Assembly – Introduction

What is Assembly Language?

Each personal computer has a microprocessor that manages the computer”s arithmetical, logical, and control activities.

Each family of processors has its own set of instructions for handling various operations such as getting input from keyboard, displaying information on screen and performing various other jobs. These set of instructions are called ”machine language instructions”.

A processor understands only machine language instructions, which are strings of 1”s and 0”s. However, machine language is too obscure and complex for using in software development. So, the low-level assembly language is designed for a specific family of processors that represents various instructions in symbolic code and a more understandable form.

Advantages of Assembly Language

Having an understanding of assembly language makes one aware of −

• How programs interface with OS, processor, and BIOS;
• How data is represented in memory and other external devices;
• How the processor accesses and executes instruction;
• How instructions access and process data;
• How a program accesses external devices.

Other advantages of using assembly language are −

• It requires less memory and execution time;

• It allows hardware-specific complex jobs in an easier way;

• It is suitable for time-critical jobs;

• It is most suitable for writing interrupt service routines and other memory resident programs.

Basic Features of PC Hardware

The main internal hardware of a PC consists of processor, memory, and registers. Registers are processor components that hold data and address. To execute a program, the system copies it from the external device into the internal memory. The processor executes the program instructions.

The fundamental unit of computer storage is a bit; it could be ON (1) or OFF (0) and a group of 8 related bits makes a byte on most of the modern computers.

So, the parity bit is used to make the number of bits in a byte odd. If the parity is even, the system assumes that there had been a parity error (though rare), which might have been caused due to hardware fault or electrical disturbance.

The processor supports the following data sizes −

• Word: a 2-byte data item
• Doubleword: a 4-byte (32 bit) data item
• Quadword: an 8-byte (64 bit) data item
• Paragraph: a 16-byte (128 bit) area
• Kilobyte: 1024 bytes
• Megabyte: 1,048,576 bytes

Binary Number System

Every number system uses positional notation, i.e., each position in which a digit is written has a different positional value. Each position is power of the base, which is 2 for binary number system, and these powers begin at 0 and increase by 1.

The following table shows the positional values for an 8-bit binary number, where all bits are set ON.

The value of a binary number is based on the presence of 1 bits and their positional value. So, the value of a given binary number is −

1 + 2 + 4 + 8 +16 + 32 + 64 + 128 = 255

which is same as 2⁸ – 1.

Hexadecimal Number System

Hexadecimal number system uses base 16. The digits in this system range from 0 to 15. By convention, the letters A through F is used to represent the hexadecimal digits corresponding to decimal values 10 through 15.

Hexadecimal numbers in computing is used for abbreviating lengthy binary representations. Basically, hexadecimal number system represents a binary data by dividing each byte in half and expressing the value of each half-byte. The following table provides the decimal, binary, and hexadecimal equivalents −

To convert a binary number to its hexadecimal equivalent, break it into groups of 4 consecutive groups each, starting from the right, and write those groups over the corresponding digits of the hexadecimal number.

Example − Binary number 1000 1100 1101 0001 is equivalent to hexadecimal – 8CD1

To convert a hexadecimal number to binary, just write each hexadecimal digit into its 4-digit binary equivalent.

Example − Hexadecimal number FAD8 is equivalent to binary – 1111 1010 1101 1000

Binary Arithmetic

The following table illustrates four simple rules for binary addition −

Rules (iii) and (iv) show a carry of a 1-bit into the next left position.

Example

A negative binary value is expressed in two”s complement notation. According to this rule, to convert a binary number to its negative value is to reverse its bit values and add 1.

Example

To subtract one value from another, convert the number being subtracted to two”s complement format and add the numbers.

Example

Subtract 42 from 53

Overflow of the last 1 bit is lost.

Addressing Data in Memory

The process through which the processor controls the execution of instructions is referred as the fetch-decode-execute cycle or the execution cycle. It consists of three continuous steps −

• Fetching the instruction from memory
• Decoding or identifying the instruction
• Executing the instruction

The processor may access one or more bytes of memory at a time. Let us consider a hexadecimal number 0725H. This number will require two bytes of memory. The high-order byte or most significant byte is 07 and the low-order byte is 25.

The processor stores data in reverse-byte sequence, i.e., a low-order byte is stored in a low memory address and a high-order byte in high memory address. So, if the processor brings the value 0725H from register to memory, it will transfer 25 first to the lower memory address and 07 to the next memory address.

x: memory address

When the processor gets the numeric data from memory to register, it again reverses the bytes. There are two kinds of memory addresses −

• Absolute address – a direct reference of specific location.

• Segment address (or offset) – starting address of a memory segment with the offset value.

Assembly – Environment Setup

Local Environment Setup

Assembly language is dependent upon the instruction set and the architecture of the processor. In this tutorial, we focus on Intel-32 processors like Pentium. To follow this tutorial, you will need −

• An IBM PC or any equivalent compatible computer
• A copy of Linux operating system
• A copy of NASM assembler program

There are many good assembler programs, such as −

• Microsoft Assembler (MASM)
• Borland Turbo Assembler (TASM)
• The GNU assembler (GAS)

We will use the NASM assembler, as it is −

• Free. You can download it from various web sources.
• Well documented and you will get lots of information on net.
• Could be used on both Linux and Windows.

Installing NASM

If you select “Development Tools” while installing Linux, you may get NASM installed along with the Linux operating system and you do not need to download and install it separately. For checking whether you already have NASM installed, take the following steps −

• Open a Linux terminal.

• Type whereis nasm and press ENTER.

• If it is already installed, then a line like, nasm: /usr/bin/nasm appears. Otherwise, you will see just nasm:, then you need to install NASM.

To install NASM, take the following steps −

• Check The netwide assembler (NASM) website for the latest version.

• Download the Linux source archive nasm-X.XX.ta.gz, where X.XX is the NASM version number in the archive.

• Unpack the archive into a directory which creates a subdirectory nasm-X. XX.

• cd to nasm-X.XX and type ./configure. This shell script will find the best C compiler to use and set up Makefiles accordingly.

• Type make to build the nasm and ndisasm binaries.

• Type make install to install nasm and ndisasm in /usr/local/bin and to install the man pages.

This should install NASM on your system. Alternatively, you can use an RPM distribution for the Fedora Linux. This version is simpler to install, just double-click the RPM file.

Assembly – Basic Syntax

An assembly program can be divided into three sections −

• The data section,

• The bss section, and

• The text section.

The data Section

The data section is used for declaring initialized data or constants. This data does not change at runtime. You can declare various constant values, file names, or buffer size, etc., in this section.

The syntax for declaring data section is −

section.data

The bss Section

The bss section is used for declaring variables. The syntax for declaring bss section is −

section.bss

The text section

The text section is used for keeping the actual code. This section must begin with the declaration global _start, which tells the kernel where the program execution begins.

The syntax for declaring text section is −

section.text
   global _start
_start:

Comments

Assembly language comment begins with a semicolon (;). It may contain any printable character including blank. It can appear on a line by itself, like −

; This program displays a message on screen

or, on the same line along with an instruction, like −

add eax, ebx     ; adds ebx to eax

Assembly Language Statements

Assembly language programs consist of three types of statements −

• Executable instructions or instructions,
• Assembler directives or pseudo-ops, and
• Macros.

The executable instructions or simply instructions tell the processor what to do. Each instruction consists of an operation code (opcode). Each executable instruction generates one machine language instruction.

The assembler directives or pseudo-ops tell the assembler about the various aspects of the assembly process. These are non-executable and do not generate machine language instructions.

Macros are basically a text substitution mechanism.

Syntax of Assembly Language Statements

Assembly language statements are entered one statement per line. Each statement follows the following format −

[label]   mnemonic   [operands]   [;comment]

The fields in the square brackets are optional. A basic instruction has two parts, the first one is the name of the instruction (or the mnemonic), which is to be executed, and the second are the operands or the parameters of the command.

Following are some examples of typical assembly language statements −

INC COUNT        ; Increment the memory variable COUNT

MOV TOTAL, 48    ; Transfer the value 48 in the 
                 ; memory variable TOTAL
					  
ADD AH, BH       ; Add the content of the 
                 ; BH register into the AH register
					  
AND MASK1, 128   ; Perform AND operation on the 
                 ; variable MASK1 and 128
					  
ADD MARKS, 10    ; Add 10 to the variable MARKS
MOV AL, 10       ; Transfer the value 10 to the AL register

The Hello World Program in Assembly

The following assembly language code displays the string ”Hello World” on the screen −

section	.text
   global _start     ;must be declared for linker (ld)
	
_start:	            ;tells linker entry point
   mov	edx,len     ;message length
   mov	ecx,msg     ;message to write
   mov	ebx,1       ;file descriptor (stdout)
   mov	eax,4       ;system call number (sys_write)
   int	0x80        ;call kernel
	
   mov	eax,1       ;system call number (sys_exit)
   int	0x80        ;call kernel

section	.data
msg db ''Hello, world!'', 0xa  ;string to be printed
len equ $ - msg     ;length of the string

Hello, world!

segment .text	   ;code segment
   global _start    ;must be declared for linker 
	
_start:	           ;tell linker entry point
   mov edx,len	   ;message length
   mov ecx,msg     ;message to write
   mov ebx,1	   ;file descriptor (stdout)
   mov eax,4	   ;system call number (sys_write)
   int 0x80	   ;call kernel

   mov eax,1       ;system call number (sys_exit)
   int 0x80	   ;call kernel

segment .data      ;data segment
msg	db ''Hello, world!'',0xa   ;our dear string
len	equ	$ - msg          ;length of our dear string

Hello, world!

section	.text
   global _start	 ;must be declared for linker (gcc)
	
_start:	         ;tell linker entry point
   mov	edx,len  ;message length
   mov	ecx,msg  ;message to write
   mov	ebx,1    ;file descriptor (stdout)
   mov	eax,4    ;system call number (sys_write)
   int	0x80     ;call kernel
	
   mov	edx,9    ;message length
   mov	ecx,s2   ;message to write
   mov	ebx,1    ;file descriptor (stdout)
   mov	eax,4    ;system call number (sys_write)
   int	0x80     ;call kernel
	
   mov	eax,1    ;system call number (sys_exit)
   int	0x80     ;call kernel
	
section	.data
msg db ''Displaying 9 stars'',0xa ;a message
len equ $ - msg  ;length of message
s2 times 9 db ''*''

Displaying 9 stars
*********

mov	eax,1		; system call number (sys_exit)
int	0x80		; call kernel

mov	edx,4		; message length
mov	ecx,msg		; message to write
mov	ebx,1		; file descriptor (stdout)
mov	eax,4		; system call number (sys_write)
int	0x80		; call kernel

section .data                           ;Data segment
   userMsg db ''Please enter a number: '' ;Ask the user to enter a number
   lenUserMsg equ $-userMsg             ;The length of the message
   dispMsg db ''You have entered: ''
   lenDispMsg equ $-dispMsg                 

section .bss           ;Uninitialized data
   num resb 5
	
section .text          ;Code Segment
   global _start
	
_start:                ;User prompt
   mov eax, 4
   mov ebx, 1
   mov ecx, userMsg
   mov edx, lenUserMsg
   int 80h

   ;Read and store the user input
   mov eax, 3
   mov ebx, 2
   mov ecx, num  
   mov edx, 5          ;5 bytes (numeric, 1 for sign) of that information
   int 80h
	
   ;Output the message ''The entered number is: ''
   mov eax, 4
   mov ebx, 1
   mov ecx, dispMsg
   mov edx, lenDispMsg
   int 80h  

   ;Output the number entered
   mov eax, 4
   mov ebx, 1
   mov ecx, num
   mov edx, 5
   int 80h  
    
   ; Exit code
   mov eax, 1
   mov ebx, 0
   int 80h

Please enter a number:
1234  
You have entered:1234

MOV DX, TAX_RATE   ; Register in first operand
MOV COUNT, CX	   ; Register in second operand
MOV EAX, EBX	   ; Both the operands are in registers

BYTE_VALUE  DB  150    ; A byte value is defined
WORD_VALUE  DW  300    ; A word value is defined
ADD  BYTE_VALUE, 65    ; An immediate operand 65 is added
MOV  AX, 45H           ; Immediate constant 45H is transferred to AX

ADD	BYTE_VALUE, DL	; Adds the register in the memory location
MOV	BX, WORD_VALUE	; Operand from the memory is added to register

BYTE_TABLE DB  14, 15, 22, 45      ; Tables of bytes
WORD_TABLE DW  134, 345, 564, 123  ; Tables of words

MOV CL, BYTE_TABLE[2]	; Gets the 3rd element of the BYTE_TABLE
MOV CL, BYTE_TABLE + 2	; Gets the 3rd element of the BYTE_TABLE
MOV CX, WORD_TABLE[3]	; Gets the 4th element of the WORD_TABLE
MOV CX, WORD_TABLE + 3	; Gets the 4th element of the WORD_TABLE

MY_TABLE TIMES 10 DW 0  ; Allocates 10 words (2 bytes) each initialized to 0
MOV EBX, [MY_TABLE]     ; Effective Address of MY_TABLE in EBX
MOV [EBX], 110          ; MY_TABLE[0] = 110
ADD EBX, 2              ; EBX = EBX +2
MOV [EBX], 123          ; MY_TABLE[1] = 123

MOV  destination, source

MOV  register, register
MOV  register, immediate
MOV  memory, immediate
MOV  register, memory
MOV  memory, register

MOV  EBX, [MY_TABLE]  ; Effective Address of MY_TABLE in EBX
MOV  [EBX], 110	      ; MY_TABLE[0] = 110

section	.text
   global _start     ;must be declared for linker (ld)
_start:             ;tell linker entry point
	
   ;writing the name ''Zara Ali''
   mov	edx,9       ;message length
   mov	ecx, name   ;message to write
   mov	ebx,1       ;file descriptor (stdout)
   mov	eax,4       ;system call number (sys_write)
   int	0x80        ;call kernel
	
   mov	[name],  dword ''Nuha''    ; Changed the name to Nuha Ali
	
   ;writing the name ''Nuha Ali''
   mov	edx,8       ;message length
   mov	ecx,name    ;message to write
   mov	ebx,1       ;file descriptor (stdout)
   mov	eax,4       ;system call number (sys_write)
   int	0x80        ;call kernel
	
   mov	eax,1       ;system call number (sys_exit)
   int	0x80        ;call kernel

section	.data
name db ''Zara Ali ''

Zara Ali Nuha Ali

[variable-name]    define-directive    initial-value   [,initial-value]...

choice		DB	''y''
number		DW	12345
neg_number	DW	-12345
big_number	DQ	123456789
real_number1	DD	1.234
real_number2	DQ	123.456

section .text
   global _start          ;must be declared for linker (gcc)
	
_start:                   ;tell linker entry point
   mov	edx,1		  ;message length
   mov	ecx,choice        ;message to write
   mov	ebx,1		  ;file descriptor (stdout)
   mov	eax,4		  ;system call number (sys_write)
   int	0x80		  ;call kernel

   mov	eax,1		  ;system call number (sys_exit)
   int	0x80		  ;call kernel

section .data
choice DB ''y''

y

choice	  DB 	''Y'' 		 ;ASCII of y = 79H
number1	  DW 	12345 	 ;12345D = 3039H
number2    DD  12345679  ;123456789D = 75BCD15H

marks  TIMES  9  DW  0

section	.text
   global _start        ;must be declared for linker (ld)
	
_start:                 ;tell linker entry point
   mov	edx,9		;message length
   mov	ecx, stars	;message to write
   mov	ebx,1		;file descriptor (stdout)
   mov	eax,4		;system call number (sys_write)
   int	0x80		;call kernel

   mov	eax,1		;system call number (sys_exit)
   int	0x80		;call kernel

section	.data
stars   times 9 db ''*''

*********

CONSTANT_NAME EQU expression

TOTAL_STUDENTS equ 50

mov  ecx,  TOTAL_STUDENTS 
cmp  eax,  TOTAL_STUDENTS

LENGTH equ 20
WIDTH  equ 10
AREA   equ length * width

SYS_EXIT  equ 1
SYS_WRITE equ 4
STDIN     equ 0
STDOUT    equ 1
section	 .text
   global _start    ;must be declared for using gcc
	
_start:             ;tell linker entry point
   mov eax, SYS_WRITE         
   mov ebx, STDOUT         
   mov ecx, msg1         
   mov edx, len1 
   int 0x80                
	
   mov eax, SYS_WRITE         
   mov ebx, STDOUT         
   mov ecx, msg2         
   mov edx, len2 
   int 0x80 
	
   mov eax, SYS_WRITE         
   mov ebx, STDOUT         
   mov ecx, msg3         
   mov edx, len3 
   int 0x80
   
   mov eax,SYS_EXIT    ;system call number (sys_exit)
   int 0x80            ;call kernel

section	 .data
msg1 db	''Hello, programmers!'',0xA,0xD 	
len1 equ $ - msg1			

msg2 db ''Welcome to the world of,'', 0xA,0xD 
len2 equ $ - msg2 

msg3 db ''Linux assembly programming! ''
len3 equ $- msg3

Hello, programmers!
Welcome to the world of,
Linux assembly programming!

%assign TOTAL 10

%assign  TOTAL  20

%define PTR [EBP+4]

INC destination

INC EBX	     ; Increments 32-bit register
INC DL       ; Increments 8-bit register
INC [count]  ; Increments the count variable

DEC destination

segment .data
   count dw  0
   value db  15
	
segment .text
   inc [count]
   dec [value]
	
   mov ebx, count
   inc word [ebx]
	
   mov esi, value
   dec byte [esi]

ADD/SUB	destination, source

SYS_EXIT  equ 1
SYS_READ  equ 3
SYS_WRITE equ 4
STDIN     equ 0
STDOUT    equ 1

segment .data 

   msg1 db "Enter a digit ", 0xA,0xD 
   len1 equ $- msg1 

   msg2 db "Please enter a second digit", 0xA,0xD 
   len2 equ $- msg2 

   msg3 db "The sum is: "
   len3 equ $- msg3

segment .bss

   num1 resb 2 
   num2 resb 2 
   res resb 1    

section	.text
   global _start    ;must be declared for using gcc
	
_start:             ;tell linker entry point
   mov eax, SYS_WRITE         
   mov ebx, STDOUT         
   mov ecx, msg1         
   mov edx, len1 
   int 0x80                

   mov eax, SYS_READ 
   mov ebx, STDIN  
   mov ecx, num1 
   mov edx, 2
   int 0x80            

   mov eax, SYS_WRITE        
   mov ebx, STDOUT         
   mov ecx, msg2          
   mov edx, len2         
   int 0x80

   mov eax, SYS_READ  
   mov ebx, STDIN  
   mov ecx, num2 
   mov edx, 2
   int 0x80        

   mov eax, SYS_WRITE         
   mov ebx, STDOUT         
   mov ecx, msg3          
   mov edx, len3         
   int 0x80

   ; moving the first number to eax register and second number to ebx
   ; and subtracting ascii ''0'' to convert it into a decimal number
	
   mov eax, [num1]
   sub eax, ''0''
	
   mov ebx, [num2]
   sub ebx, ''0''

   ; add eax and ebx
   add eax, ebx
   ; add ''0'' to to convert the sum from decimal to ASCII
   add eax, ''0''

   ; storing the sum in memory location res
   mov [res], eax

   ; print the sum 
   mov eax, SYS_WRITE        
   mov ebx, STDOUT
   mov ecx, res         
   mov edx, 1        
   int 0x80

exit:    
   
   mov eax, SYS_EXIT   
   xor ebx, ebx 
   int 0x80

Enter a digit:
3
Please enter a second digit:
4
The sum is:
7

section	.text
   global _start    ;must be declared for using gcc
	
_start:             ;tell linker entry point
   mov	eax,''3''
   sub     eax, ''0''
	
   mov 	ebx, ''4''
   sub     ebx, ''0''
   add 	eax, ebx
   add	eax, ''0''
	
   mov 	[sum], eax
   mov	ecx,msg	
   mov	edx, len
   mov	ebx,1	;file descriptor (stdout)
   mov	eax,4	;system call number (sys_write)
   int	0x80	;call kernel
	
   mov	ecx,sum
   mov	edx, 1
   mov	ebx,1	;file descriptor (stdout)
   mov	eax,4	;system call number (sys_write)
   int	0x80	;call kernel
	
   mov	eax,1	;system call number (sys_exit)
   int	0x80	;call kernel
	
section .data
   msg db "The sum is:", 0xA,0xD 
   len equ $ - msg   
   segment .bss
   sum resb 1

The sum is:
7

MUL/IMUL multiplier

MOV AL, 10
MOV DL, 25
MUL DL
...
MOV DL, 0FFH	; DL= -1
MOV AL, 0BEH	; AL = -66
IMUL DL

section	.text
   global _start    ;must be declared for using gcc
	
_start:             ;tell linker entry point

   mov	al,''3''
   sub     al, ''0''
	
   mov 	bl, ''2''
   sub     bl, ''0''
   mul 	bl
   add	al, ''0''
	
   mov 	[res], al
   mov	ecx,msg	
   mov	edx, len
   mov	ebx,1	;file descriptor (stdout)
   mov	eax,4	;system call number (sys_write)
   int	0x80	;call kernel
	
   mov	ecx,res
   mov	edx, 1
   mov	ebx,1	;file descriptor (stdout)
   mov	eax,4	;system call number (sys_write)
   int	0x80	;call kernel
	
   mov	eax,1	;system call number (sys_exit)
   int	0x80	;call kernel

section .data
msg db "The result is:", 0xA,0xD 
len equ $- msg   
segment .bss
res resb 1

The result is:
6

DIV/IDIV	divisor

section	.text
   global _start    ;must be declared for using gcc
	
_start:             ;tell linker entry point
   mov	ax,''8''
   sub     ax, ''0''
	
   mov 	bl, ''2''
   sub     bl, ''0''
   div 	bl
   add	ax, ''0''
	
   mov 	[res], ax
   mov	ecx,msg	
   mov	edx, len
   mov	ebx,1	;file descriptor (stdout)
   mov	eax,4	;system call number (sys_write)
   int	0x80	;call kernel
	
   mov	ecx,res
   mov	edx, 1
   mov	ebx,1	;file descriptor (stdout)
   mov	eax,4	;system call number (sys_write)
   int	0x80	;call kernel
	
   mov	eax,1	;system call number (sys_exit)
   int	0x80	;call kernel
	
section .data
msg db "The result is:", 0xA,0xD 
len equ $- msg   
segment .bss
res resb 1

The result is:
4

             Operand1: 	0101
             Operand2: 	0011
----------------------------
After AND -> Operand1:	0001

AND	BL,   0FH   ; This sets BL to 0000 1010

AND	AL, 01H     ; ANDing with 0000 0001
JZ    EVEN_NUMBER

section .text
   global _start            ;must be declared for using gcc
	
_start:                     ;tell linker entry point
   mov   ax,   8h           ;getting 8 in the ax 
   and   ax, 1              ;and ax with 1
   jz    evnn
   mov   eax, 4             ;system call number (sys_write)
   mov   ebx, 1             ;file descriptor (stdout)
   mov   ecx, odd_msg       ;message to write
   mov   edx, len2          ;length of message
   int   0x80               ;call kernel
   jmp   outprog

evnn:   
  
   mov   ah,  09h
   mov   eax, 4             ;system call number (sys_write)
   mov   ebx, 1             ;file descriptor (stdout)
   mov   ecx, even_msg      ;message to write
   mov   edx, len1          ;length of message
   int   0x80               ;call kernel

outprog:

   mov   eax,1              ;system call number (sys_exit)
   int   0x80               ;call kernel

section   .data
even_msg  db  ''Even Number!'' ;message showing even number
len1  equ  $ - even_msg 
   
odd_msg db  ''Odd Number!''    ;message showing odd number
len2  equ  $ - odd_msg

Even Number!

mov  ax, 9h                  ; getting 9 in the ax

Odd Number!

             Operand1:     0101
             Operand2:     0011
----------------------------
After OR -> Operand1:    0111

OR BL, 0FH                   ; This sets BL to  0011 1111

OR AL, BL

section .text
   global _start            ;must be declared for using gcc
	
_start:                     ;tell linker entry point
   mov    al, 5             ;getting 5 in the al
   mov    bl, 3             ;getting 3 in the bl
   or     al, bl            ;or al and bl registers, result should be 7
   add    al, byte ''0''      ;converting decimal to ascii
	
   mov    [result],  al
   mov    eax, 4
   mov    ebx, 1
   mov    ecx, result
   mov    edx, 1 
   int    0x80
    
outprog:
   mov    eax,1             ;system call number (sys_exit)
   int    0x80              ;call kernel
	
section    .bss
result resb 1

7

             Operand1:     0101
             Operand2:     0011
----------------------------
After XOR -> Operand1:    0110

XOR     EAX, EAX

TEST    AL, 01H
JZ      EVEN_NUMBER

             Operand1:    0101 0011
After NOT -> Operand1:    1010 1100

CMP destination, source

CMP DX,	00  ; Compare the DX value with zero
JE  L7      ; If yes, then jump to label L7
.
.
L7: ...  

INC	EDX
CMP	EDX, 10	; Compares whether the counter has reached 10
JLE	LP1     ; If it is less than or equal to 10, then jump to LP1

JMP	label

MOV  AX, 00    ; Initializing AX to 0
MOV  BX, 00    ; Initializing BX to 0
MOV  CX, 01    ; Initializing CX to 1
L20:
ADD  AX, 01    ; Increment AX
ADD  BX, AX    ; Add AX to BX
SHL  CX, 1     ; shift left CX, this in turn doubles the CX value
JMP  L20       ; repeats the statements

CMP	AL, BL
JE	EQUAL
CMP	AL, BH
JE	EQUAL
CMP	AL, CL
JE	EQUAL
NON_EQUAL: ...
EQUAL: ...

section	.text
   global _start         ;must be declared for using gcc

_start:	                 ;tell linker entry point
   mov   ecx, [num1]
   cmp   ecx, [num2]
   jg    check_third_num
   mov   ecx, [num2]
   
	check_third_num:

   cmp   ecx, [num3]
   jg    _exit
   mov   ecx, [num3]
   
	_exit:
   
   mov   [largest], ecx
   mov   ecx,msg
   mov   edx, len
   mov   ebx,1	;file descriptor (stdout)
   mov   eax,4	;system call number (sys_write)
   int   0x80	;call kernel
	
   mov   ecx,largest
   mov   edx, 2
   mov   ebx,1	;file descriptor (stdout)
   mov   eax,4	;system call number (sys_write)
   int   0x80	;call kernel
    
   mov   eax, 1
   int   80h

section	.data
   
   msg db "The largest digit is: ", 0xA,0xD 
   len equ $- msg 
   num1 dd ''47''
   num2 dd ''22''
   num3 dd ''31''

segment .bss
   largest resb 2  

The largest digit is: 
47

MOV	CL, 10
L1:
<LOOP-BODY>
DEC	CL
JNZ	L1

LOOP 	label

mov ECX,10
l1:
<loop body>
loop l1

section	.text
   global _start        ;must be declared for using gcc
	
_start:	                ;tell linker entry point
   mov ecx,10
   mov eax, ''1''
	
l1:
   mov [num], eax
   mov eax, 4
   mov ebx, 1
   push ecx
	
   mov ecx, num        
   mov edx, 1        
   int 0x80
	
   mov eax, [num]
   sub eax, ''0''
   inc eax
   add eax, ''0''
   pop ecx
   loop l1
	
   mov eax,1             ;system call number (sys_exit)
   int 0x80              ;call kernel
section	.bss
num resb 1

123456789:

section	.text
   global _start        ;must be declared for using gcc
	
_start:	                ;tell linker entry point
   mov	eax,''3''
   sub     eax, ''0''
	
   mov 	ebx, ''4''
   sub     ebx, ''0''
   add 	eax, ebx
   add	eax, ''0''
	
   mov 	[sum], eax
   mov	ecx,msg	
   mov	edx, len
   mov	ebx,1	         ;file descriptor (stdout)
   mov	eax,4	         ;system call number (sys_write)
   int	0x80	         ;call kernel
	
   mov	ecx,sum
   mov	edx, 1
   mov	ebx,1	         ;file descriptor (stdout)
   mov	eax,4	         ;system call number (sys_write)
   int	0x80	         ;call kernel
	
   mov	eax,1	         ;system call number (sys_exit)
   int	0x80	         ;call kernel
	
section .data
msg db "The sum is:", 0xA,0xD 
len equ $ - msg   
segment .bss
sum resb 1

The sum is:
7

31	32	33	34H

section	.text
   global _start        ;must be declared for using gcc
	
_start:	                ;tell linker entry point
   sub     ah, ah
   mov     al, ''9''
   sub     al, ''3''
   aas
   or      al, 30h
   mov     [res], ax
	
   mov	edx,len	        ;message length
   mov	ecx,msg	        ;message to write
   mov	ebx,1	        ;file descriptor (stdout)
   mov	eax,4	        ;system call number (sys_write)
   int	0x80	        ;call kernel
	
   mov	edx,1	        ;message length
   mov	ecx,res	        ;message to write
   mov	ebx,1	        ;file descriptor (stdout)
   mov	eax,4	        ;system call number (sys_write)
   int	0x80	        ;call kernel
	
   mov	eax,1	        ;system call number (sys_exit)
   int	0x80	        ;call kernel
	
section	.data
msg db ''The Result is:'',0xa	
len equ $ - msg			
section .bss
res resb 1  

The Result is:
6

01	02	03	04H

12	34H

section	.text
   global _start        ;must be declared for using gcc

_start:	                ;tell linker entry point

   mov     esi, 4       ;pointing to the rightmost digit
   mov     ecx, 5       ;num of digits
   clc
add_loop:  
   mov 	al, [num1 + esi]
   adc 	al, [num2 + esi]
   aaa
   pushf
   or 	al, 30h
   popf
	
   mov	[sum + esi], al
   dec	esi
   loop	add_loop
	
   mov	edx,len	        ;message length
   mov	ecx,msg	        ;message to write
   mov	ebx,1	        ;file descriptor (stdout)
   mov	eax,4	        ;system call number (sys_write)
   int	0x80	        ;call kernel
	
   mov	edx,5	        ;message length
   mov	ecx,sum	        ;message to write
   mov	ebx,1	        ;file descriptor (stdout)
   mov	eax,4	        ;system call number (sys_write)
   int	0x80	        ;call kernel
	
   mov	eax,1	        ;system call number (sys_exit)
   int	0x80	        ;call kernel

section	.data
msg db ''The Sum is:'',0xa	
len equ $ - msg			
num1 db ''12345''
num2 db ''23456''
sum db ''     ''

The Sum is:
35801

msg  db  ''Hello, world!'',0xa ;our dear string
len  equ  $ - msg            ;length of our dear string

msg db ''Hello, world!'',0xa ;our dear string
len equ 13                 ;length of our dear string

message DB ''I am loving it!'', 0

MONTHS	DW	12
MONTHS	DW	0CH
MONTHS	DW	0110B

NUMBERS	DW  34,  45,  56,  67,  75, 89

INVENTORY   DW  0
            DW  0
            DW  0
            DW  0
            DW  0
            DW  0
            DW  0
            DW  0

INVENTORY   DW  0, 0 , 0 , 0 , 0 , 0 , 0 , 0

INVENTORY TIMES 8 DW 0

section	.text
   global _start   ;must be declared for linker (ld)
	
_start:	
 		
   mov  eax,3      ;number bytes to be summed 
   mov  ebx,0      ;EBX will store the sum
   mov  ecx, x     ;ECX will point to the current element to be summed

top:  add  ebx, [ecx]

   add  ecx,1      ;move pointer to next element
   dec  eax        ;decrement counter
   jnz  top        ;if counter not 0, then loop again

done: 

   add   ebx, ''0''
   mov  [sum], ebx ;done, store result in "sum"

display:

   mov  edx,1      ;message length
   mov  ecx, sum   ;message to write
   mov  ebx, 1     ;file descriptor (stdout)
   mov  eax, 4     ;system call number (sys_write)
   int  0x80       ;call kernel
	
   mov  eax, 1     ;system call number (sys_exit)
   int  0x80       ;call kernel

section	.data
global x
x:    
   db  2
   db  4
   db  3

sum: 
   db  0

9

proc_name:
   procedure body
   ...
   ret

CALL proc_name

section	.text
   global _start        ;must be declared for using gcc
	
_start:	                ;tell linker entry point
   mov	ecx,''4''
   sub     ecx, ''0''
	
   mov 	edx, ''5''
   sub     edx, ''0''
	
   call    sum          ;call sum procedure
   mov 	[res], eax
   mov	ecx, msg	
   mov	edx, len
   mov	ebx,1	        ;file descriptor (stdout)
   mov	eax,4	        ;system call number (sys_write)
   int	0x80	        ;call kernel
	
   mov	ecx, res
   mov	edx, 1
   mov	ebx, 1	        ;file descriptor (stdout)
   mov	eax, 4	        ;system call number (sys_write)
   int	0x80	        ;call kernel
	
   mov	eax,1	        ;system call number (sys_exit)
   int	0x80	        ;call kernel
sum:
   mov     eax, ecx
   add     eax, edx
   add     eax, ''0''
   ret
	
section .data
msg db "The sum is:", 0xA,0xD 
len equ $- msg   

segment .bss
res resb 1

The sum is:
9

PUSH    operand
POP     address/register

; Save the AX and BX registers in the stack
PUSH    AX
PUSH    BX

; Use the registers for other purpose
MOV	AX, VALUE1
MOV 	BX, VALUE2
...
MOV 	VALUE1, AX
MOV	VALUE2, BX

; Restore the original values
POP	BX
POP	AX

section	.text
   global _start        ;must be declared for using gcc
	
_start:	                ;tell linker entry point
   call    display
   mov	eax,1	        ;system call number (sys_exit)
   int	0x80	        ;call kernel
	
display:
   mov    ecx, 256
	
next:
   push    ecx
   mov     eax, 4
   mov     ebx, 1
   mov     ecx, achar
   mov     edx, 1
   int     80h
	
   pop     ecx	
   mov	dx, [achar]
   cmp	byte [achar], 0dh
   inc	byte [achar]
   loop    next
   ret
	
section .data
achar db ''0''  

0123456789:;?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[]^_`abcdefghijklmnopqrstuvwxyz{|}
...
...

Fact (n) = n * fact (n-1) for n > 0

section	.text
   global _start         ;must be declared for using gcc
	
_start:                  ;tell linker entry point

   mov bx, 3             ;for calculating factorial 3
   call  proc_fact
   add   ax, 30h
   mov  [fact], ax
    
   mov	  edx,len        ;message length
   mov	  ecx,msg        ;message to write
   mov	  ebx,1          ;file descriptor (stdout)
   mov	  eax,4          ;system call number (sys_write)
   int	  0x80           ;call kernel

   mov   edx,1            ;message length
   mov	  ecx,fact       ;message to write
   mov	  ebx,1          ;file descriptor (stdout)
   mov	  eax,4          ;system call number (sys_write)
   int	  0x80           ;call kernel
    
   mov	  eax,1          ;system call number (sys_exit)
   int	  0x80           ;call kernel
	
proc_fact:
   cmp   bl, 1
   jg    do_calculation
   mov   ax, 1
   ret
	
do_calculation:
   dec   bl
   call  proc_fact
   inc   bl
   mul   bl        ;ax = al * bl
   ret

section	.data
msg db ''Factorial 3 is:'',0xa	
len equ $ - msg			

section .bss
fact resb 1

Factorial 3 is:
6

%macro macro_name  number_of_params
<macro body>
%endmacro

mov	edx,len	    ;message length
mov	ecx,msg	    ;message to write
mov	ebx,1       ;file descriptor (stdout)
mov	eax,4       ;system call number (sys_write)
int	0x80        ;call kernel

; A macro with two parameters
; Implements the write system call
   %macro write_string 2 
      mov   eax, 4
      mov   ebx, 1
      mov   ecx, %1
      mov   edx, %2
      int   80h
   %endmacro
 
section	.text
   global _start            ;must be declared for using gcc
	
_start:                     ;tell linker entry point
   write_string msg1, len1               
   write_string msg2, len2    
   write_string msg3, len3  
	
   mov eax,1                ;system call number (sys_exit)
   int 0x80                 ;call kernel

section	.data
msg1 db	''Hello, programmers!'',0xA,0xD 	
len1 equ $ - msg1			

msg2 db ''Welcome to the world of,'', 0xA,0xD 
len2 equ $- msg2 

msg3 db ''Linux assembly programming! ''
len3 equ $- msg3

Hello, programmers!
Welcome to the world of,
Linux assembly programming!

section	.text
   global _start         ;must be declared for using gcc
	
_start:                  ;tell linker entry point
   ;create the file
   mov  eax, 8
   mov  ebx, file_name
   mov  ecx, 0777        ;read, write and execute by all
   int  0x80             ;call kernel
	
   mov [fd_out], eax
    
   ; write into the file
   mov	edx,len          ;number of bytes
   mov	ecx, msg         ;message to write
   mov	ebx, [fd_out]    ;file descriptor 
   mov	eax,4            ;system call number (sys_write)
   int	0x80             ;call kernel
	
   ; close the file
   mov eax, 6
   mov ebx, [fd_out]
    
   ; write the message indicating end of file write
   mov eax, 4
   mov ebx, 1
   mov ecx, msg_done
   mov edx, len_done
   int  0x80
    
   ;open the file for reading
   mov eax, 5
   mov ebx, file_name
   mov ecx, 0             ;for read only access
   mov edx, 0777          ;read, write and execute by all
   int  0x80
	
   mov  [fd_in], eax
    
   ;read from file
   mov eax, 3
   mov ebx, [fd_in]
   mov ecx, info
   mov edx, 26
   int 0x80
    
   ; close the file
   mov eax, 6
   mov ebx, [fd_in]
   int  0x80 
	
   ; print the info 
   mov eax, 4
   mov ebx, 1
   mov ecx, info
   mov edx, 26
   int 0x80
       
   mov	eax,1             ;system call number (sys_exit)
   int	0x80              ;call kernel

section	.data
file_name db ''myfile.txt''
msg db ''Welcome to Tutorials Point''
len equ  $-msg

msg_done db ''Written to file'', 0xa
len_done equ $-msg_done

section .bss
fd_out resb 1
fd_in  resb 1
info resb  26

Written to file
Welcome to Tutorials Point

section	.text
   global _start         ;must be declared for using gcc
	
_start:	                 ;tell linker entry point

   mov	eax, 45		 ;sys_brk
   xor	ebx, ebx
   int	80h

   add	eax, 16384	 ;number of bytes to be reserved
   mov	ebx, eax
   mov	eax, 45		 ;sys_brk
   int	80h
	
   cmp	eax, 0
   jl	exit	;exit, if error 
   mov	edi, eax	 ;EDI = highest available address
   sub	edi, 4		 ;pointing to the last DWORD  
   mov	ecx, 4096	 ;number of DWORDs allocated
   xor	eax, eax	 ;clear eax
   std			 ;backward
   rep	stosd            ;repete for entire allocated area
   cld			 ;put DF flag to normal state
	
   mov	eax, 4
   mov	ebx, 1
   mov	ecx, msg
   mov	edx, len
   int	80h		 ;print a message

exit:
   mov	eax, 1
   xor	ebx, ebx
   int	80h
	
section	.data
msg    	db	"Allocated 16 kb of memory!", 10
len     equ	$ - msg

Allocated 16 kb of memory!