- 79 Views
- Uploaded on
- Presentation posted in: General

Practical Session 2

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Practical Session 2

- A flagis a single bit of information whose meaning is independent from any other bit

- Each flag has a two-letter symbol

- CF - Carry Flag
- If the result of an arithmetic or shift operation "carries out" a bit from the operand, CF becomes set11111111 + 00000001 = 100000000 CF = 1
- The carry (borrow) flag is also set if the subtraction of two numbers requires a borrow into MSB subtracted00000000 - 00000001 = 11111111 CF = 1
- In unsigned arithmetic, watch the carry flag to detect errors.
- In signed arithmetic, the carry flag tells you nothing interesting.
- 00000000 - 00000001 = 11111111
- unsigned arithmetic => 11111111 = 255 (decimal) => got the wrong answer
- signed arithmetic => 11111111 = -1 (decimal) => got right answer, do not care about value of CF

OF – Overflow Flag

- If the sum of two numbers with the sign bits off yields a result number with the sign bit on, the "overflow" flag is turned on. 0100 + 0100 = 1000 OF = 1
- If the sum of two numbers with the sign bits on yields a result number with the sign bit off, the "overflow" flag is turned on. 1000 + 1000 = 0000 OF = 1
- Otherwise, the overflow flag is turned off. 0100 + 0001 = 0101 OF = 0
- 0110 + 1001 = 1111 OF = 0
- 1000 + 0001 = 1001 OF = 0
- 1100 + 1100 = 1000 OF = 0
- In signed arithmetic, overflow flag on means the answer is wrong - you added two positive numbers and got a negative, or you added two negative numbers and got a positive.
- In unsigned arithmetic, the overflow flag tells you nothing interesting.

ZF, SF, PF, and AF Flags

- ZF – Zero Flag - set if result is zero; cleared otherwise
- add 0, 0 ZF = 1

- SF – Sign Flag becomes set when the result of an operation is negative (i.e. MSB of the result is 1)
- all arithmetic operations except multiplication and division
- compare instructions - CMP
- logical instructions - XOR, AND, OR
- TEST instructions (equivalent to AND instructions without storing the result)

- sub00000000, 00000001 11111111 SF = 1
- PF – Parity Flag - set if low-order eight bits of result contain an even number of "1" bits; cleared otherwiseadd 11010, 1 (result is 11011 4 bits are ‘1’ PF = 1)
- add 11000, 1 (result is 11001 3 bits are ‘1’ PF = 0)
- AF–Auxiliary Carry Flag (or Adjust Flag) is set if there is a carry from low nibble (4 bits) to high nibble or a borrow from a high nibble to low nibble
- 1111 + 0001 = 10000 AF = 1

first nibble

second nibble

Jcc: Conditional Branch

- Every process consists of sections that are accessible to the process when it is running
- Each sections holds the bulk of object file information
Data

.bss - holds uninitialized data; occupies no file space

.data and .data1 - hold initialized data

.rodata and .rodata1 - hold read-only data

Text

.text - holds the ‘‘text’’ (executable instructions) of a program

Stack - is used for local variables, information that is saved each time a function is called

Heap - contains the dynamically allocated memory

Example

int a,b,c = 1; ----> .data

char *str; ----> .bss

const int i = 10; ----> .rodata

main()

{

int ii,a=1,b=2,c; ----> local variables on Stack

char * ptr = malloc(4); ----> allocated memory in Heap c= a+b+i; ----> .text

}

Memory layout for Linux

USER Space Virtual Addresses

- program(.exe file) creates its own memory space in the RAM process
- 4GB of virtual address space on 32-bit architecture
- 3GB is accessible to the user space
- 1GB is accessible to the kernel space (kernel code, data, heap and stack)

.bss

.data

loaded from .exe file

Read-only Data Segment

.text

.rodata

Examples:

1. buffer: resb 64 ; reserve 64 bytes

2. wordVar: resw 1 ; reserve a word

3. realArray: resq 10 ; array of ten real numbers

Note: you can not make any assumption about values of a storage space cells.

Pseudo-instructionsRESB, RESW, RESD, RESQ and rest:declaring uninitialized storage space

- TIMES prefix causes the instruction to be assembled multiple times
zeroBuf: times 64 db 0; 64 bytes initialized to 0

- TIMES can be applied to ordinary instructions, so you can code trivial loopsmov EAX, 0times 100 inc EAX; EAX =100 => loop
- the argument to TIMES is not just a numeric constant, but a numeric expression
buffer: db ‘go’

times 64-$+buffer db ‘!’

; (64 - $ + buffer = 64 – 35 + 33 = 64 – 2 = 62)

RAM

buffer + 64

$ (current position

in section)

buffer

$$ (start of the current

section) - start of section .data

- EQU defines a symbol to a given constant value
- when EQU is used, the source line must contain a label
- this definition cannot changed later.
Example:

Foo: EQU 1 ; Foo = 1

- If the hardware is built so that the lowest, least significant byte of a multi-byte scalar is stored "first", at the lowest memory address, then the hardware is said to be "little-endian.
- numeric into memory reversed order
- dd 0x12345678 ; 0x78 0x56 0x34 0x12
- numeric into register source ordermov EAX, 0x12345678 ; 0x12 0x34 0x56 0x78
- characters into memory source order
- dw ‘ab’; 0x61 0x62
- dw ‘abc’ ; 0x61 0x62 0x63 0x00
- characters into register reversed order
- mov eax, ‘abc’; 0x00 0x63 0x62 0x61
- memory into register reversed order
- mov [buffer], eax; 0x61 0x62 0x63 0x64
- register into memory reversed order
- mov eax, [buffer]; 0x64 0x63 0x62 0x61

- Effective address is any operand to an instruction which references memory.
- Effective address syntax: consists of an expression evaluating to the desired address, enclosed in square brackets.
Example

wordvar: dw 0x5A, 0x39; a request for two words 0x005A and 0x0039

mov ax, [wordvar] ; ax = 0x005A (in little-endian format)

mov ax, [wordvar+1] ; ax = 0x3900

In memory we get the following:

[wordvar]

[wordvar+1]

[wordvar+2]

[wordvar+3]

NASM understands four different types of constant: numeric, character, string and floating points.

1. Numeric Constants

A numeric constant is simply a number. NASM allows to specify numbers in a variety of number bases, in a variety of ways: suffix H, Q and B for hex, octal and binary, etc.

Examples:

mov ax,100 ; decimal

mov ax,0A2h ; hex

mov ax,777q ; octal

mov ax,10010011b ; binary

2. Character Constants

- A character constant consists of up to 4 characters enclosed in either single or double quotes.
- A character constant with more than one character will be arranged with little-endian order in mind.
Examples:

- mov eax,'abcd'
The constant generated is not 0x61626364, but 0x64636261, so that if you were then to store the value into memory, it would read ‘abcd’ rather than ‘dcba’. That way, it is stored “backwards” in eax.

3. String Constants

- A string constants are only acceptable to some pseudo-instructions, like DB family.
Examples:

db 'hello’; string constant

db 'h','e','l','l','o’; equivalent character constants

dw 'abc' ; 0x61 0x62 0x63 0x00

MUL r/m - unsigned integer multiplication

IMUL r/m - signed integer multiplication

MUL r/m8mov bl,5 ; multipliermov al,9; multiplicandmul bl; => ax = 2Dh

MUL r/m16mov bx, 8000h mov ax, 2000h mul bx; => dx:ax = 1000:0000h

MUL r/m32

mov ebx, 80008000h mov eax, 20002000h mul ebx; => edx:eax = 10002000:10000000h

SHL, SHR – Bitwise Logical Shifts on the first operand

- number of bits to shift by is given by the second operand
- vacated bits are filled with zero
- shifted bit enters the Carry Flag
SHL r/m8/m16/m32 1/CL/imm8

SHR r/m8/m16/m32 1/CL/imm8

Example:

mov CL , 3

mov AL ,10110111b ; AL = 10110111

shr AL, 1 ; shift right 1 AL = 01011011, CF = 1

shr AL, CL; shift right 3 AL = 00001011, CF = 0

Note: that shift indeed performs division/multiplication by 2

SAL, SAR – Bitwise Arithmetic Shifts on the first operand

- vacated bits are filled with zero for SAL
- vacated bits are filledwith copies of the original high bit of the source operand for SAR
SAL r/m8/m16/m32 1/CL/imm8

SAR r/m8/m16/m32 1/CL/imm8

Example:

mov CL , 3

mov AL ,10110111b ; AL = 10110111

sar AL, 1 ; shift right 1 AL = 11011011

sar AL, CL; shift right 3 AL = 11111011

ROL, ROR – Bitwise Rotate (i.e. moves round) on the first operand

ROL r/m8/m16/m32 1/CL/imm8

ROR r/m8/m16/m32 1/CL/imm8

Example:

mov CL, 3

mov BH ,10110111b ; BH = 10110111

rol BH, 01 ; rotate left 1 bit BH = 01101111

rol BH, CL; rotate left 3 bits BH = 01111011

RCL, RCR – Bitwise Rotate through Carry Bit on the first operand and Carry Flag

RCL r/m8/m16/m32 1/CL/imm8

RCR r/m8/m16/m32 1/CL/imm8

Example:

mov BH ,10110111b ; BH = 10110111 , CF = 0

rcl BH, 01 ; rotate left 1 bit BH = 01101110 , CF = 1

LOOP, LOOPE, LOOPZ, LOOPNE, LOOPNZ – loop with counter (CX or ECX*)

Example:mov ax, 1 mov cx, 3 my_ loop: add ax, ax loop my_ loop, cx

1. decrements its counter register (in this case it is CX register)

2. if the counter does not become zero as a result of this operation, it jumps to the given label

LOOPE ≡ LOOPZ: jumps if the counter is nonzero andZero Flag = 1

LOOPNE ≡ LOOPNZ: jumps if the counter is nonzero and Zero Flag = 0

Note: if a counter is not specified explicitly, the BITS** setting dictates which is used.** The BITS directive specifies whether NASM should generate code designed to run on a processor operating in 16-bit mode, or code designed to run on a processor operating in 32-bit mode. The syntax is BITS 16 or BITS 32.