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Introduction to NASM
Prepared By:
Muhammed Yazar Y
Updated By:
Sonia V Mathew
Govind R
Darshana Suresh - Dheeraj Mohan - Jyothsna Shaji
Lakshmi Alwin - Naveen Babu - Nikhil Sojan
Nileena P.C. - Sanju Alex Jacob - Vrindha K
B Tech
Dept: of CSE - NIT Calicut
Under The Guidance of:
Mr. Jayaraj P B
Mr. Saidalavi Kalady
Assistant Professors
Dept: of CSE
NIT Calicut

Department Of Computer Science & Engineering
NATIONAL INSTITUTE OF TECHNOLOGY CALICUT
2019
1

Contents

1 Basics of Computer Organization

3

2 Introduction to NASM

15

3 Basic I/O in NASM

30

4 Introduction to Programming in NASM

34

5 Integer Handling

38

6 Subprograms

44

7 Arrays and Strings

52

8 Floating Point Operations

74

1

Acknowledgement
I would like to express my gratitude to Saidalavy Kalady Sir and Jayaraj P B
Sir (Assistant Professors, Dept: of CSE, NIT Calicut) for guiding me throughout
while making this reference material on NASM Assembly language programming.
Without their constant support and guidance this work would not have been possible. Thanks to Lyla B Das madam (Associate Professor, Dept: of ECE, NIT
Calicut) for encouraging me to bring out an updated version of this work. I would
also wish to thank Meera Sudharman, my classmate who helped me in verifying
the contents of this document. Special thanks are due to Govind R and Sonia
V Mathew (BTech 2011-2015 Batch) for updating the contents and adding more
working examples to it. I am extremely grateful to Dheeraj Mohan, Nikhil Sojan,
Darshana Suresh, Jyothsna Shaji, Lakshmi Alwin, Vrindha K, Nileena P.C., Sanju
Alex Jacob and Naveen Babu (2016 - 2020 batch) for restructuring the manual
with different learning strategies and working examples. I want to Thanks to all
my dear batch mates and juniors who have been supporting me through the work
of this and for providing me with their valu- able suggestions.

Muhammed Yazar

2

Chapter 1
Basics of Computer Organization
Machine Language
Machine language consists of instructions in the form of 0âĂŹs and 1âĂŹs. Every
CPU has its own machine language. It is very difficult to write programs using
the combination of 0âĂŹs and 1âĂŹs. So we rely upon either assembly language
or high level language for writing programs.

Assembly Language
An assembly language is a low-level programming language for microprocessors
and other programmable devices. Assembly language uses a mnemonic to represent
each low-level machine instruction.

Why Assembly Language ?
The symbolic programming of Assembly Language is easier to understand and
saves a lot of time and effort of the programmer. When you study assembly language, you will get a better idea of computer organization and how a program
executes in a computer. A program written in assembly language will be more
efficient than the same program written in a high level language. Some portions
of Operating System (Eg: Linux kernel) and some system software are written
in assembly language. In programming languages like C, C++ we can even em-

3

bed assembly language instructions into it using functions like asm( ); (Refer to
Chapter 8 of ’The Intel Microprocessors’ by Barry B. Brey for more details)

What is NASM ?
The Netwide Assembler is an assembler and disassembler for the Intel X86 architecture (explained in subsequent section). It can be used to write 16-bit, 32-bit
(IA-32) and 64-bit (x86-64) programs. NASM is considered to be one of the most
popular assemblers for Linux.

Computer Architecture
The basic operational design of a computer is called architecture. It is a set of rules
and methods that describe the functionality, organization and implementation of
computer systems. X86 architecture follows Von Neumann architecture which is
based on stored program concept.

Von Neumann Architecture
Von Neumann architecture machine, designed by physicist and mathematician
John Von Neumann (1903 - 1957) is a theoretical design for a stored program
computer that serves as the basis for almost all modern computers. A Von Neumann machine consists of a central processor with an arithmetic/logic unit and a
control unit, a memory, mass storage, and input and output.

X86 Architecture
The x86 architecture is an Instruction Set Architecture (ISA) series for computer
processors. Developed by Intel Corporation, x86 architecture defines how a processor handles and executes different instructions passed from the operating system
(OS) and software programs. The ’x’ in x86 denotes ISA version.
Key features include:
• Provides a logical framework for executing instructions through a processor.
• Allows software programs and instructions to run on any processor in the
Intel 8086 family.
4

Figure 1.1: Von Neumann Architecture
• Provides procedures for utilizing and managing the hardware components of
a central processing unit (CPU).
Historically there have been 2 types of Computers:
Fixed Program Computers - Their function is very specific and they can’t be programmed, e.g. Calculators.
Stored Program Computers - These can be programmed to carry out many different tasks, applications are stored on them, hence the name.

Processor
Processor is the brain of the computer. It performs all mathematical, logical and
control operations of a computer. It is the main component of the computer which
executes all the instructions given to it in the form of programs. It interacts with
I/O devices, memory (RAM) and secondary storage devices and thus implements
the instructions given by the user.
The term processor is used interchangeably with the term central processing unit
(CPU), although strictly speaking, the CPU is not the only processor in a computer.

5

Figure 1.2: Intel Core i9

Registers
Registers are the most immediately accessible memory units for the processor.
They are the fastest among all the types of memory. They reside inside the processor and the processor can access the contents of any register in a single clock
cycle. It is the working memory for a processor, i.e, if we want the processor to
perform any task it needs the data to be present in any of its registers.
What is a clock cycle?
In computers, the clock cycle is the amount of time between two pulses of an
oscillator. It is a single increment of the central processing unit (CPU) clock
during which the smallest unit of processor activity is carried out. The clock cycle
helps in determining the speed of the CPU, as it is considered the basic unit of
measuring how fast an instruction can be executed by the computer processor.
height.
The series of processors released on or after 80186 like 80186, 80286, 80386, Pentium etc are referred to as x86 or 80x86 processors. The processors released on or
after 80386 are called I386 processors. They are 32 bit processors internally and
externally. So their register sizes are generally 32 bit. In this section we will go
through the i386 registers.

6

Intel maintains its backward compatibility of instruction sets, i.e, we canrun
a program designed for an old 16 bit machine in a 32bit machine.That is the reason
why we can install 32-bit OS in a 64 bit PC. The onlyproblem is that, the program
will not use the complete set of registers andother available resources and thus it
will be less efficient.
A register may hold an instruction, a storage address, or any kind of data (such as
a bit sequence or individual characters). A register must be large enough to hold
an instruction - for example, in a 64-bit computer, a register must be 64 bits in
length.
1. General Purpose Registers
General purpose registers are used to store temporary data within the microprocessor. There are eight general purpose registers. They are EAX, EBX,
ECX, EDX, EBP, ESI, EDI, ESP. We can refer to the lower 8 and 16 bits of
these registers (see image). This is to maintain the backward compatibility
of instruction sets. These registers are also known as scratchpad area as they
are used by the processor to store intermediate values in a calculation and
also for storing address locations.
The General Purpose Registers are used for :
• EAX: Accumulator Register - Contains the value of some operands in
some operations (E.g.: multiplication).
• EBX: Base Register - Pointer to some data in Data Segment.
• ECX: Counter Register - Acts as loop counter, used in string operations
etc.
• EDX: Used as pointer to I/O ports.
• ESI: Source Index - Acts as source pointer in string operations. It can
also act as a pointer in Data Segment (DS).
• EDI: Destination Index - Acts as destination pointer in string operations. It can also act as a pointer in Extra Segment (ES).
• ESP: Stack Pointer - Always points to the top of system stack.
• EBP: Base Pointer - It points to the starting of system stack (ie.bottom/base
of stack).
2. Flags and EIP
FLAGS are special purpose registers inside the CPU that contains the status
of CPU / the status of last operation executed by the CPU. Some of the bits
7

Figure 1.3: i386 Registers

8

Figure 1.4: Segment Registers
in FLAGS need special mention:
• Carry Flag: When a processor does a calculation, if there is a carry
then the Carry Flag will be set to 1.
• Zero Flag: If the result of the last operation was zero, Zero Flag will
be set to 1, else it will be zero.
• Sign Flag : If the result of the last signed operation is negative then
the Sign Flag is set to 1, else it will be zero.
• Parity Flag: If there are odd number of ones in the result of the last
operation, parity flag will be set to 1.
• Interrupt Flag: If interrupt flag is set to 1, then only it will listen to
external interrupts.
EIP:
EIP is the instruction pointer, it points to the next instruction to be executed.
In memory there are basically two classes of things stored:
• Data
• Program
When we start a program, it will be copied into the main memory and EIP
is the pointer which points to the starting of this program in memory and
execute each instruction sequentially. Branch statements like JMP, RET,
CALL, JNZ (we will see shortly) alter the value of EIP.
3. Segment Registers
In x86 processors, for accessing the memory basically there are two types of
registers used; Segment Register and Offset. Segment register contains the
9

base address of a particular data section and Offset will contain how many
bytes should be displaced from the segment register to access the particular
data. CS contains the base address of Code Segment and EIP is the offset.
It keeps on updating while executing each instruction. SS or Stack Segment
contains the address of top most part of system stack. ESP and EBP will
be the offset for that. Stack is a data structure that follows LIFO ie. LastIn-First-Out. There are two main operations associated with stack: push
and pop. If we need to insert an element into a stack, we will push it and
when we give the pop instruction, we will get the last value which we have
pushed. Stack grows downward. So SP will always points to the top of stack
and if we push an element, ESP (Stack Pointer) will get reduced by sufficient
number of bytes and the data to be stored will be pushed over there. DS,
ES, FS and GS acts as base registers for a lot of data operations like array
addressing, string operations etc. ESI, EDI and EBX can act as offsets for
them. Unlike other registers, Segment registers are still 16 bit wide in 32-bit
processors.
In modern 32 bit processor the segment address will be just an entry into a
descriptor table in memory and using the offset it will get the exact memory
locations through some manipulations. This is called segmentation.
In x86 architecture when we push or save some data in memory, the lower
bytes of it will be addressed immediately and thus it is said to follow Little
Endian Form. MIPS architecture follows Big Endian Form.

Little Endian and Big Endian
Endianness refers to the sequential order in which bytes are arranged into larger
numerical values when stored in memory or when transmitted over digital links.
Endianness is of interest in computer science because two conflicting and incompatible formats are in common use: words may be represented in big-endian or
little-endian format, depending on whether bits or bytes or other components are
ordered from the big end (most significant bit) or the little end (least significant
bit).
Big Endian Byte Order: The most significant byte (the "big end") of the data
is placed at the byte with the lowest address. The rest of the data is placed in
order in the next three bytes in memory.
Little Endian Byte Order: The least significant byte (the "little end") of the
data is placed at the byte with the lowest address. The rest of the data is placed
in order in the next three bytes in memory.

10

Suppose an integer is stored as 4 bytes(32-bits), then a variable with value 0x01234567(
Hexa-decimal representation) is stored as four bytes 0x01, 0x23, 0x45, 0x67, on
Big-endian while on Little-Endian (Intel x86), it will be stored in reverse order:

Bus
Bus is a name given to any communication medium, that transfers data between
two components. (A bus is a subsystem that is used to connect computer components and transfer data between them). A bus may be parallel or serial. Parallel
buses transmit data across multiple wires. Serial buses transmit data in bit-serial
format. We can classify the buses associated with the processor into three.
• Data Bus :
It is the bus used to transfer data between the processor and memory or any
other I/O devices (for both reading and writing). As the size of data bus
increases, it can transfer more data in a single stretch. The size of data bus
in common processors by Intel are given below.
Processor
8088, 80188
8086, 80816, 80286, 80386SX
80386DX, 80486
80586, Pentium Pro and later processors

Bus size
8 bit
16 bit
32 bit
64 bit

• Address Bus :
The address bus is used by the CPU or a direct memory access (DMA)
enabled device to locate the physical address to communicate read/write
commands. Memory management Unit (MMU) or Memory Control Unit
(MCU) is the set of electronic circuits present in the motherboard which
helps the processor in reading or writing the data to or from a location in
the RAM. All address buses are read and written by the CPU in the form
of bits. An address bus is measured by the amount of memory a system
can retrieve. A system with a 32-bit address bus can address 4 gigabytes of
11

memory space.
The maximum size of RAM which can be used in a PC is determined by the
size of the address bus. If the size of address bus is n bits, it can address a
maximum of 2n bytes of RAM. This is the reason why even if we add more
than 4 GB of RAM in a 32 bit PC, system cannot find more than 4 GB of
available memory.
Processor

Address Bus Width

8088, 8086, 80186, 80188
80386SX, 80286
80486, 80386DX, Pentium,
Pentium Override
Pentium II, Pentium Pro

20
24
32

Maximum
Addressable RAM size
1 MB
16 MB
4 GB

36

64 GB

• Control Bus :
Control bus contains information which controls the operations of processor
or any other memory or I/O device.
For example, the data bus is used for both reading and writing purpose and
how is it that the Memory or MMU knows the data has to be written to a
location or has to be read from a location, when it encounters an address in
the address bus? This ambiguity is being cleared using read and write bits
in the control bus. When the read bit is enabled, the data in the data bus
will be written to that location. When the write bit is enabled, MMU will
write the data in the data bus into the address location in the address bus.

Interrupts
Interrupts are the most critical routines executed by a processor. Interrupts may
be triggered by external sources or due to internal operations. In linux based
systems 80h is the interrupt number for OS generated interrupts and in windows
based systems it is 21h. The Operating System Interrupt is used foI would like
to express my gratitude to Saidalavy Kalady Sir and Jayaraj P B Sir (Assistant
Professors, Dept: of CSE, NIT Calicut) for guiding me throughout while making
this reference material on NASM Assembly language programming. Without their
constant support and guidance this work would not have been possible. Thanks
to Lyla B Das madam (Associate Professor, Dept: of ECE, NIT Calicut) for
encouraging me to bring out an updated version of this work. I would also wish to
thank Meera Sudharman, my classmate who helped me in verifying the contents of
this document. Special thanks are due to Govind R and Sonia V Mathew (BTech
12

2011-2015 Batch) for updating the contents and adding more working examples
to it. I am extremely grateful to Dheeraj Mohan, Nikhil Sojan, Jyothsna Shaji,
Lakshmi Alwin, Vrindha K, Nileena P.C., Sanju Alex Jacob and Naveen Babu
(2016 - 2020 batch) for restructuring the manual with different learning strategies
and working examples. I want to Thanks to all my dear batch mates and juniors
who have been supporting me through the work of this and for providing me with
their valu- able suggestions.r implementing systems calls.
Whenever an interrupt occurs, processor will stop its present work, preserve the
values in registers into memory and then execute the ISR (Interrupt Service Routine) by referring to an interrupt vector table. ISR is the set of instructions to
be executed when an interrupt occurs. By referring to the interrupt vector table,
the processor can get which ISR it should execute for the given interrupt. After
executing the ISR, processor will restore the registers to its previous state and
continue the process that it was executing before. Almost all the I/O devices work
by utilizing the interrupt requests.
Here interrupt can be seen as a signal from a device, such as the keyboard, to the
CPU, telling processor to immediately stop whatever it is currently doing and do
something else. For example, the keyboard controller sends an interrupt when a
key is pressed. To know how to call on the kernel when a specific interrupt arise,
the CPU has a vector table setup by the OS, and stored in memory. There are
256 interrupt vectors on x86 CPUs, numbered from 0 to 255 which act as entry
points into the kernel.The number of interrupt vectors or entry points supported
by a CPU differs based on the CPU architecture.

System call
System calls are Application Programmer’s Interface to the kernel space. In a
NASM program, input has to be taken from the standard input device (Keyboard) and output has to be given to the standard output device (monitor). This
is implemented using the Operating System’s read and write system call respectively. Interrupt number 80h(in hexadecimal) is given to the software generated
interrupt in Linux Systems. Applications invoke the System Calls using this interrupt number. When an application triggers int 80h, then OS will understand that
it is a request for a system call and it will refer the general purpose registers to
find out and execute the exact Interrupt Service Routine (ie. System Call here).
The standard convention to use a system call is,
• System call number is stored in EAX register.

13

• Other parameters needed to implement the system call is stored in other
general purpose registers. 24
• Trigger the 80h interrupt using the instruction INT 80h.
Then OS will implement the system call.
System memory in Linux can be divided into two distinct regions: kernel space
and user space. Kernel space is where the kernel (i.e., the core of the operating
system) executes (i.e., runs) and provides its services.

14

Chapter 2
Introduction to NASM
Sections in NASM
A NASM program is divided into three sections.
1. section .text :
This section contains the executable code from where execution starts. It is
analogous to the main() function in C.
2. section .bss :
Here, variables are declared without initialisation.
3. section .data :
Variables are declared and initialised in this section.
For declaring space in the memory the following directives are used,
(a) RESx:
Reserve just space in memory for a variable without giving any initial
values.
(b) Dx:
Declaring space in the memory for any variable and also providing the
initial values at that moment. Where x can be replaced with different
characters as shown in the below table.

15

x
Meaning
b
BYTE
w
WORD
d DOUBLE WORD
q
QUAD WORD
t
TEN BYTE

Bytes
1
2
4
8
20

Examples:
section .data
var1: db 10 ;Reserve one byte in memory for storing var1 and var1=10
var2 : db 1,2,3,4
string: db ’Hello’
string2: db ’H’,’e’,’l’,’l’,’o’
Here both string and string2 are identical. They are 5 bytes long and stores the
string Hello. Each character in the string will be first converted to ASCII code
and that numeric value will be stored in each byte location.
section .bss
var1: resb 1
var2: resq 1
var3: resw 1

TIMES
It is used to create and initialize large arrays with a common
initial value for all its elements.
Eg: var: times 100 db 1
Creates an array of 100 bytes and each element
will be initialized with the value 1.

Dereferencing in NASM
To access the data stored at an address, the dereferencing operator used is ’[ ]’.
Examples:
mov eax, [var] ;Value at address location var would be copied to eax

16

mov eax,var ;Address location var is copied to eax

Type casting
It is required for the operands for which the assembler cannot predict the number
of memory locations to dereference to get the data(like INC , MOV etc). For
other instructions (like ADD, SUB etc) it is not mandatory. The directives used
for specifying the datatype are: BYTE, WORD, DWORD, QWORD, TWORD.
Eg:
MOV dword[ebx], 1
INC BYTE[label]
ADD eax, dword[label]

x86 Instruction Set
1. MOV Move/Copy
Copy the content of one register/memory to another or change the value of
a register/ memory variable to an immediate value.
Syntax: mov dest, src
• src should be a register/memory operand.
• Both src and dest cannot together be memory operands.
Eg:
mov eax, ebx
;Copy the content of ebx to eax
mov ecx, 109
;Changes the value of ecx to 109
mov al, bl
mov byte[var1], al ;Copy the content of al register to the variable var1 in
memory
mov word[var2], 200
mov eax, dword[var3]
2. MOVZX Move and Extend
Copy and extend a variable from a lower spaced memory / register location

17

to a higher one
syntax : mov src, dest
• size of dest should be greater than or equal to size of src.
• src should be a register / memory operand.
• Both src and dest cannot together be memory operands.
• Works only with signed numbers.
Eg:
movzx eax, ah
movzx cx, al
3. ADD - Addition
Sytax : add dest, src
dest = dest + src;
Used to add the values of two registers/memory variables and store the result
in the first operand.
• src should be a register / memory operand.
• Both src and dest cannot together be memory operands.
• Both the operands should have the same size.
Eg:
add
add
add
add

eax, ecx
al, ah
ax, 5
edx, 31h

; eax = eax + ecx
; al = al + ah

4. SUB - Subtraction
Sytax : sub dest, src
dest = dest - src;
Used to subtract the values of two registers/memory variables and store the
result in the first operand.
• src should be a register/memory operand.
• Both src and dest cannot together be memory operands.
• Both the operands should have the same size.

18

Eg:
sub
sub
sub
sub

eax, ecx
al, ah
ax, 5
edx, 31h

; eax = eax - ecx
; al = al - ah

5. INC - Increment operation
Used to increment the value of a registers/memory variables by 1
Eg:
INC eax
; eax++
INC byte[var]
INC al
6. DEC
Used
Eg:
DEC
DEC
DEC

- Decrement operation
to decrement the value of a registers/memory variables by 1
eax
; eax- byte[var]
al

7. MUL - Multiplication
Syntax : mul src
Used to multiply the value of a registers/memory variables with the EAX/AX/AL
reg. MUL works according to the following rules.
• If src is 1 byte then AX = AL * src.
• If src is 1 word (2 bytes) then DX:AX = AX * src (ie. Upper 16 bits of
the result (AX*src) will go to DX and the lower 16 bits will go to AX).
• If src is 2 words long(32 bit) then EDX:EAX = EAX * src (ie. Upper
32 bits of the result will go to EDX and the lower 32 bits will go to
EAX).
8. IMUL - Multiplication of signed numbers
IMUL instruction works with the multiplication of signed numbers. It can
be used mainly in three different forms.
Syntax :
(a) imul src

19

(b) imul dest, src
(c) imul dest, src1, src2
• If we use imul as in (a) then its working follows the same rules of MUL.
• If we use that in (b) form then dest = dest * src.
• If we use that in (c) form then dest = src1 * scr2.
9. DIV - Division
Synatx : div src
Used to divide the value of EDX:EAX or DX:AX or AX register with registers/memory variables in src. DIV works according to the following rules.
• If src is 1 byte then AX will be divide by src, remainder will go to AH
and quotient will go to AL.
• If src is 1 word (2 bytes) then DX:AX will be divided by src, remainder
will go to DX and quotient will go to AX.
• If src is 2 words long(32 bit) then EDX:EAX will be divide by src,
remainder will go to EDX and quotient will go to EAX.
10. NEG - Negation of Signed numbers.
Sytax : NEG op1
NEG Instruction negates a given registers/memory variables.
11. CLC - Clear Carry
This instruction clears the carry flag bit in CPU FLAGS.
12. ADC - Add with Carry
Syntax : ADC dest, src
ADC is used for the addition of large numbers. Suppose we want to add two
64 bit numbers. We keep the first number in EDX:EAX (ie. most significant
32 bits in EDX and the others in EAX) and the second number in EBX:ECX.
Then we perform addition as follows
Eg:
clc
; Clearing the carry FLAG
add eax, ecx ; Normal addition of eax with ecx
adc edx, ebx ; Adding with carry for the higher bits.
13. SBB - Subtract with Borrow

20

Syntax : SBB dest, src
SBB is analogous to ADC and it is used for the subtraction of large numbers.
Suppose we want to subtract two 64 bit numbers. We keep the first numbers in EDX:EAX and the second number in EBX:ECX. Then we perform
subtraction as follows
Eg:
clc ; Clearing the carry FLAG
sub eax, ecx ; Normal subtraction of ecx from eax
sbb edx, ebx ; Subtracting with carry for the higher bits.

Branching In x86
14. JMP - Unconditionally Jump to label
JMP is similar to the goto label statements in C/C++. It is used to jump
control to any part of our program without checking any conditions.
15. CMP - Compares the Operands
Syntax :CMP op1, op2
When we apply CMP instruction over two operands say op1 and op2, it will
perform the operation op1 − op2 and will not store the result. Instead it will
affect the CPU FLAGS. It is similar to the SUB operation, without saving
the result. For example if op1 6= op2 then the Zero Flag(ZF) will be set to 1.
NB: For generating conditional jumps in X86 programming we will first perform the CMP operation between two registers/memory operands and then
we use the following jump operations which checks the CPU FLAGS.
Conditional Jump Instructions:

21

Instruction

Working

JZ

Jump If Zero Flag is Set

JNZ

Jump If Zero Flag is Unset

JC

Jump If Carry Flag is Set

JNC

Jump If Carry Flag is Unset

JP

Jump If Parity Flag is Set

JNP

Jump If Parity Flag is Unset

JO

Jump If Overflow Flag is Set

JNO

Jump If Overflow Flag is Unset

Advanced Conditional Jump Instructions:
In 80x86 processors Intel has added some enhanced versions of the conditional
operations which are much more easier than traditional Jump instructions.
They are easy to perform comparison between two variables.
First we need to use CMP op1, op2 before even using these set of Jump
instructions. There is separate class for comparing the signed and unsigned
numbers.
(a) For Unsigned numbers:
Instruction

Working

JE

Jump if op1 = op2

JNE

Jump if op1 6= op2

JA (jump if above)

Jump if op1 >op2

JNA

Jump if op1 ≤ op2

JB (jump if below)

Jump if op1 op2

JNG

Jump if op1 ≤ op2

JL (jump if below)

Jump if op1 = 0 )
sum = sum + ecx;
ecx − −;

x86 equivalent
mov dword[sum], 0
mov ecx, dword[n]
addition:
add [sum], ecx
loop addition ; Decrements ecx and
checks if ecx is not equal to 0 , if
so it will jump to addition

Boolean Operators
17. AND - (Bitwise Logical AND)
Syntax : AND op1, op2
Performs bitwise logical AND operation of op1 and op2, assign the result to
op1.
op1 = op1&op2; //Equivalent C Statement
Let x = 10101001b and y = 10110010b be two 8-bit binary numbers. Then
x&y
x
1 0 1 0 1 0 0 1
y
1 0 1 1 0 0 1 0
x AND y 1 0 1 0 0 0 0 0
18. OR - (Bitwise Logical OR)
Syntax: OR op1, op2
Performs bitwise logical OR operation of op1 and op2, assign the result to
op1.
op1 = op1||op2; //Equivalent C Statement
Let x = 10101001b and y = 10110010b be two 8-bit binary numbers. Then
xky
x
1 0 1 0 1 0 0 1
y
1 0 1 1 0 0 1 0
x OR y 1 0 1 1 1 0 1 1
19. XOR - (Bitwise Logical Exclusive OR)
Syntax: XOR op1, op2
24

Performs bitwise logical XOR operation of op1 and op2, assign the result to
op1.
op1 = op1 ˆ op2; //Equivalent C Statement
Let x = 10101001b and y = 10110010b be two 8-bit binary numbers. Then
x⊕y
x
y
x XOR y

1 0 1 0 1 0 0 1
1 0 1 1 0 0 1 0
0 0 0 1 1 0 1 1

20. NOT - (Bitwise Logical Negation)
Syntax: NOT op1
Performs bitwise logical NOT of op1 and assign the result to op1.
op1 = ¬op1; //Equivalent C Statement
x 1 0 1 0 1 0 0 1
¬x 0 1 0 1 0 1 1 0
21. TEST - (Logical AND, affects only CPU FLAGS)
Syntax: TEST op1, op2
• It performs the bitwise logical AND of op1 and op2 but it wont save the
result to any registers. Instead, the result of the operation will affect
CPU FLAGs.
• It is similar to the CMP instruction in usage.
22. SHL - Shift Left
Syntax : SHL op1, op2
op1 = op1 «op2; //Equivalent C Statement
• SHL performs the bitwise left shift. op1 should be a registers/memory
variables but op2 must be an immediate(constant) value.
• It will shift the bits of op1, op2 number of times towards the left and
put the rightmost op2 number of bits to 0.
Example:
shl al,3
al
1 0 1 0 1 0 0 1
al«3 0 1 0 0 1 0 0 0

25

23. SHR - Shift Right
Syntax : SHR op1, op2
op1 = op1»op2; //Equivalent C Statement
• SHR performs the bitwise right shift. op1 should be a registers/memory
variables but op2 must be an immediate(constant) value.
• It will shift the bits of op1, op2 number of times towards the right and
put the leftmost op2 number of bits to 0.
Example:
shr al,3
al
1 0 1 0 1 0 0 1
al»3 0 0 0 1 0 1 0 1
24. ROL - Rotate Left
Syntax : ROL op1, op2
ROL performs the bitwise cyclic left shift. op1 could be a registers/memory
variables but op2 must be an immediate(constant) value.
Example : rol al,3
al
rol al,3

1 0 1 0 1 0 0 1
0 1 0 0 1 1 0 1

25. ROR - Rotate Right
Syntax : ROR op1, op2
ROR performs the bitwise cyclic right shift. op1 could be a registers/memory variables but op2 must be an immediate(constant) value.
Example: ror eax, 5
al
ror al,3

1 0 1 0 1 0 0 1
0 0 1 1 0 1 0 1

26. RCL - Rotate Left with Carry
Syntax : RCL op1, op2
Its working is same as that of rotate left except it will consider the carry bit
as its left most extra bit and then perform the left rotation.
27. RCR - Rotate Right with Carry
Syntax : RCR op1, op2
26

Its working is same as that of rotate right except it will consider the carry
bit as its left most extra bit and then perform the right rotation.

Stack Operations
28. PUSH
Pushes a value into system stack. It decreases the value of ESP and copies
the value of a register/constant into the system stack
Syntax: PUSH register/constant
Examples
PUSH ax
; ESP = ESP − 2 and copies value of ax to [ESP]
PUSH eax ; ESP = ESP − 4 and copies value of ax to [ESP]
PUSH ebx
PUSH dword 5
PUSH word 258
29. POP
Pops off a value from the system stack.POP Instruction takes the value stored
in the top of system stack to a register and then increases the value of ESP.
Examples:
POP bx
; ESP= ESP + 2 and copies value from stack to bx
POP ebx
; ESP= ESP + 4
POP eax
30. PUSHA
Pushes the value of all general purpose registers. PUSHA is used to save the
value of general purpose registers especially when calling some subprograms
which will modify their values.
31. POPA
Pops off the value of all general purpose registers which we have pushed before using PUSHA instruction
32. PUSHF Pushes all the CPU FLAGS
33. POPF
27

POP off and restore the values of all CPU Flags which have been pushed
before using PUSHF instructions.

NB: It is important to pop off the values pushed into the stack properly.
Even a minute mistake in any of the PUSH / POP instruction could make
the program not working.
34. Pre-processor Directives in NASM: In NASM %define acts similar to the C 0 s
preprocessor directive define. This can be used to declare constants. Eg:
%define SIZE 100

Comments in NASM
Only single line comments are available in NASM. A semicolon (;) is inserted in
front of the line to be commented.
Ex: ; A program to find the largest of 2 numbers

NASM Installation
NASM is freely available on internet. You can visit www.nasm.us . It’s documentation is also available there. In order to install NASM in windows you can
download it as an installation package from the site and install it easily.
In Ubuntu Linux you can give the command :
sudo apt-get install nasm
and in fedora you can use the command:
su -c yum install nasm
in a terminal and easily install nasm.

Compilation
1. Assembling the source file
nasm -f elf filename.asm
This creates an object file, filename.o in the current working directory.
28

2. Creating an executable file
For a 32 bit machine
ld filename.o -o output filename
For 64 bit machine
ld -melf_i386 filename.o -o output filename
This creates an executable file of the name output filename.
3. Program execution
./output filename
For example, if the program to be run is first.asm
nasm -f elf first.asm
ld first.o -o output
./output

29

Chapter 3
Basic I/O in NASM
In this chapter, we will learn how to obtain user input and how to return the output to the output device. The input from the standard input device (Keyboard)
and Output to the standard output device (monitor) in a NASM Program is implemented using the Operating SystemâĂŹs read and write system call. Interrupt no:
80h is given to the software generated interrupt in Linux Systems. Applications
implement the System Calls using this interrupt. When an application triggers
int 80h, then OS will understand that it is a request for a system call and it will refer the general purpose registers to find out and execute the exact Interrupt Service
Routine (ie. System Call here).The standard convention to use the software 80h
interrupt is, we will put the system call no: in eax register and other parameters
needed to implement the system calls in the other general purpose registers. Then
we will trigger the 80h interrupt using the instruction âĂŸINT 80hâĂŹ. Then OS
will implement the system call.
1. Exit System Call
• This system call is used to exit from the program
• System call number for exit is 1, so it is copied to eax reg.
• Output of a program if the exit is successful is 0 and it is being passed
as a parameter for exit( ) system call. We need to copy 0 to ebx reg.
• Then we will trigger INT 80h
mov eax, 1
mov ebx, 0
int 80h

; System Call Number
; Parameter
; Triggering OS Interrupt
30

2. Read System Call
• Using this we could read only string/character
• System Call Number for Read is 3. It is copied to eax.
• The standard Input device(keyboard) is having the reference number 0
and it must be copied to ebx reg.
• We need to copy the pointer in memory, to which we need to store the
input string to ecx reg.
• We need to copy the number of characters in the string to edx reg.
• Then we will trigger INT 80h.
• We will get the string to the location which we copied to ecx reg.

mov
mov
mov
mov
int

eax,
ebx,
ecx,
edx,
80h

3
0
var
dword[size]

; Sys_call number for read
; Source Keyboard
; Pointer to memory location
; Size of the string
; Triggering OS Interrupt

• This method is also used for reading integers and it is a bit tricky. If we
need to read a single digit, we will read it as a single character and then
subtract 30h from it(ASCII of 0 = 30h). Then we will get the actual
value of that number in that variable.
(a) Reading a single digit number
mov
mov
mov
mov
int
sub

eax, 3
ebx, 0
ecx, digit1
edx, 1
80h
byte[digit1], 30h

;Now we have the actual number in [digit1]

(b) Reading a two digit number
31

;Reading
mov eax,
mov ebx,
mov ecx,
mov edx,
int 80h

first digit
3
0
digit1
1

;Reading
mov eax,
mov ebx,
mov ecx,
mov edx,
int 80h

second digit
3
0
digit2
2

;Here we put 2 because we need to read and
omit enter key press as well

sub byte[digit1], 30h
sub byte[digit2], 30h
;Getting the number from ASCII
; num = (10* digit1) + digit2
mov al, byte[digit1]
mov bl, 10
mul bl
movzx bx, byte[digit2]
add ax, bx

; Copying first digit to al
; Multiplying al with 10
; Copying digit2 to bx

mov byte[num], al
; We are sure that no less than 256, so we can
omit higher 8 bits of the result.
3. Write System Call
• This system call can be used to print the output to the monitor
• Using this we could write only string / character
• System Call Number for Write is 4. It is copied to eax.
• The standard Output device(Monitor) is having the reference number
1 and it must be copied to ebx reg.
• We need to copy the pointer in memory, where the output sting resides

32

to ecx reg.
• We need to copy the number of characters in the string to edx reg.
• Then we will trigger INT 80h.
See the below given example which is equivalent to the C statement
printf()
Eg:
mov
mov
mov
mov
int

eax,
ebx,
ecx,
edx,
80h

4
1
msg1
size1

;Sys_call number
;Standard Output device
;Pointer to output string
;Number of characters
;Triggering interrupt.

• This method is even used to output numbers. If we have a number we
will break that into digits. Then we keep each of that digit in a variable
of size 1 byte. Then we add 30h (ASCII of 0) to each, doing so we will
get the ASCII of character to be print.

33

Chapter 4
Introduction to Programming in
NASM
Now that we have gone through all the basics required for creating a program, let’s
begin. Like all programming lessons, we will first learn how to create a Hello World
program. Programming in NASM is easy if you understand how each construct
is used and implemented. A transition from C like language can be frustrating at
first, but let’s take it step by step. We will learn to program by translating code
snippets in C language to assembly code.

Hello World program
As discussed earlier, the whole program is divided into 3 sections, namely code
section (section .text), section for uninitalised variables (section .bss) and section
for initialised variables (section .data).
section .text is the place from where the execution starts in NASM program,
analogous to the main() function in C-Programming.
1. Let’s store the string "Hello World" into a variable named str. We are going
to print that string as the output
Pseudo Code:
char str[13] = "Hello World";
NASM Code:
section .data

;For Storing Initialized Variables
34

string: db ’Hello World’, 0Ah
newline character
length: equ 13

;String Hello World followed by a
; Length of the string stored to a constant.

NB: Using equ we declare constants, ie. their value wonâĂŹt change during
execution. $ -string will return the length of string variables in bytes (ie.
number of characters)
2. Now we need to print this variable onto the screen. For that we use the write
system call.
Pseudo Code:
printf("%s",str);
NASM Code:
mov
mov
mov
mov
int

eax,
ebx,
ecx,
edx,
80h

4
1
string
length

; Using int 80h to implement write() sys_call

3. Now since we have printed the text, let’s exit the program. We use the exit
system call for the same
Pseudo Code:
return 0;
NASM Code:
;Exit System Call
mov eax, 1
mov ebx, 0
int 80h

; Length of the string stored to a constant.

4. section .text is the place from where the execution starts in NASM program,
analogous to the main() function in C-Programming. So let’s put the print
and exit parts into the section
The final program looks like this:
section .text
global _start:

; Code Section

35

_start:
mov eax,
mov ebx,
mov ecx,
mov edx,
int 80h

4
1
string
length

; Using int 80h to implement write() sys_call

;Exit System Call
mov eax, 1
mov ebx, 0
int 80h
section
string:
newline
length:

.data
db ’Hello World’, 0Ah
character
equ 13

;For Storing Initialized Variables
;String Hello World followed by a
; Length of the string stored to a constant.

Compilation
1. Assembling the source file
nasm -f elf filename.asm
This creates an object file, filename.o in the current working directory.
2. Creating an executable file
For a 32 bit machine
ld filename.o -o output filename
For 64 bit machine
ld -melf_i386 filename -o output filename
This creates an executable file of the name output filename.
3. Program execution
./output filename
For example, if the program to be run is first.asm
nasm -f elf first.asm
ld first.o -o output
./output

36

It doesn’t seem as difficult as you felt in the beginning, right? In the same pattern, we’ll see how both simple and complex constructs are formed in assembly
language.

37

Chapter 5
Integer Handling
Integers are stored as characters in NASM. This chapter describes how to read and
print integers. We have already skimmed through the basic operations in integer
handling (add, sub, mul, div) in Chapter 3. Here, we will look at sample programs
that will make use of these operations.
1. Declaring an integer
section.bss
var: resb/resw/resd 10
Here we have reserved 10 bytes/words/double words for the memory location
’var’
2. Reading a single digit
To read a single digit, we will read it as a single character and then subtract
30h from it (ASCII of 0 = 30h). This is to get the actual value of that
number in that variable.
mov
mov
mov
mov
int
sub

eax, 3 ; Syscall number for read
ebx, 0 ; Source Keyboard
ecx, var ; Pointer to memory location âĂŹvarâĂŹ
edx, byte[size] ; Size of the input
80h ; Triggering OS Interrupt
byte[var], 30h ; Subtracting 30h to get integer value

Note that the single digit taken as input here is stored in the first byte of
’var’. If ’var’ was declared as a word/double word, the size of the input stored

38

in the edx register should be changed accordingly to word[size] or dword[size].
3. Reading a multi-digit number
To read a multi-digit number, we read the input character by character (here,
digit by digit) until we obtain the enter-key. Each digit is multiplied with 10
and added to the next digit to obtain the original number.
- Pseudo Code
while(temp!=enterkey)
num=0
read(temp)
num=numx10+temp
endwhile
NASM Code:
read_num:
;;push all the used registers into the stack using pusha
pusha
;;store an initial value 0 to variable âĂŹnumâĂŹ
mov word[num], 0

loop_read:
;; read a digit
mov eax, 3
mov ebx, 0
mov ecx, temp
mov edx, 1
int 80h
;;check if the read digit is the end of number, i.e, the enter-key whose ASCII
cmp byte[temp], 10
je end_read
mov ax, word[num]
mov bx, 10
mul bx
mov bl, byte[temp]
sub bl, 30h
mov bh, 0
add ax, bx

39

mov word[num], ax
jmp loop_read
end_read:
;;pop all the used registers from the stack using popa
popa
ret
4. Printing a number
Unlike when we’re reading a number, there is no end character like the enterkey marking the number’s last digit. Here, we extract each digit from the
number and push it to the stack. We also keep a count of the number of
digits in the number. Using this count, the digits are popped out of the stack
and printed in order.
- Pseudo Code
count=0
while (num!=0)
temp=num%10
count++
push(temp)
num=num/10
endwhile
while(count!=0)
temp=pop()
print(temp)
count- endwhile
NASM Code:
print_num:
mov byte[count],0
pusha
extract_no:
cmp word[num], 0
je print_no
inc byte[count]
mov dx, 0
mov ax, word[num]

40

mov bx, 10
div bx
push dx
mov word[num], ax
jmp extract_no
print_no:
cmp byte[count], 0
je end_print
dec byte[count]
pop dx
mov byte[temp], dl
add byte[temp], 30h
mov eax, 4
mov ebx, 1
mov ecx, temp
mov edx, 1
int 80h
jmp print_no
end_print:
mov eax,4
mov ebx,1
mov ecx,newline
mov edx,1
int 80h
;;The memory location âĂŹnewlineâĂŹ should be declared with the ASCII key for
popa
ret
Example Program:
Read 2 numbers and find their GCD.
Pseudo Code:
read(num1)
read(num2)
while(1)
if(num1%num2==0)
goto endloop

41

temp=num1%num2
num1=num2
num2=temp
endwhile
endloop:
print(num2)
NASM Code
section .data
newline: db 10
section .bss
num1:resw 10
num2 resw 10
temp:resb 10
num:resw 10
nod:resb 10
count:resb 10
section .text
global _start
_start:
call read_num
mov cx,word[num]
mov word[num1],cx
call read_num
mov cx,word[num]
mov word[num2],cx
mov ax,word[num1]
mov bx,word[num2]
loop1:
mov dx,0
div bx
cmp dx,0
je end_loop
mov ax,bx
mov bx,dx
jmp loop1

42

end_loop:
mov word[num],bx
call print_num
exit:
mov eax,1
mov ebx,0
int 80h
Do not forget to add the read_num, print_num functions

43

Chapter 6
Subprograms
We have now covered the basics of programming in assembly language. Often
we might have to repeat the same code in the program and that is when we use
functions in languages like C. Let’s see how it is done in assembly language.
The concept of subprograms can be used heavily in assembly language, since input
and output operations require more than one line of code. Making them into
subprograms not only makes the code smaller, but also helps you understand and
debug the code better. It is highly reccommended that you start using subprograms
in your code from the very start of programming in assembly language.
CALL & RET Statements:
In NASM, subprograms are implemented using the call and ret instructions. The
general syntax is as follows:

;main code....
------------------------call function_name
------------------------;rest of the code....

function_name:

;Label for subprogram
44

------------------;function code....
------------------ret
• When we use the CALL instruction, address of the next instruction will be
copied to the system stack and it will jump to the subprogram. ie. ESP
will be decreased by 4 units and address of the next instruction will go over
there.
• When we call the ret towards the end of the sub-program then the address
being pushed to the top of the stack will be restored and control will jump
to that.
Calling Conventions:
The set of rules that the calling program and the subprogram should obey to pass
parameters are called calling conventions. Calling conventions allow a subprogram
to be called at various parts of the code with different data to operate on. The
data may be passed using system registers/memory variables/system stack. If we
are using system stack, parameters should be pushed to system stack before the
CALL statement and remember to pop off, preserve and to push back the return
address within the sub program which will be in the top of the stack.
Now, let’s see an example on how to use subprograms effectively to simplify your
code. Below is a program which uses subprograms to calculate the sum of 10 input
integers.
NB: One major advantage of subprograms that you can exploit is by creating subprograms for input and output, as those components appear in almost all the programs
you are going to write. As the codes are quite lengthy, repeating the code for each
input and output will make your code even lengthier, making debugging and understanding the code a tedious task. In the program given below, three subprograms
are used, one for input, one for output and the last one for calculating the sum. It
is suggested that you use subprograms for input and ouput from now on.
section .data
msg1: db 0Ah,"Enter a number"
size1: equ $-msg1
msg2: db 0Ah,"the sum= "

45

size2: equ $-msg2
newline: db 0Ah
section .bss
num: resw 1
sum: resw 1
digit1: resb 1
digit0: resb 1
digit2: resb 1
temp: resb 1
counter:resb 1
count: resb 1
section .text
global _start
_start:
mov byte[counter],0
mov byte[sum],0
read_and_add:
call read_num ;; Instead of the code for reading a number,
;; we just call the subprogram
call add
;loop condition checking
inc byte[counter]
cmp byte[counter],10
jl read_and_add

mov ax, word[sum]
mov word[num], ax
call print_num ;; Instead of the code for printing a number,
;; we just call the subprogram

46

;exit
mov eax,1
mov ebx,0
int 80h
;sub-program add to add a number to the existing sum
add:
mov ax,word[num]
add word[sum],ax
ret
;;subprogram to print a number
print_num:
mov byte[count],0
pusha
extract_no:
cmp word[num], 0
je print_no
inc byte[count]
mov dx, 0
mov ax, word[num]
mov bx, 10
div bx
push dx
mov word[num], ax
jmp extract_no
print_no:
cmp byte[count], 0
je end_print
dec byte[count]
pop dx
mov byte[temp], dl
add byte[temp], 30h
mov eax, 4
mov ebx, 1
mov ecx, temp
mov edx, 1
int 80h
jmp print_no

47

end_print:
mov eax,4
mov ebx,1
mov ecx,newline
mov edx,1
int 80h
;;The memory location âĂŹnewlineâĂŹ should be declared with the ASCII key for
;;new line in section.data.
popa
ret
;; subprogram for inputting a number

read_num:
;;push all the used registers into the stackusing pusha
pusha
;;store an initial value 0 to variable âĂŹnumâĂŹ
mov word[num], 0
loop_read:
;; read a digit
mov eax, 3
mov ebx, 0
mov ecx, temp
mov edx, 1
int 80h
;;check if the read digit is the end of number,i.e, the enter-key whose ASCII key i
cmp byte[temp], 10
je end_read
mov ax, word[num]
mov bx, 10
mul bx
mov bl, byte[temp]
sub bl, 30h
mov bh, 0
add ax, bx
mov word[num], ax
jmp loop_read
end_read:

48

;;pop all the used registers from the stackusing popa
popa
ret
Recursive Sub-routine:
A subprogram which calls itself again and again recursively to calculate the return
value is called recursive sub-routine. We could implement recursive sub-routine for
operations like calculating factorial of a number very easily in NASM. A program
to print fibonacci series up to a number using recursive subprogram is given in the
appendix.
Using C Library functions in NASM:
We can embed standard C library functions into our NASM Program especially
for I/O operations. If we would like to use that then we have to follow Cś calling
conventions given below:
• parameters are passed to the system stack from left to right order.
Eg: printf(âĂIJ%dâĂİ,x)
Here value of x must be pushed to system stack and then the format string.
• C-Function wonâĂŹt pop out the parameters automatically, so it is our duty
to restore the status of stack pointers(ESP and EBP) after the function being
called.
Eg:

Reading an integer using the C-Functions...

section .text
global main
;Declaring the external functions to be used in the program.......
extern scanf
extern printf
;Code to read an integer using the scanf function
getint:
push ebp
;Steps to store the stack pointers
mov ebp , esp

49

;scanf(âĂIJ\%dâĂİ,&x)
;Creating a space of 2 bytes on top of stack to store the int value
sub esp , 2
lea eax , [ ebp-2]
push eax
push fmt1
call scanf
mov ax, word [ebp-2]
mov word[num], ax

; Pushing the address of that location
; Pushing format string
; Calling scanf function

;Restoring the stack registers.
mov esp , ebp
pop ebp
ret
putint:
push ebp
mov ebp , esp
;printf(âĂIJ%dâĂİ,x)
sub esp , 2
mov ax, word[num]
mov word[ebp-2], ax
push fmt2
call printf
mov esp , ebp
pop ebp
ret

; Steps to store the stack pointers

; Creating a space of 2 bytes and storing the int value t

; Pushing pointer to format string
; Calling printf( ) function
; Restoring stack to initial values

main: ; Main( ) Function
mov eax, 4
mov ebx, 1
mov ecx, msg1
mov edx, size1
int 80h
call getint
mov ax, word[num]

50

mov bx, ax
mul bx
mov word[num], ax
call putint
exit:
mov ebx , 0
mov eax, 1
int 80h
section .data
fmt1 db "\%d",0
fmt2 db "Square of the number is : %\d",10
msg1: âĂIJEnter an integer : "
size1: db $-msg1
section .bss
num: resw 1
NB: Assembling and executing the code...
• First we have to assemble the code to object code with NASM Assembler,
then we have to use gcc compiler to make the executable code.
• nasm -f elf âĂŞo int.o int.asm
• gcc int.o -o int
• ./int

51

Chapter 7
Arrays and Strings
An Array is a continuous storage block in memory. Each element of the array have
the same size. We access each element of the array using:
i) Base address/address of the first element of the array.
ii) Size of each element of the array.
iii) Index of the element we want to access.

7.1

One Dimensional Arrays

In NASM there is no array element accessing/dereferencing operator like [ ] in C
/ C++ / Java using which we can access each element. Here we compute the
address of each element using an iterative control structure and traverse though
the elements of the array.
1. Declaring/ Initializing an array
section .bss
arr: resb/resw/resd 50
array1: db 2, 5, 8, 10, 12, 15
array2: dw 191, 122, 165, 165
array3: dd 111, 111, 111

52

;Declares an array of size 50
;Initializes array
;{2,5,8,10,12,15}
;An array of 4 words
;An array of 4 dwords

We can also use TIMES keyword to repeat each element with a given value
and thus easily create array elements:
Eg:
array1:
TIMES
with each element=1
array2:
TIMES

100 db

1

; An array of 100 bytes

20

2

; An array of 20 dwords

dw

2. Reading an n-size array
Pseudo Code:
i=0
while(i
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