CS 225 Assemply Lab Manual -To Be Completed

7/23/2019 CS 225 Assemply Lab Manual -To Be Completed

http://slidepdf.com/reader/full/cs-225-assemply-lab-manual-to-be-completed 1/20

1

CS225 Lab Manual

Course Description:

This course concentrates on the practical part of Computer Organization by using

Assembly language. This course allows students to practice writing programs based on the

concepts the will learn through the course by giving the students different types ofproblems to be solved using MASM (Macro Assembler)

Course Objectives:

1. Teach students basic principles about computer architecture, machine language,

and low-level programming.

2. Teach students enough assembly language to enhance their knowledge on today's

most widely used microcomputer family.

3. Improving students systems programming skills through programming exercises

carried out by students.

4.

Students are expected to implement solutions to problems using the concepts theywill take through the course.

Learning Outcomes:

After completing this course the student expect to:

Solve a problem and write a program for this problem using assembly language by using

MASM and the concepts they will take through the course.



2

Course Plan:1. Basic Concepts

2. Assembly Language Fundamentals

3. Data Transfers, Addressing, and Arithmetic

4. Procedures

5.

Conditional Processing6. Integer Arithmetic

7. Strings and Arrays

8. 16-Bit MS-DOS Programming



3

Introduction:

What Is Assembly Language?

Assembly language is a level lays between high level language and machine code, it reflects the

instruction set as well as the processor architecture. In this course we are focusing on Intel 32-bit processors. We will show how to code in assembly language which will be translated in to

machine code later. Many types of assemblers are used for Intel processors, such free assembler

as: NASM (Netwide Assembler), MASM (Microsoft Assembler), and TASM (Borland TurboAssembler).

Coding with assembly language provides time-efficient and space-efficient codes. Assembly

language allows to access computer hardware easily. Assembly language composed assembleand linker. An assembler is responsible to translate assembly program into object file. A linker

is responsible to aggregates multiple object files generated by an assembler into an executable

process.

The following figure shows P6 Family Processor Basic Execution Environment



4

general purpose registers[2]

Segment register[2]



5

Segment registers [2]

Data sizes [2]



6

2- Number Systems Used in Computers

We will talk about numbering system such as binary system, hexadecimal system, decimal

system and octal system. Binary numbers mainly deals with two digits, which are zero and one.

Example of binary number is (10110011)2 , 2 means base 2, and it is equivalent to (179)10 , 10

means base 10.

1. Converting from decimal number into binary number

To convert the whole number portion to binary, use successive division by 2 until the quotient

is 0. The remainders form each successive division compose the answer, with the first

remainder as the least significant bit (LSB) and the last as the most significant bit (MSB).

To Convert 201210 to binary number, use repeated division by 2 and collect reminders until the

quotient becomes 0. The collected reminders are the answer on condition the first reminder is

the least significant bit and the last reminder is the most significant bit. See the following

example.

2012 / 2 = 1006 remainder 0 (LSB)

/ 2 = 503 remainder 0






/ 2 = 7 remainder 1

/ 2 = 3 remainder 1

/ 2 = 1 remainder 1

/ 2 = 0 remainder 1 (MSB)

201210 = 111110111002

To convert decimal fractions to binary, multiply it by 2, collect non-fraction 0 or 1, then replace

it by zero, repeat this actions until the whole value becomes 0. The collected 0s and 1s are the

answer on condition the first value is the most significant bit and the last 0 or 1 is the least

significant bit. See the following example.

Example: Convert 0.62510 to binary



7

.625 2 = 1.25 1 (MSB)

.25 2 = 0.50 0

.5 2 = 1.0 1 (LSB)

0.62510 = 0.1012

2. Converting from decimal number into hexadecimal number

Hexadecimal system contains the following digits: 0, 1,2, 3, 4, 5, 6, 7, 8, 9. A, B, C, D, E and F.

the following table shows some binary numbers and their equivalence hexadecimal values.

Table 1: binary and hexadecimal values

BinaryHexadicimal

BinaryHexadicimal

BinaryHexadicimal

BinaryHexadicimal

0000 0 0001 1 0010 2 0011 3

0100 1 0101 5 0110 6 0111 7

1000 2 1001 9 1010 A 1011 B

1100 C 1101 D 1110 E 1111 F

To convert from decimal value into binary value, use successive division by 16 until the

quotient becomes 0. The remainders form each successive division compose the answer, with

the first remainder as the least significant bit (LSB) and the last as the most significant bit

(MSB).

To Convert 201210 to hexadicimal number, use repeated division by 16 and collect reminders

until the quotient becomes 0. The collected reminders are the answer on condition the first

reminder is the least significant bit (LSB) and the last reminder is the most significant bit

(MSB). See the following example.

2012 / 16 = 125 remainder C (LSB)

125 / 2 = 7 remainder D

7 / 2 = 0 remainder 7 (MSB)

201210 = 7DC16

An easier way to convert from decimal to hexadecimal is get binary value first and then

convert the binary number into hexadecimal using table 1.

201210 = 111110111002



8

To do this, divide the binary number into multiple of four binary digits starting from right to

left: (111 1101 1100)2 . if the last part is less than four digits, then pad it with zeros: (0111 1101

1100)2 = ( 7DC)16

To convert decimal fractions to hexadecimal value, multiply it by 16, and collect non-fraction

value, then replace it by zero. Repeat this actions until the whole value becomes 0. The collected

numbers are the answer on condition the first value is the most significant bit and the last

number is the least significant bit. See the following example.

Example: Convert 0.039062510 to hexadecimal

0. 0.0390625 16 = 0.625 0 (MSB)

0.625 x 16 = 10.0 A (LSB)

0.039062510 = .0A16

3. Converting from decimal number into octal number

To Convert 201210 to octal number, use repeated division by 8 and collect the reminders until

the quotient becomes 0. The collected reminders are the answer on condition the first

reminder is the least significant bit (LSB) and the last reminder is the most significant bit

(MSB). See the following example.

2012 / 8 = 251 remainder 4 (LSB)

251 / 8 = 31 remainder 3

31 / 8 = 3 remainder 7

3 / 8 = 0 remainder 3 (MSB)

201210 = 43338

An easier way to convert from decimal to hexadecimal is get binary value first and then

convert the binary number into octal using table 2.

Table 2: binary and equivalent octal values

Binary Hexadicimal Binary Hexadicimal Binary Hexadicimal Binary Hexadicimal

000 0 001 1 010 2 011 3

100 1 101 5 110 6 111 7

201210 = 111110111002



9

To do this, divide the binary number into multiple of three binary digits starting from right to

left: (11 111 011 100)2 . if the last part is less than three digits, then pad it with zeros:

(011 111 011 100)2 = ( 3734)16

To convert decimal fractions to hexadecimal value, multiply it by 8, and collect non-fraction

value, then replace it by zero. Repeat this actions until the whole value becomes 0. The collected

numbers are the answer on condition the first value is the most significant bit and the last

number is the least significant bit. See the following example.

Example: Convert 0.039062510 to octal

0. 0.0390625 8 = 0.03125 0 (MSB)

0.03125 x 8 = 0.25 0

0.25 x 8 2.0 2 (LSB)

0.039062510 = .00216

4. Converting binary value into decimal value

To convert a binary value into decimal value, multiplies each digit by its corresponding power

of 2 and then add them. Starting from right to left, power starts with 0, and then 1, and so on.

For example, (100101000101)2 = 1 * 20 + 0 * 2

1 + 1 * 2

2 + 0 * 2

3 + 0 * 2

4 + 0 * 2

5 + 1 * 2

6 + 0 *

27 + 1 * 28 + 0 * 29 + 0 * 210 + 1 * 211 = 1 + 4 + 64 + 256 + 2048 = 2373

To convert fractioned binary value into decimal value, multiplies each digit by its

corresponding power of 2 and then add them. Starting from left to right after the fraction point

“.”, power starts with -1, and then -2, and so on. For example, (0.1001)2 = 1 * 2-1 + 0 * 2-1 + 0 *

2-2 + 1 * 2-3 = .5 + 0.1250= 0.625.

5. Converting hexadecimal value into decimal value

To convert a hexadecimal value into decimal value, multiplies each digit by its corresponding

power of 16 and then adds them. Starting from right to left, power starts with 0, and then 1,

and so on. For example, (A13F)16 = F * 160 + 3 * 16

1 + 1 * 16

2 + A * 16

3 = 15 * 16 + 3 * 16 + 1

* 256 + 10 * 4096 = 41279.

To convert fractioned hexadecimal value into decimal value, multiplies each digit by its

corresponding power of 16 and then adds them. Starting from left to right after the fraction



10

point “.”, power starts with -1, and then -2, and so on. For example, (0.EF1) 16 = E * 16-1 + F *

16-1

+ 1 * 16-2

= (14/16) + (15/256) + (1/4096)= 0.933837890625.

6. Converting octal value into decimal value

To convert a octal value into decimal value, multiplies each digit by its corresponding power of

8 and then adds them. Starting from right to left, power starts with 0, and then 1, and so on.

For example, (A13F)16 = F * 160 + 3 * 161 + 1 * 162 + A * 163 = 15 * 16 + 3 * 16 + 1 * 256 + 10

* 4096 = 41279.

To convert fractioned octal value into decimal value, multiplies each digit by its corresponding

power of 8 and then adds them. Starting from left to right after the fraction point “.”, power

starts with -1, and then -2, and so on. For example, (0.734)8 = 7 * 8-1 + 3 * 8-1 + 4 * 8-2 = (7/8)

+ (3/64) + (4/512) = 0.9296875.



11

3- Binary Arithmetic Operations

a. Unsigned addition

Decimal number addition is done by adding each pair of digits. The carry is added with the

next pair of digits and so on. Adding two binary numbers are done in the same way by addingeach pair of bits together, the resulted carry is added with the next pair.

1 1 1 0 1 0 1 1 1 Carry

X 211 0 1 1 0 1 0 0 1 1 +

Y 343 1 0 1 0 1 0 1 1 1

554 0 0 0 1 0 1 0 1 0

Hexadecimal and octal addition is done by the same procedure. However we can convert each

hexadecimal and octal numbers into corresponding binary number. By then we can do simpler

addition.

1 1 0 1 0 1 1 1 Carry

( D3) 16 + 1 1 0 1 0 0 1 1 +

( 57) 16 0 1 0 1 0 1 1 1

( 12A) 16 0 0 1 0 1 0 1 0

b. Signed Integers

Signed integers are represented by reserving the most significant bit. If the MSB is equal to 1,

then the signed integer number is negative, otherwise, it is positive. In this section the student

need to learn about the range of signed fixed number of bits. See the figure below. Table 3

shows the code representation for signed and unsigned numbers for four bits. You will notice

that signed four-bit ranges between -7 and +7 for 1

st

complement, and ranges from -8 to +7 for2

nd complement.



12

Table 3 the code representation for signed and unsigned numbers

Signed number is represented either by one’s complement or by two’s complement. To

get the first complement of any binary number is accomplished by negating each bit of

the corresponding binary number. So, the 1st complement of (10001010010)2 is equal to

(01110001101)2, the 2nd complement is represented by adding one to the 1st complement.

Then (01110001101)2 + 1 = (01110001110)2. as a result (01110001110)2 is the second

complement of (10001010010)2.

c. Convert Signed Binary to Decimal

4-bit binary

value

Unsigned

decimal value

Signed decimal value

1st complement 2nd complement

0000 0 +0 0

0001 1 1 10010 2 2 2

0011 3 3 3

0100 4 4 4

0101 5 5 5

0110 6 6 6

0111 7 7 7

1000 8 -7 -8

1001 9 -6 -7

1010 10 -5 -6

1011 11 -4 -5

1100 12 -3 -41101 13 -2 -3

1110 14 -1 -2

1111 15 -0 -1



13

If the MSB is equal to 0, then the sign of the binary code is positive, in this case we

follow the same procedures we did in section “Converting binary value into decimal

value” to get it\s value.

If the MSB bit is 1 then the sign of the binary number is negative. To get its positive

decimal equivalent we need to convert it into two’s complemented, afterward; convertthe result into decimal value. Ex, if the number (1110 0101)2 is represented in the 2nd

complement, the binary positive equivalent value is (0001 1011 + 1 = 0001 1011) 2. The

decimal value for (0001 1011)2 is (27)10, so the decimal equivalent of (1110 0101)2 is (-

27)10.

There are basically two methods summarized in table 3 for representing signed

numbers: one's complement, and two's complement.

d.

signed addition

Suppose we want to subtract (00101011)2 from (01110001)2

So subtracting [(00101011)2 which is equal 43] from [(01110001)2 which is equal 113] yield

to [(01000110) 2 which is equal 70]. The result is correct if the carry-in to sign bit is equal to

carry-out from the sign bit, otherwise an overflow bit is set to one.

To accomplish the operation we need to

convert (00101011)2 into negative value by

getting the 2nd complement of the binary

number, by then adding the two values

leads to the target



14

e. Number of bits versus number of states

One bits is able to store either 0 or 1, in another words, we can say that 1 bit is able to

address (21) 2 states. 2 bits is able to address (2

2) 4 states, which are 00, 01, 10 and 11. 8 bits

forms a byte, which is the smallest addressable data item. A byte is able to address (28)

256

states. The range of unsigned values that can be derived from 8 bits are numbers between0 and (2

8-1) 255. The range of signed values using 2

nd complement that can be derived from

8 bits are numbers between (-(27)) -128 and (27-1) +127, since that the MSB bit is reserved

for the sign and the rest of 7 bits are reserved for the value. Table 4 shows the range of

unsigned and signed numbers.

Table 4 Range of unsigned and signed numbers

Try this:

What is the largest unsigned integer that may be stored in 30 bits?

What is the largest signed integer using 2nd complement that may be stored in 30 bits?

Byte size Range of

Unsigned

decimal values

Range of Signed decimal values

1st complement 2nd complement

1 byte 0 to (2

8

-1) -(2

7

-1) to +(2

7

-1) -(2

7

) to +(2

7

-1)2 bytes 0 to (2

16-1) -(2

15-1) to +(2

15-1) -(2

15) to +(2

15-1)

4 bytes 0 to (232-1) -(231-1) to +(231-1) -(231) to +(231-1)

8 bytes 0 to (264-1) -(263-1) to +(263-1) -(263) to +(263-1)

10 bytes 0 to (280-1) -(279-1) to +(279-1) -(279) to +(279-1)



15

Section 2:

a) The Intel 80386 Processor

Intel 8286 Processor consists of general purpose registers eax, ebx, and so on, see the

figure below. CPU registers are very fast memory. Eax, ebx, ecx and edx are trated as 8

bits, 16 bits and 32 bits.

EAX, AX and AL or AH can be reached individually, EBX, ECX, and EDX have the

same structure. AX means the accumulator register, BX means the base address register,

CX means the counter register and DX means the data register. EAX means extended

AX, and so on for EBX, ECX, and EDX. Extended instruction pointer register (EIP)

contains the address of the next instruction to execute.

b)

The x86 Instruction Set

a. MOV instruction

mov instruction used to transfer data between registers, from memory to a register or

from register to memory; but not between memoirs. The mov instruction moves the value

stored in operand2 to the operand1, so the first operand is the destination and the second

operand is the source.

mov operand1, operand2 operand2 operand1

mov reg8,16,32, reg8,16,32

mov reg8,16,32, memory 8,16,32

mov reg8,16,32, constant

mov memory8,16,32, reg8,16,32

mov memory8,16,32, constant

Ex:

CS

SS

DS

ES

EIP

EFLAGS

16-bit Segment Registers

EAX

EBX

ECX

EDX

32-bit General-Purpose Registers

FS

GS

EBP

ESP

ESI

EDI



16

Mov ax, bx ; move the value stored in register bx into register ax, note that both the

; source and the destination storage must have the same size.

Mov ebx, ecx ; move the value stored in register ecx into register ebx.

Mov dl, dh ; move the value stored in register dh into register dl.

Mov al, cl ; move the value stored in register cl into register al.

b. The add instruction

The add instruction adds the value stored in operand2 into the value stored in operand1,

storing the result in operand1.

add operand1, operand2 operand1 = operand1 + operand2

add reg8,16,32, reg8,16,32

add reg8,16,32, memory 8,16,32

add reg8,16,32, constant

add memory8,16,32, reg8,16,32

add memory8,16,32, constant

Ex:

add ax, bx ; add the value stored in register bx into register ax, note that both the

; source and the destination storage must have the same size.

add ebx, ecx ; add the value stored in register ecx into register ebx.

add dl, dh ; add the value stored in register dh into register dl.

add al, cl ; add the value stored in register cl into register al.

c. The sub instruction

The sub instruction subtracts the value of the operand2 from the value stored in operand1,

storing the result in operand1.

sub operand1, operand2 operand1 = operand1 - operand2

sub reg8,16,32, reg8,16,32

sub reg8,16,32, memory 8,16,32

sub reg8,16,32, constant

sub memory8,16,32, reg8,16,32

sub memory8,16,32, constant

Ex:

sub ax, bx ; subtract the value stored in register bx from the value stored in register

; ax, and then store the result in ax. Note that both the source and the



17

; destination storage must have the same size.

sub ebx, ecx ; subtract the value stored in register ecx from the value stored in register

; ebx value.

sub dl, dh ; subtract the value stored in register dh from the value stored in register

; dl.

sub al, cl ; subtract the value stored in register cl from the value stored from register

; al.

d. The sub instruction

The cmp instruction is used to compare between operand1 and operand2. The result of

the comparison affect the flag register for subsequent use for conditional jump

instruction.

cmp operand1, operand2 compare operand1with operand2

Flag-reg

cmp reg8,16,32, reg8,16,32

cmp reg8,16,32, memory 8,16,32

cmp reg8,16,32, constant

cmp memory8,16,32, reg8,16,32

cmp memory8,16,32, constant

Ex:

cmp ax, bx ; compare the value stored in register bx with the value stored in register

; ax. Note that both the source and the destination storage must have the; same size.

cmp ebx, ecx ; compare the value stored in register ecx with the value stored in register

; ebx.

cmp dl, dh ; compare the value stored in register dh with the value stored in register

; dl.

cmp al, cl ; compare the value stored in register cl with the value stored in register al.

e. The and instruction

The and instructions is used to bitwise and between operand1 and operand2, storing the

result into operand1.

and operand1, operand2 operand1 = operand1 & operand2

and reg8,16,32, reg8,16,32

and reg8,16,32, memory 8,16,32



18

and reg8,16,32, constant

and memory8,16,32, reg8,16,32

and memory8,16,32, constant

Ex:

and ax, bx ; and the value stored in register bx with the value stored in register

; ax, storing the result in register ax. Note that both the source and the


and ebx, ecx ; and the value stored in register ecx with the value stored in register

; ebx, storing the result in register ebx.

and dl, dh ; and the value stored in register dh with the value stored in register

; dl, storing the result in register dl.

and al, cl ; and the value stored in register cl with the value stored in register al,

; storing the result in register ax.

f. The or instruction

The and instructions is used to bitwise and between operand1 and operand2, storing the

result into operand1.

or operand1, operand2 operand1 = operand1 | operand2

or reg8,16,32, reg8,16,32

or reg8,16,32, memory 8,16,32

or reg8,16,32, constant

or memory8,16,32, reg8,16,32 or memory8,16,32, constant

Ex:

or ax, bx ; or the value stored in register bx with the value stored in register

; ax, storing the result in register ax. Note that both the source and the


or ebx, ecx ; or the value stored in register ecx with the value stored in register

; ebx, storing the result in register ebx.

or dl, dh ; or the value stored in register dh with the value stored in register

; dl, storing the result in register dl.

or al, cl ; or the value stored in register cl with the value stored in register al,

; storing the result in register ax.



19

g. The not instruction

The not instruction is used to get the 1st complement of the value stored in operand1.

not operand1 operand1 = first-complement(operand1)

not reg8,16,32

not memory8,16,32

Ex:

not ax ; get the first complement of ax.

not ebx ; get the first complement of ebx.

not dl ; get the first complement of dl.

not al ; get the first complement of al.

Not [mem] ; get the first complement of memory location. The size of the location

; must be specified.

h. Conditional jump instruction

ja dest -- Jump if above

jae dest -- Jump if above or equal

jb dest -- Jump if below

jbe dest -- Jump if below or equal

je dest -- Jump if equal

jne dest -- Jump if not equal

jmp dest -- Unconditional jumpiret -- Return from an interrupt



20

References:

[1] http://www.asmirvine.com/

[2] Intel Architecture Software Developer’s Manual Volume 1: Basic Architecture

[3] Guide to Assembly Language Programming in Linux book.

[4] http://www.intel.com/content/dam/www/public/us/en/documents/datasheets/3rd-gen-core-

desktop-vol-1-datasheet.pdf

[5] The art of assembly language book. http://flint.cs.yale.edu/cs422/doc/art-of-asm/pdf/

http://www.asmirvine.com/





http://www.intel.com/content/dam/www/public/us/en/documents/datasheets/3rd-gen-core-desktop-vol-1-datasheet.pdf




http://flint.cs.yale.edu/cs422/doc/art-of-asm/pdf/








Date post:	18-Feb-2018
Category:	Documents
Upload:	mohit-su
View:	216 times
Download:	0 times

CS 225 Assemply Lab Manual -To Be Completed

Documents