1. Computer Architecture and Assembly Language
Practical Session 1

2. Data Representation Basics
Bit – basic information unit: (1/0)
Byte – sequence of 8 bits: 7 6 5 4 3 2 1
MSB (Most Significant Bit) LSB (Least Significant Bit)
232-1
2K-1 1 …
Main Memory is an array of bytes, addressed by 0 to 232-1=0xFFFFFFFF
232 bytes = 4∙210∙3 bytes = 4 G bytes
address space
physical memory
Word – a sequence of bits addressed as a single entity by the computer
byte
byte
16 bit word

3. Registers Register file - CPU unit which contains (32 bit) registers.
general purpose registers EAX, EBX, ECX, EDX (Accumulator, Base, Counter, Data)
index registers ESP, EBP, ESI, EDI (Stack pointer - contains the address of last used dword in the stack, Base pointer, Source index, Destination Index)
flag register / status register EFLAGS
Instruction Pointer / Program Counter EIP / EPC
- contains address (offset) of the next instruction that is going to be executed (at run time) - changed by unconditional jump, conditional jump, procedure call, and return instructions
Register file
High byte
Low byte
Extended
16-bit register
Note that the list of registers above is partial. The full list can be found here.

4. Assembly Language Program
consists of a series of processor instructions, meta-statements, comments, and data
translated by assembler into machine language instructions (binary code) that can be loaded into memory and executed
NASM - Netwide Assembler - is assembler and for x86 architecture
Example:
assembly code:
MOV AL, 61h ; load AL with 97 decimal (61 hex)
binary code:
a binary code (opcode) of instruction 'MOV'
specifies if data is byte (‘0’) or full size 16/32 bits (‘1’)
a binary identifier for a register 'AL'
a binary representation of 97 decimal (97d = (int)(97/16)*10 + (97%16 converted to hex digit) = 61h)

5. label: (pseudo) instruction operands ; comment
Basic Assembly Instruction Structure
label: (pseudo) instruction operands ; comment
either required or forbidden by an instruction
optional fields
RAM
each instruction has its offset (address)
we mark an instruction with a label to refer it in the code
(non-local) labels have to be unique
an instruction that follows a label can be at the same / next line
colon is optional …
buffer
2048
64 bytes
Examples:
mov ax, ; moves constant 2 to the register ax buffer: resb ; reserves 64 bytes 
Notes:
- backslash (\) : if a line ends with backslash, the next line is considered to be a part of the backslash-ended line - no restrictions on white space within a line
mov buffer, 2
= mov 2048, 2 
mov [buffer], 2
= mov [2048], 2 
mov byte [buffer], 2
= mov byte [2048], 2

6. Instruction Arguments
A typical instruction has 2 operands - target operand (left) - source operand (right)
3 kinds of operands exists - immediate : value - register : AX,EBP,DL etc. - memory location : variable or pointer
Examples:
mov ax, mov [buffer], ax
target operand
source operand
target operand
source operand
register
immediate
memory location
register
! Note that x86 processor does not allow both operands be memory locations.
mov [var1],[var2]

7. MOV - Move Instruction – copies source to destination
mov reg8/mem8(16,32),reg8/imm8(16,32) (copies content of register / immediate (source) to register / memory location (destination))
mov reg8(16,32),reg8/mem8(16,32) (copies content of register / memory location (source) to register (destination))
operands have to be of the same size
Examples:
mov eax, 0x2334AAFF mov [buffer], ax mov word [var], 2
reg32
imm32
mem16
reg16
mem16
imm16
Note that NASM doesn’t remember the types of variables you declare . It will deliberately remember nothing about the symbol var except where it begins, and so you must explicitly code mov word [var], 2.

8. Basic Arithmetical Instruction
<instruction> reg8/mem8(16,32),reg8/imm8(16,32) (source - register / immediate, destination- register / memory location)
<instruction> reg8(16,32),reg8/mem8(16,32) (source - register / immediate, destination - register / memory location)
ADD - add integers
Example: add AX, BX ;(AX gets a value of AX+BX)
SUB - subtract integers
Example: sub AX, BX ;(AX gets a value of AX-BX)
ADC - add integers with carry (value of Carry Flag)
Example: adc AX, BX ;(AX gets a value of AX+BX+CF)
SBB - subtract with borrow (value of Carry Flag)
Example: sbb AX, BX ;(AX gets a value of AX-BX-CF)

9. Basic Arithmetical Instruction
<instruction> reg8/mem8(16,32) (source / destination - register / memory location)
INC - increment integer
Example: inc AX ;(AX gets a value of AX+1)
DEC - increment integer
Example: dec byte [buffer] ;([buffer] gets a value of [buffer] -1)

10. Basic Logical Instructions
<instruction> reg8/mem8(16,32) (source / destination - register / memory location)
NOT – one’s complement negation – inverts all the bits
Example: mov al, b not al ;(AL gets a value of b) ;( b b = b)
NEG – two’s complement negation – inverts all the bits, and adds 1
Example: mov al, b neg al ;(AL gets a value of not( b)+1= b+1= b) ;( b b = b = 0)

11. Basic Logical Instructions
<instruction> reg8/mem8(16,32),reg8/imm8(16,32) (source - register / immediate, destination- register / memory location)
<instruction> reg8(16,32),reg8/mem8(16,32) (source - register / immediate, destination - register / memory location)
OR – bitwise or – bit at index i of the destination gets ‘1’ if bit at index i of source or destination are ‘1’; otherwise ‘0’
Example: mov al, b mov bl, b or AL, BL ;(AL gets a value b)
AND– bitwise and – bit at index i of the destination gets ‘1’ if bits at index i of both source and destination are ‘1’; otherwise ‘0’
Example: or AL, BL ;(with same values of AL and BL as in previous example, AL gets a value 0)

12. CMP – Compare Instruction – compares integers
CMP performs a ‘mental’ subtraction - affects the flags as if the subtraction had taken place, but does not store the result of the subtraction.
cmp reg8/mem8(16,32),reg8/imm8(16,32) (source - register / immediate, destination- register / memory location)
cmp reg8(16,32),reg8/mem8(16,32) (source - register / immediate, destination - register / memory location)
Examples:
mov al, b mov bl, b cmp al, bl ;(ZF (zero flag) gets a value 0)
mov al, b mov bl, b cmp al, bl ;(ZF (zero flag) gets a value 1)

13. JMP – unconditional jump
jmp label
JMP tells the processor that the next instruction to be executed is located at the label that is given as part of jmp instruction.
Example:
mov eax, 1 inc_again: inc eax jmp inc_again mov ebx, eax
this is infinite loop !
this instruction is never reached from this code

14. J<Condition> – conditional jump
j<cond> label
execution is transferred to the target instruction only if the specified condition is satisfied
usually, the condition being tested is the result of the last arithmetic or logic operation
mov eax, 1 inc_again: inc eax cmp eax, 10 jne inc_again ; if eax ! = 10, go back to loop
Example:
mov eax, 1 inc_again: inc eax cmp eax, 10 je end_of_loop ; if eax = = 10, jump to end_of_loop jmp inc_again ; go back to loop end_of_loop:

15. Jcc: Conditional Branch
Instruction
Description
Flags JO
Jump if overflow
OF = 1
JNO
Jump if not overflow
OF = 0 JS
Jump if sign
SF = 1
JNS
Jump if not sign
SF = 0
JE  JZ
Jump if equal  Jump if zero
ZF = 1
JNE  JNZ
Jump if not equal  Jump if not zero
ZF = 0
JB  JNAE  JC
Jump if below  Jump if not above or equal  Jump if carry
CF = 1
JNB  JAE  JNC
Jump if not below  Jump if above or equal  Jump if not carry
CF = 0
JBE  JNA
Jump if below or equal  Jump if not above
CF = 1 or ZF = 1
JA  JNBE
Jump if above  Jump if not below or equal
CF = 0 and ZF = 0
JL  JNGE
Jump if less  Jump if not greater or equal
SF <> OF
JGE  JNL
Jump if greater or equal  Jump if not less
SF = OF
JLE  JNG
Jump if less or equal  Jump if not greater
ZF = 1 or SF <> OF
JG  JNLE
Jump if greater  Jump if not less or equal
ZF = 0 and SF = OF
JP  JPE
Jump if parity  Jump if parity even
PF = 1
JNP  JPO
Jump if not parity  Jump if parity odd
PF = 0
JCXZ  JECXZ
Jump if CX register is 0  Jump if ECX register is 0
CX = 0  ECX = 0

16. d<size> – declare initialized data
d<size> initial value
<size> value
<size> filed
Pseudo-instruction
1 byte
byte DB
2 bytes
word DW
4 bytes
double word DD
8 bytes
quadword DQ
10 bytes
tenbyte DT
16 bytes
double quadword
DDQ
octoword DO
Examples: var: db x55 ; define a variable ‘var’ of size byte, initialized by 0x55
var: db x55,0x56,0x57 ; three bytes in succession
var: db 'a‘ ; character constant 0x61 (ascii code of ‘a’)
var: db 'hello',13,10,'$‘ ; string constant
var: dw x ; 0x34 0x12
var: dw ‘A' ; 0x41 0x00 – complete to word
var: dw ‘AB‘ ; 0x41 0x42
var: dw ‘ABC' ; 0x41 0x42 0x43 0x00 – complete to word
var: dd x ; 0x78 0x56 0x34 0x12

17. Assignment 0
You get a simple program that receives a string from a user. Then, this program calls to a function (that you should implement in assembly) that receives a string as an argument and does the following:
let n - the fourth digit in the smaller id (of the partners’ ids), if n = 0 then you take the next digit in the id. (1<=n<=9)
Add n to each character of the input string to get the result string.
Count the number of letters in the input string that are converted to non-letter character in the result string.
The function returns the result string and the counter.
The characters conversion should be in-place.
examples:
> abcd
> efgh
> 0
> stuvwxyz
> wxyz{|}~
> 4

18. main.c #include <stdio.h>
# define MAX_LEN // Maximal line size
extern int strToLeet (char*);
int main(void) {
char str_buf[MAX_LEN];
int count= 0;
fgets(str_buf, MAX_LEN, stdin); // Read user's command line string
count = add_Str_N (str_buf); // Your assembly code function
printf("%s\n%d\n",str_buf,count); }

19. myasm.s section .data ; data section, read-write
an: DD ; this is a temporary var
section .text ; our code is always in the .text section
global add_Str_N ; makes the function appear in global scope
extern printf ; tell linker that printf is defined elsewhere ; (not used in the program)
add_Str_N: ; functions are defined as labels
push ebp ; save Base Pointer (bp) original value
mov ebp, esp ; use base pointer to access stack contents
pushad ; push all variables onto stack
mov ecx, dword [ebp+8] ; get function argument
;;;;;;;;;;;;;;;; FUNCTION EFFECTIVE CODE STARTS HERE ;;;;;;;;;;;;;;;;
mov dword [an], 0 ; initialize answer
label_here:
; Your code goes somewhere around here...
inc ecx ; increment pointer
cmp byte [ecx], 0 ; check if byte pointed to is zero
jnz label_here ; keep looping until it is null terminated
;;;;;;;;;;;;;;;; FUNCTION EFFECTIVE CODE ENDS HERE ;;;;;;;;;;;;;;;;
popad ; restore all previously used registers
mov eax,[an] ; return an (returned values are in eax)
mov esp, ebp
pop ebp
ret

20. Running NASM To assemble a file, you issue a command of the form
> nasm -f <format> <filename> [-o <output>] [ -l listing]
Example:
> nasm -f elf myasm.s -o myelf.o
It would create myelf.o file that has elf format (executable and linkable format). We use main.c file (that is written in C language) to start our program, and sometimes also for input / output from a user. So to compile main.c with our assembly file we should execute the following command:
gcc –m32 main.c myelf.o -o myexe.out
The -m32 option is being used to comply with 32- bit environment
It would create executable file myexe.out. In order to run it you should write its name on the command line:
> myexe.out

21. Example (n=4)

22. How to run Linux from Window
Go to
Run the following executable
Use “lvs.cs.bgu.ac.il” or “lace.cs.bgu.ac.il” host name and click ‘Open’
Use your Linux username and password to login lace server
Go to
Choose any free Linux computer
Connect to the chosen computer by using “ssh –X cs302six1-4” (maybe you would be asked for your password again)
cd (change directory) to your working directory

23. Ascii table