Practical Session 1

Computer Architecture and Assembly Language

• Byte – sequence of 8 bits:

16 35 4 07 2

MSB (Most Significant Bit) LSB (Least Significant Bit)

• Bit – basic information unit: (1/0)

• Word – a sequence of bits addressed as a single entity by the computer

byte byte 16 bit word

• Main Memory is an array of bytes, addressed by 0 to 232-1=0xFFFFFFFF

232 bytes = 4∙210∙3 bytes = 4 G bytes

…

…

232-1

2K-1

10

address space

physical memory

Data Representation BasicsData Representation Basics

Register file - CPU unit which contains (32 bit) registers.

general purpose registersEAX, EBX, ECX, EDX EAX, EBX, ECX, EDX (Accumulator, Base, Counter, Data)

index registersESP, EBP, ESI, EDIESP, EBP, ESI, EDI(Stack pointer - contains the address of last used dword in the stack, Base pointer, Source index, Destination Index)

flag register / status registerEFLAGSEFLAGS

Instruction Pointer / Program Counter EIP / EPCEIP / EPC- - contains address (offset) of the next instruction that is going to be executed (at run timeat run time)- - changed by unconditional jump, conditional jump, procedure call, and return instructions

RegistersRegisters

Note that the list of registers above is partial. The full list can be found here.

Register file

High byte

Low byte

Extended

16-bit register

Assembly Language ProgramAssembly 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 codeassembly code:

MOV AL, 61h ; load AL with 97 decimal (61 hex)

binary codebinary code:

10110000 01100001

1011 a binary code (opcode) of instruction 'MOV'

0 specifies if data is byte (‘0’) or full size 16/32 bits (‘1’)

000 a binary identifier for a register 'AL'

01100001 a binary representation of 97 decimal (97d = (int)(97/16)*10 + (97%16 converted to hex digit) = 61h)

label: (pseudo) instruction operands ; comment

optional fieldseither required or forbidden by an instruction

Notes:

- backslash (\) uses as the line continuation character: 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- a colon after a label is optional

Examples:

mov ax, 2 ; moves constant 2 to the register axbuffer: resb 64 ; reserves 64 bytes

Basic Assembly Instruction StructureBasic Assembly Instruction Structure

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

Instruction ArgumentsInstruction Arguments

mov [var1],[var2]! ! Note that x86 processor does not Note that x86 processor does not allow both operands be memory locations.allow both operands be memory locations.

Examples:

mov ax, 2 mov [buffer], ax

source operand target operand immediateregister

source operand

target operand registermemory location

mov reg8/mem8mov reg8/mem8(16,32),,reg8/imm8reg8/imm8(16,32)

(copies content of register / immediate (source) to register / memory location (destination))

mov reg8mov reg8(16,32),,reg8/reg8/mem8mem8(16,32)

(copies content of register / memory location (source) to register (destination))

MOV - MOV - Move Instruction – Move Instruction – copies source to destinationcopies source to destination

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.

operands have to be of the same size

Examples:

mov eax, 0x2334AAFF mov [buffer], ax mov wordword [var], 2

imm32reg32 reg16mem16 imm16mem16

ADDADD - add integers

Example:add AX, BX ;(AX gets a value of AX+BX)

ADCADC - add integers with carry (value of Carry Flag)

Example:adc AX, BX ;(AX gets a value of AX+BX+CF)

Basic Arithmetical InstructionBasic Arithmetical Instruction

<instruction> reg8/mem8<instruction> reg8/mem8(16,32),,reg8/imm8reg8/imm8(16,32)

(source - register / immediate, destination- register / memory location)

<instruction> reg8<instruction> reg8(16,32),,reg8/mem8reg8/mem8(16,32)

(source - register / immediate, destination - register / memory location)

SUB - subtract integers

Example:sub AX, BX ;(AX gets a value of AX-BX)

SBB - subtract with borrow (value of Carry Flag)

Example:sbb AX, BX ;(AX gets a value of AX-BX-CF)

Basic Arithmetical InstructionBasic Arithmetical Instruction

<instruction> reg8/mem8<instruction> reg8/mem8(16,32) (source / destination - register / memory location)

INC INC - increment integer

Example: inc AX ;(AX gets a value of AX+1)

DEC DEC - increment integer

Example: dec byte [buffer] ;([buffer] gets a value of [buffer] -1)

Basic Logical InstructionsBasic Logical Instructions

<instruction> reg8/mem8<instruction> reg8/mem8(16,32) (source / destination - register / memory location)

NOT NOT – one’s complement negation – inverts all the bits

Example: mov al, 11111110b not al ;(AL gets a value of 00000001b) ;(11111110b + 00000001b = 11111111b)

NEG NEG – two’s complement negation – inverts all the bits, and adds 1

Example: mov al, 11111110b neg al ;(AL gets a value of not(11111110b)+1=00000001b+1=00000010b) ;(11111110b + 00000010b = 1100000000b = 0)

Basic Logical InstructionsBasic Logical Instructions

OR 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, 11111100 b

mov bl, 00000010b

or AL, BL ;(AL gets a value 11111110b)

<instruction> reg8/mem8<instruction> reg8/mem8(16,32),,reg8/imm8reg8/imm8(16,32)

(source - register / immediate, destination- register / memory location)

<instruction> reg8<instruction> reg8(16,32),,reg8/mem8reg8/mem8(16,32)

(source - register / immediate, destination - register / memory location)

ANDAND– 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 11000000)

cmp reg8/mem8cmp reg8/mem8(16,32),,reg8/imm8reg8/imm8(16,32)

(source - register / immediate, destination- register / memory location)

cmp reg8cmp reg8(16,32),,reg8/mem8reg8/mem8(16,32)

(source - register / immediate, destination - register / memory location)

CMP – CMP – Compare Instruction – Compare Instruction – compares integerscompares integers

Examples:

mov al, 11111100b mov bl, 00000010b

cmp al, bl ;(ZF (zero flag) gets a value 0)

CMP performs a ‘mental’ subtraction - affects the flags as if the subtraction had taken place, but does not store the result of the subtraction.

mov al, 11111100b mov bl, 11111100 b

cmp al, bl ;(ZF (zero flag) gets a value 1)

• each instruction / data has its offset (address)

• if we want to refer to the specific instruction / data in the code, we should mark it with a label

• (non-local) labels have to be unique• an instruction that follows a label can be at the same / next

line• colon is optional

LabelLabel – specifies instruction’s offset (address) – specifies instruction’s offset (address)

Examples:

my_instructionmy_instruction: add ax, ax ;(my_instruction is an address of ‘add ax, ax’ instruction)

bufferbuffer: db 0add byte [bufferbuffer], 2 ;([buffer] gets a value of [buffer]+2)

JMP tells the processor that the next instruction to be executed is located at the label that is given as part of jmp instruction.

JMPJMP – unconditional jump – unconditional jump

jmp label jmp label

Example:

mov eax, 1inc_again:

inc eaxjmpjmp inc_againmov ebx, eax

this is infinite loop !

this instruction is never reached from this code!

• 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

J<Condition>J<Condition> – conditional jump – conditional jump

j<cond> label j<cond> label

Example:read_char:

… ; get a character into AL cmp al, ‘a’ ; compare the character to ‘a’

jeje a_received ; if value of al register equals to ‘a’, jump to a_received

jmpjmp read_char ; go back to read another

a_received:…

Instruction Description FlagsJO Jump if overflow OF = 1JNO Jump if not overflow OF = 0JS Jump if sign SF = 1JNS Jump if not sign SF = 0JE 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

Jcc: Conditional BranchJcc: Conditional Branch

D<Size>D<Size> – declare initialized data – declare initialized data

d<size> initial valued<size> initial value

Examples:var: db 0x55 ; define a variable ‘var’ of size byte, initialized by 0x55

var: db 0x55,0x56,0x57 ; three bytes in succession

var: db 'a‘ ; character constant 0x61 (ascii code of ‘a’)

var: db 'hello',13,10,'$‘ ; string constant

var: dw 0x1234 ; 0x34 0x12

var: dw 'a' ; 0x41 0x00 – complete to word0x00 – complete to word

var: dw 'ab‘ ; 0x41 0x42

var: dw 'abc' ; 0x41 0x42 0x43 0x00 – complete to word0x00 – complete to word

var: dd 0x12345678 ; 0x78 0x56 0x34 0x12

Pseudo-instruction <size> filed <size> value

DB byte 1 byte

DW word 2 bytes

DD double word 4 bytes

DQ quadword 8 bytes

DT tenbyte 10 bytes

DDQ double quadword 16 bytes

DO octoword 16 bytes

Assignment 0Assignment 0

Letter Leet symbolA 4B 8C (E 3G 6H #I !L 1O 0S 5T 7Z 2

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:

1.Convert the uppercase letters to their equivalent Leet symbol, according to the table below.

2.All other uppercase letters should be converted to lowercase letters

3.Return the number of letters converted to Leets in the input string

Example:

>Enter a string: HELLO WorlD!

>Result string: #3110 world!

>Number of letters converted to Leet: 5

The function returns the number of characters which aren’t uppercase or lowercase letter (the output should be just the number) . The characters conversion should be in-place.

#include <stdio.h>

# define MAX_LEN 100 // Maximal line size

extern int strToLeet (char*);

int main(void) {

char str_buf[MAX_LEN];

int str_len = 0;

printf("Enter a string: ");

fgets(str_buf, MAX_LEN, stdin); // Read user's command line string

str_len = strToLeetstrToLeet (str_buf); // Your assembly code function

printf("\nResult string:%s\nNumber of letters converted to Leet: %d\n",str_buf,str_len);

}

main.cmain.c

section .data ; data section, read-write an: DD 0 ; this is a temporary var

section .text ; our code is always in the .text section global strToLeetstrToLeet ; makes the function appear in global scope extern printf ; tell linker that printf is defined elsewhere

; (not used in the program)

strToLeetstrToLeet: ; 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 answerlabel_here:

; Your code goes somewhere around here...; Your code goes somewhere around here...

inc ecx ; increment pointercmp byte [ecx], 0 ; check if byte pointed to is zerojnz 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

myasm.smyasm.s

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

Running NASMRunning NASM

How to run Linux from Window Go to http://www.chiark.greenend.org.uk/~sgtatham/putty/download.html

Run the following executable

Use “lvs.cs.bgu.ac.il” or “lace.cs.bgu.ac.il” host nameand click ‘Open’

Use your Linux username and password to login lace server

Go to http://www.cs.bgu.ac.il/facilities/labs.html

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

Ascii tableAscii table