Operand
INC Increments by 1 register or variable register / variable in memory -
DEC decrements by 1 register or variable register / variable in memory -
ADD binary addition, adds value to 1st operand register / variable to increment quantity to add
SUB binary subtraction, subtracts value from 1st operand register / variable to decrement quantity to subtract

Example 1

This example:
  1. saves in memory the decimal value 65, that is ASCII character 'A'
  2. increments it to 66 (that is ASCII 'B') and prints it
The output of the program is simply 'AB'.
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; section .data val1 db 65 ; that is ASCII character 'A' ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; section .text global _start _start: ; Print value at address val1 mov eax, 4 ; sys_write mov ebx, 1 ; stdout mov ecx, val1 ; value in memory mov edx, 1 ; 1 byte long int 0x80 ; increment val1. New value should be 66, that is 'B' ; in ascii code. ; I move the value at val1 to ah rgister, since has ; the same dimensions as my data mov ah, [val1] ; moving value into ah inc ah ; increment ah mov [val1], ah ; move value at ah into val1 ; Print value at val1 mov eax, 4 mov ebx, 1 mov ecx, val1 mov edx, 1 int 0x80 ; exit mov eax, 1 mov ebx, 0 int 0x80

Example 2: horsing around with numbers

Well, since sys_write interprets its input as an ASCII code, printing the result of operations between numbers is not trivial. Since all we can do is entering characters from keyboard, it's useful to see a way to convert the ASCII character '1' to the actual number 1.
In this example we enter a number (a single digit, from 0 to 9) and sum it to another digit. In order to do this:
  1. we take a character as keyboard input (say, '1', that is ASCII 0x31, decimal 49)
  2. we convert it to the corresponding digit (say, 1)
  3. we perform the operation
  4. we transform the result back to ASCII and print it
The result must be smaller than 9 or the program won't work... :-(
Take a look at the ascii table: since the number 0 corresponds to decimal 48 and subsequent numbers follow in order, one can convert the ASCII value to the actual value by subtracting the value 48.
;;;;;; MACROS ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; ; ---- print_msg macro ---- %macro print_msg 2 mov edx, %2 ; 2nd argument, message length mov ecx, %1 ; 1st argument, memory position mov ebx, 1 ; stdout mov eax, 4 ; sys_write int 0x80 ; perform system call %endmacro ; ---- get_key macro ---- %macro get_key 1 mov edx, 2 ; length: 1 byte mov ecx, %1 ; destination mov ebx, 0 ; stdin mov eax, 3 ; sys_read int 0x80 %endmacro ; ---- exit macro ---- %macro exit 1 mov ebx, %1 ; error status mov eax, 1 ; sys_exit int 0x80 ; perform system call %endmacro ;;;;;; Data Section ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; section .data msg1 db 'Enter first digit: ' len1 equ $ - msg1 msg2 db 'You have entered: ' len2 equ $ - msg2 msg3 db 0xA, 0xA, 'Enter second digit: ' len3 equ $ - msg3 msg4 db 0xA, 0xA, 'The sum is: ' len4 equ $ - msg4 ;;;;;; bss Section ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; section .bss ; Please note that if you enter a character and then ; press ENTER, sys_read will get TWO characters (also ; ENTER is included!!!) Thus digit1 and digit2 must be ; 2 bytes long. digit1 resb 2 ; 1 byte is *NOT* enough! digit2 resb 2 ; 1 byte is *NOT* enough, again. sumresult resb 1 ; result of sum ;;;;;; text Section ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; section .text global _start _start: ; Print message: 'Enter a digit:' and read keyboard print_msg msg1, len1 get_key digit1 ; Print message: 'You have entered:' and print digit print_msg msg2, len2 print_msg digit1, 1 ; Print message: 'Now enter another one:' and read keyboard print_msg msg3, len3 get_key digit2 ; Print message: 'You have entered:' and print digit print_msg msg2, len2 print_msg digit2, 1 ; SUM THE DIGITS. ; Since we are dealing with 1 digit numbers (saved in 1 byte of ; memory), using the ah accumulator is enough (it is 1 byte ; long). ; STEP 1) put data into registers ; NOTE THAT I RESERVED 2 BYTES FOR digit1 AND digit2 ; BUT I NOW PUT THEM INTO 1 BYTE REGISTER! THIS IS OK mov ah, [digit1] mov bh, [digit2] ; STEP 2) subtract decimal 48 to data since it is in ASCII form ; note that subtracting character '0' is the same since ; its ascii value is precisely 48 sub ah, '0' sub bh, '0' ; STEP 3) sum the results into the accumulator ah for example add ah, bh ; STEP 4) sum again the value decimal 48 so it becomes ASCII ; again add ah, 48 ; STEP 5) save it to variable mov [sumresult], ah ; Print the variable print_msg msg4, len4 print_msg sumresult, 1 ; exit exit 0
There are some CAVEATs in the code above.